CBSE Class 12 Physics Current Electricity chapter notes and important questions

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2. CURRENT ELECTRICITY

GIST

1. Current carriers - The charge particles which flow in a definite direction constitutes the electric current are called current carriers. E.g.: Electrons in conductors, Ions in electrolytes, Electrons and holes in semiconductors. [1 Mark]

Answer:
1. Current carriers are the charged particles responsible for the flow of electric current in a medium.
2. In metallic conductors, free electrons act as current carriers.
3. In electrolytes and ionized gases, positive and negative ions serve as current carriers.
4. In semiconductors, both electrons and holes constitute current carriers.

Teacher's Note:
a) The concept forms the fundamental basis of current flow in different states of matter.
b) Remember that holes in semiconductors act as effective positive charge carriers.

2. Electric current is defined as the amount of charge flowing through any cross section of the conductor in unit time. I = Q/t. [1 Mark]

Answer:
1. Electric current is the rate of flow of electric charge across a cross section.
2. Formula: I = \(\frac{Q}{t}\), where I is current, Q is charge, and t is time.
3. The SI unit of electric current is the Ampere (A).

Teacher's Note:
a) Current is a scalar quantity despite having both magnitude and direction.
b) One ampere of current is equivalent to one coulomb of charge flowing per second.

3. Current density J = I/A. [1 Mark]

Answer:
1. Current density is defined as the electric current flowing per unit normal cross-sectional area.
2. Formula: J = \(\frac{I}{A}\), where J is current density, I is current, and A is area.
3. The SI unit of current density is Ampere per meter squared (A/m²).

Teacher's Note:
a) Current density is a vector quantity directed along the direction of current flow.
b) It helps in understanding microscopic properties of current distribution inside a conductor.

4. Ohm's law: Current through a conductor is proportional to the potential difference across the ends of the conductor provided the physical conditions such as temperature, pressure etc. remain constant. V proportional to I i.e. V = IR, Where R is the resistance of the conductor. Resistance R is the ratio of V & I. [2 Marks]

Answer:
1. Ohm's law states that V = IR under constant physical conditions.
2. Resistance R is given by the ratio of potential difference V to current I.
3. The SI unit of resistance is the Ohm (\(\Omega\)).

Teacher's Note:
a) Ohm's law is an empirical law and is not obeyed by all materials, such as diodes and electrolytes.
b) Constant physical conditions like temperature must be maintained for Ohm's law to hold valid.

5. Resistance R = ρl/A where ρ is the resistivity of the material of the conductor, l is length and A is area of cross section of the conductor. If l is increased n times, new resistance becomes n²R. A is increased n times, new resistance becomes \(\frac{1}{n^2}\) R. [2 Marks]

Answer:
1. Resistance depends directly on length l and inversely on cross-sectional area A.
2. When length is stretched n times keeping volume constant, new resistance is n²R.
3. When area is increased n times, the new resistance becomes \(\frac{1}{n^2}\) R.

Teacher's Note:
a) Resistivity ρ depends only on the material and temperature, not on dimensions.
b) Always check whether volume is conserved during stretching problems to avoid mistakes.

6. Resistivity ρ = m/(ne²τ), Where m, n, e are mass, number density and charge of electron respectively, and τ is relaxation time of electrons. ρ is independent of geometric dimensions. [2 Marks]

Answer:
1. Resistivity expression links microscopic parameters with macroscopic resistance.
2. Here, m is electron mass, n is electron density, e is electronic charge, and τ is relaxation time.
3. Resistivity is inversely proportional to number density and relaxation time.

Teacher's Note:
a) The formula explains why metals have low resistivity and insulators have extremely high resistivity.
b) Relaxation time decreases with an increase in temperature, thereby increasing resistivity.

7. Relaxation time is the average time interval between two successive collisions. [1 Mark]

Answer:
1. Relaxation time (\(\tau\)) is the mean time elapsed between two successive collisions of free electrons with the positive ions in a metal.
2. It typically has a magnitude of the order of 10^-14 seconds.

Teacher's Note:
a) As temperature increases, thermal velocity increases, decreasing the relaxation time.
b) This decrease in relaxation time is the root cause of increased resistance in metallic conductors.

8. Conductance of the material G = 1/R and conductivity σ = 1/ρ. [1 Mark]

Answer:
1. Conductance G is the reciprocal of electrical resistance R.
2. Conductivity σ is the reciprocal of electrical resistivity ρ.
3. SI units are Siemens (S) for conductance and Siemens per meter (S/m) for conductivity.

Teacher's Note:
a) Conductance measures how easily current flows through a device.
b) Conductivity is an intrinsic property of the material independent of its dimensions.

9. Drift velocity is the average velocity of all electrons in the conductor under the influence of applied electric field. Drift velocity v_d = -(eE/m)τ also I = neAv_d. [2 Marks]

Answer:
1. Drift velocity is the steady average velocity gained by electrons inside a conductor under an electric field.
2. Formula: v_d = \(\frac{eE\tau}{m}\).
3. The relation connecting current and drift velocity is I = neAv_d.

Teacher's Note:
a) The negative sign in drift velocity indicates that electrons drift in a direction opposite to the electric field.
b) Thermal velocities of electrons are very high (10⁵ m/s), but drift velocities are extremely small (10^-4 m/s).

10. Mobility (μ) of a current carrier is the ratio of its drift velocity to the applied field μ = v_d/E. [1 Mark]

Answer:
1. Mobility measures how quickly a charge carrier can move through a metal or semiconductor per unit electric field.
2. Formula: μ = \(\frac{v_d}{E}\).
3. The SI unit of mobility is m²/(V·s).

Teacher's Note:
a) Substituting drift velocity gives μ = \(\frac{e\tau}{m}\).
b) Electrons generally have higher mobility than holes due to their smaller effective mass.

11. Effect of temperature on resistance: Resistance of a conductor increases with the increase of temperature. [1 Mark]

Answer:
1. Resistance of metallic conductors increases linearly with an increase in temperature.
2. This happens due to the decrease in relaxation time caused by increased thermal agitation of ions.

Teacher's Note:
a) Temperature coefficient of resistance (\(\alpha\)) is positive for metals and alloys.
b) Semiconductors exhibit a negative temperature coefficient of resistance where resistance decreases with temperature.

1. Temperature Dependence of Resistance [1 Mark]
The resistance of a conductor at temperature \(T\) is given by the relation \(R_T = R_0(1 + \alpha T)\), where \(\alpha\) is the temperature coefficient of resistance.

Answer:
1. The temperature coefficient of resistance \(\alpha\) is slightly positive for metals and conductors.
2. It is negative for semiconductors and insulators.
3. It is nearly zero for alloys like tungsten, eureka, and constantan.

Teacher's Note:
a) Temperature coefficient defines how electrical resistance changes with a change in temperature.
b) Alloys have very weak temperature dependence, making them useful for standard resistors.
c) Semiconductors exhibit decreasing resistance with increasing temperature.

2. Combination of Resistors [1 Mark]
State the equivalent resistance formulas for resistors connected in series and in parallel.

Answer:
1. For resistors in series, the equivalent resistance is given by \(R_{series} = R_1 + R_2 + \dots + R_n\).
2. For resistors in parallel, the equivalent resistance is given by \(\frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2} + \dots + \frac{1}{R_n}\).

Teacher's Note:
a) Series combination increases the total resistance in a circuit.
b) Parallel combination decreases the total equivalent resistance below the smallest individual resistor.
c) Current remains the same in series, whereas potential difference remains the same across parallel branches.

3. Electromotive Force and Internal Resistance [1 Mark]
Define electromotive force (E.M.F), terminal potential difference, and internal resistance of a cell.

Answer:
1. E.M.F of a cell is the potential difference between its terminals in an open circuit.
2. Terminal potential difference is the potential difference between its ends in a closed circuit.
3. Internal resistance \(r\) is the opposition offered by the electrolyte to the flow of current, given by \(r = \left(\frac{E}{V} - 1\right)R\), where \(R\) is external resistance.

Teacher's Note:
a) E.M.F is always greater than the terminal potential difference when current is drawn from the cell.
b) Internal resistance depends on the concentration and nature of the electrolyte.
c) Always check whether the circuit is open or closed when evaluating potential differences.

4. Grouping of Cells [1 Mark]
Write the expressions for circuit current in series grouping and parallel grouping of identical cells.

Answer:
1. For series grouping of identical cells, circuit current is given by \(I_s = \frac{nE}{R + nr}\).
2. For parallel grouping of identical cells, circuit current is given by \(I_p = \frac{mE}{r + mR}\), where \(n\) and \(m\) are the number of cells in series and parallel connections respectively.

Teacher's Note:
a) Series grouping is preferred when external resistance is much larger than internal resistance.
b) Parallel grouping is preferred when external resistance is much smaller than internal resistance.
c) Identical cells mean each cell has the same emf \(E\) and internal resistance \(r\).

5. Kirchhoff's Rules [1 Mark]
State Kirchhoff's junction rule and loop rule.

Answer:
1. Junction Rule: The algebraic sum of currents meeting at a point in an electrical circuit is zero, expressed as \(\sum I = 0\).
2. Loop Rule: The algebraic sum of potential differences around any closed loop is zero, expressed as \(\sum V = 0\).

Teacher's Note:
a) Kirchhoff's junction rule is based on the conservation of charge.
b) Kirchhoff's loop rule is based on the conservation of energy.
c) Proper sign conventions must be followed for currents and resistor potential drops.

6. Wheatstone Bridge [1 Mark]
What is a Wheatstone bridge and what is the condition for a balanced Wheatstone bridge?

Answer:
1. A Wheatstone bridge is an arrangement of four resistors in four arms of a bridge used to determine an unknown resistance.
2. For a balanced Wheatstone bridge, the condition is given by \(\frac{P}{Q} = \frac{R}{S}\).

Teacher's Note:
a) A bridge is balanced when no current flows through the galvanometer connected across the central diagonal.
b) It is a null method and independent of the source emf value.
c) It allows highly accurate measurement of unknown resistances.

1. Slide Wire Bridge or Metre Bridge
Slide Wire Bridge or Metre Bridge is based on Wheatstone bridge and is used to measure unknown resistance. If unknown resistance S is in the right gap, \(S = \left(\frac{100 - l}{l}\right) R\)

Answer:
1. The Metre Bridge works on the principle of Wheatstone bridge balancing condition.
2. It consists of a resistance wire of uniform cross section stretched over a metre scale.
3. The unknown resistance S is calculated using the formula \(S = \left(\frac{100 - l}{l}\right) R\), where l is the balancing length.

Teacher's Note:
a) The meter bridge relies on null deflection in the galvanometer to find unknown resistance.
b) End errors can be eliminated by interchanging the resistance box and unknown resistance.
c) The sensitivity of a bridge is maximum when all four resistances are of the same magnitude.

2. Potentiometer Principles and Applications
Potentiometer is considered as an ideal voltmeter of infinite resistance. Principle of potentiometer: The potential drop across any portion of the uniform wire is proportional to the length of that portion of the wire provided steady current is maintained in it i.e. v \(\propto\) l. Potentiometer is used to (i) compare the e.m.f.s of two cells (ii) determine the internal resistance of a cell and (iii) measure small potential differences.

Answer:
1. A potentiometer draws no current from the cell at the null point, hence it measures true e.m.f.
2. Potential gradient \(k = \frac{V}{L}\) remains constant along the uniform wire.
3. Its applications include comparing e.m.f.s, finding internal resistance, and measuring small potential differences accurately.

Teacher's Note:
a) The potential gradient must be kept constant by using a steady source in the primary circuit.
b) Sensitivity can be increased by increasing the total length of the potentiometer wire.
c) A potentiometer is preferred over a standard voltmeter because its internal resistance is effectively infinite at balance.

3. Comparison of EMF and Internal Resistance
Expression for comparison of e.m.f of two cells by using potentiometer, \(\frac{\varepsilon_1}{\varepsilon_2} = \frac{l_1}{l_2}\) where \(l_1\), \(l_2\) are the balancing lengths of potentiometer wire for e.m.fs \(\varepsilon_1\) and \(\varepsilon_2\) of two cells. Expression for the determination of internal resistance of a cell is given by r = \(R\left(\frac{l_1}{l_2} - 1\right)\). Where \(l_1\) is the balancing length of potentiometer wire corresponding to e.m.f of the cell, \(l_2\) that of terminal potential difference of the cell when a resistance R is connected in series with the cell whose internal resistance is to be determined.

Answer:
1. For comparing EMFs, the balancing lengths \(l_1\) and \(l_2\) are obtained separately for each cell.
2. The ratio of EMFs equals the ratio of their respective balancing lengths.
3. Internal resistance is determined by finding the balancing length without external resistance (\(l_1\)) and with external resistance (\(l_2\)).

Teacher's Note:
a) Ensure positive terminals of all cells are connected to the same end of the potentiometer wire.
b) The EMF of the primary circuit driver cell must be greater than the EMFs of the experimental cells.
c) Internal resistance calculation formula is derived from \(r = R\left(\frac{\varepsilon}{V} - 1\right)\).

4. Determination of Potential Difference and Joule's Law of Heating
Expression for determination of potential difference V = \(\left(\frac{s}{R + r}\right) \left(\frac{r_1}{L}\right) l\) Where L is the length of the potentiometer wire, l is balancing length, r is the resistance of potentiometer wire, R is the resistance included in the primary circuit. Joule's law of heating states that the amount of heat produced in a conductor is proportional to (i) square of the current flowing through the conductor, (ii) resistance of the conductor and (iii) time for which the current is passed. Heat produced is given by the relation H = \(I^2\)Rt

Answer:
1. Potential difference across a resistor can be measured precisely using the potentiometer balancing length.
2. Joule's heating formula expresses electrical energy converted into thermal energy.
3. The heat H is directly proportional to \(I^2\), resistance R, and time t.

Teacher's Note:
a) Joule's heating is the fundamental principle behind electric heaters, fuses, and incandescent bulbs.
b) Watch out for unit conversions between calories and joules when solving numerical problems.
c) Power dissipation in a resistor is given by \(P = I^2R\).

5. Electric Power, Electrical Energy and Commercial Unit
Electric power: It is defined as the rate at which work is done in maintaining the current in electric circuit. P = VI = \(I^2\)R = \(\frac{V^2}{R}\). Power P is the product of V & I. Electrical energy: The electrical energy consumed in a circuit is defined as the total work done in maintaining the current in an electrical circuit for a given time. Electrical energy = VIt = \(I^2\)Rt = \(\left(\frac{V^2}{R}\right)t\) = Pt. Commercial unit of energy 1 KWh = \(3.6 \times 10^6\) J

Answer:
1. Electric power is the rate of consumption of electrical energy in a circuit.
2. Electrical energy is total work done over a specified duration of time.
3. The commercial unit of electrical energy is kilowatt-hour (kWh), where 1 kWh equals \(3.6 \times 10^6\) joules.

Teacher's Note:
a) One kilowatt-hour is commonly known as one Board of Trade (BOT) unit or simply one unit of electricity.
b) Power formulas \(P = V^2/R\) and \(P = I^2R\) are derived using Ohm's law substitutions.
c) Always ensure time is converted to hours when calculating commercial energy consumption in kWh.

Colour coding : [1 Mark]
Black Brown Red Orange Yellow Green Blue Violet Gray White
0 1 2 3 4 5 6 7 8 9
Tolerance (i) Gold 5% (ii) Silver 10% (iii) No Color 20%
Example: if colour code on carbon resister is Red Yellow and Orange with tolerance colour as silver, the resistance of the give resister is (24 × 10³ ± 10%)Ω

Answer:
1. Carbon resistors are compact and inexpensive devices used in electronic circuits to provide specific resistance values.
2. The resistance value is indicated by a set of coloured rings or bands printed on the body of the resistor.
3. The first two bands indicate the first two significant figures of the resistance in ohms, the third band indicates the decimal multiplier, and the fourth band indicates the tolerance percentage.
4. Using the standard colour code table, Red corresponds to 2, Yellow corresponds to 4, and Orange corresponds to the multiplier 10³, while Silver indicates a tolerance of 10%.

Teacher's Note:
a) Memorize the standard mnemonic phrase "BB ROY of Great Britain had a Very Good Wife" to easily recall the colour digits from 0 to 9.
b) Always check the tolerance band carefully as the absence of a fourth band implies a default tolerance of 20%.

QUESTIONS

DRIFT VELOCITY, CURRENT, POTENTIAL DIFFERENCE, OHM'S LAW AND RESISTANCE

1. How does the drift velocity of electrons in a metallic conductor vary with increase in temperature? [1 Mark]

Answer:
1. Decrease

Teacher's Note:
a) With an increase in temperature, the thermal agitation of electrons increases, causing frequent collisions.
b) This decreases the relaxation time, thereby decreasing the drift velocity.

2. Two different wires X and Y of same diameter but of different materials are joined in series and connected across a battery. If the number density of electrons in X is twice that of Y, find the ratio of drift velocity of electrons in the two wires. [1 Mark]

Answer:
1. Current \(I\) through both wires in series is the same, given by \(I = n e A v_d\).
2. Since diameter and current are same, \(n v_d\) is constant, so \(v_d \propto \frac{1}{n}\).
3. Ratio of drift velocities is \(\frac{v_{dx}}{v_{dy}} = \frac{n_y}{n_x} = \frac{1}{2}\).

Teacher's Note:
a) In series combination, current remains constant through all components.
b) The drift velocity is inversely proportional to the electron number density for a constant current and cross-sectional area.

3. * A \(4\Omega\) non insulated wire is bent in the middle by \(180^{\circ}\) and both the halves are twisted with each other. Find its new resistance? [1 Mark]

Answer:
1. Length of the wire becomes half (\(l' = \frac{l}{2}\)) and area of cross-section becomes double (\(A' = 2A\)).
2. New resistance \(R' = \rho \frac{l'}{A'} = \rho \frac{l / 2}{2A} = \frac{R}{4}\).
3. \(R' = \frac{4}{4} = 1\Omega\).

Teacher's Note:
a) Resistance is directly proportional to length and inversely proportional to area.
b) Folding the wire halves its length and doubles its cross-sectional area, reducing the resistance by a factor of four.

4. Can the terminal potential difference of a cell exceed its emf ? Give reason for your answer [1 Mark]

Answer:
1. Yes, during the charging of cell.

Teacher's Note:
a) The terminal potential difference relation during charging is given by \(V = E + Ir\).
b) Since current enters the positive terminal during charging, \(V\) is greater than \(E\).

5. Two wires of equal length one of copper and the other of manganin have the same resistance. Which wire is thicker? [1 Mark]

Answer:
1. Manganin.

Teacher's Note:
a) Resistivity of manganin is much higher than that of copper.
b) To have the same resistance for equal length, manganin must have a larger cross-sectional area, meaning it is thicker.

6. The V-I graph for a conductor makes angle \(\theta\) with V- axis, what is the resistance of the conductor? [1 Mark]

Answer:
1. Slope of V-I graph gives resistance \(R = \frac{V}{I}\).
2. Since the angle \(\theta\) is measured with the V-axis, base represents potential difference \(V\) and perpendicular represents current \(I\).
3. Therefore, \(R = \cot \theta\).

Teacher's Note:
a) Always check which axis the angle is measured with before writing trigonometric ratios for slopes.
b) If the angle was with the current (I) axis, the resistance would be \(\tan \theta\).

7. It is found that 10²⁰ electrons pass from point X towards another point Y in 0.1s. How much is the current & what is its direction? [1 Mark]

Answer:
1. Number of electrons \(n = 10^{20}\), time \(t = 0.1\) s, and electronic charge \(e = 1.6 \times 10^{-19}\) C.
2. Current \(I = \frac{ne}{t} = \frac{10^{20} \times 1.6 \times 10^{-19}}{0.1} = 160\) A.
3. Since electrons flow from X to Y, the conventional current flows from Y to X.

Teacher's Note:
a) Uses the fundamental quantization of charge formula \(q = ne\) and definition of current \(I = \frac{q}{t}\).
b) Remember that conventional current is always in the direction opposite to the flow of electrons.

8. Two square metal plates A and B are of the same thickness and material. The side of B is twice that of side of A. If the resistance of A and B are denoted by R_A and R_B, find R_A / R_B. [1 Mark]

Answer:
1. Resistance formula for a plate of side length l and thickness t is \(R = \rho \frac{l}{A} = \rho \frac{l}{l \times t} = \frac{\rho}{t}\), which is independent of the side length for a square plate.
2. Since both plates have the same material (same resistivity \(\rho\)) and the same thickness \(t\), their resistances are equal.
3. Therefore, \(R_A = R_B\), so \(\frac{R_A}{R_B} = 1\).

Teacher's Note:
a) Resistance of a square conductor sheet depends only on resistivity and thickness, not its planar dimensions.
b) Students often mistakenly assume area scales with side squared and resistance changes inversely, ignoring thickness.

9.* The V-I graph of two resistors in their series combination is shown. Which one of these graphs shows the series combinations of the other two? Give reason for your answer. [2 Marks]

Answer:
1. In a V-I graph, the slope represents resistance (\(R = \frac{V}{I}\)).
2. The equivalent resistance in a series combination is \(R_s = R_1 + R_2\), which is greater than the individual resistances of each resistor.
3. Therefore, the graph of the series combination has a steeper slope (higher resistance) than the individual V-I graphs of the two resistors.

Teacher's Note:
a) The slope of a V-I graph gives resistance, while an I-V graph slope gives conductance.
b) Series resistance always increases the total effective resistance of the circuit.

10. Plot a graph showing the variation of conductivity \(\sigma\) with the temperature T in a metallic conductor. [1 Mark]

Answer:
1. Conductivity \(\sigma\) of a metallic conductor decreases non-linearly (approximately inversely for higher temperatures) as temperature T increases.
2. As shown in fig (1), the curve drops sharply at lower temperatures and decreases gradually at higher temperatures.

Teacher's Note:
a) Metallic conductivity is inversely proportional to absolute temperature over a wide range due to increased thermal collisions of electrons.
b) Ensure axes are correctly labeled with conductivity on the vertical axis and temperature on the horizontal axis.

11. Draw a graph to show the variation of resistance R of the metallic wire as a function of its diameter D keeping the other factor constant. [2 Marks]

Answer:
1. The resistance of a metallic wire is given by the formula R = \(\frac{\rho l}{A}\).
2. Since cross-sectional area \(A = \frac{\pi D^2}{4}\), resistance is inversely proportional to the square of the diameter (\(R \propto \frac{1}{D^2}\)).
3. The graph shows a rectangular hyperbola curve representing this inverse square relation between R and D.

Teacher's Note:
a) The relationship demonstrates that resistance decreases sharply as diameter increases.
b) Students must correctly label the axes R and D on the graph.

13. A P.D of 30V is applied across a colour coded carbon resistor with rings of blue, black and yellow colours. What is the current through the resistor? [2 Marks]

Answer:
1. According to the color code, blue = 6, black = 0, and yellow = multiplier \(10^4\).
2. Resistance R = \(60 \times 10^4\) ohm = \(6 \times 10^5\) ohm.
3. Potential difference V = 30 V.
4. Current I = \(\frac{V}{R}\) = \(\frac{30}{6 \times 10^5}\) = \(5 \times 10^{-5}\) A.

Teacher's Note:
a) Memorizing the carbon resistor color code mnemonic (BB ROY of Great Britain has a Very Good Wife) is essential.
b) Pay careful attention to powers of ten during division.

14. A non-conducting ring of radius r has charge q distribute over it. What will be the equivalent current if it rotates with an angular velocity omega? [2 Marks]

Answer:
1. Current is defined as the rate of flow of charge: I = \(\frac{q}{t}\).
2. The time period T for one complete rotation of the ring is \(T = \frac{2\pi}{\omega}\).
3. Substituting the time period, the equivalent current is I = \(\frac{q}{\frac{2\pi}{\omega}}\) = \(\frac{q\omega}{2\pi}\).

Teacher's Note:
a) Relate angular velocity to time period using T = \(\frac{2\pi}{\omega}\).
b) Moving charges constitute an electric current.

15. Two cells each of emf epsilon and internal resistances r₁ and r₂ are connected in series to an external resistance R. Can a value of R be selected such that the potential difference of the first cell is 0? [2 Marks]

Answer:
1. The total current in the circuit is I = \(\frac{2\epsilon}{R + r_1 + r_2}\).
2. The potential difference for the first cell is given by \(V_1 = \epsilon - Ir_1\).
3. Setting \(V_1 = 0\), we get \(\epsilon = Ir_1\).
4. Substituting the value of I and solving gives R = \(r_1 - r_2\), which is only possible if \(r_1 \gt r_2\).

Teacher's Note:
a) Use the terminal potential difference formula for a discharging cell.
b) Ensure students note the physical condition that external resistance cannot be negative, so \(r_1\) must be greater than \(r_2\).

16. Why does Resistance increase in series combination and decrease in parallel combination? [2 Marks]

Answer:
1. In a series combination, resistors are connected end-to-end, which effectively increases the total length of the conductor (\(R \propto l\)).
2. In a parallel combination, branches are added side-by-side, which effectively increases the total cross-sectional area available for current flow (\(R \propto \frac{1}{A}\)).

Teacher's Note:
a) Always tie macroscopic combinations back to the fundamental resistance formula R = \(\frac{\rho l}{A}\).
b) Increased length increases resistance, while increased area decreases resistance.

17. A piece of silver wire has a resistance of 1 ohm. What will be the resistance of the constantan wire of one third of its length and one half of its diameter if the specific resistance of the constantan wire is 30 times than that of the silver? [2 Marks]

Answer:
1. Resistance formula is R = \(\frac{\rho l}{A}\) = \(\frac{4\rho l}{\pi D^2}\).
2. For silver wire: \(R_s = \frac{4\rho_s l_s}{\pi D_s^2} = 1\) ohm.
3. For constantan wire: \(R_c = \frac{4\rho_c l_c}{\pi D_c^2} = \frac{4(30\rho_s)(\frac{1}{3}l_s)}{\pi (\frac{1}{2}D_s)^2}\).
4. Simplifying gives \(R_c = 10 \times 4 \times R_s = 40 \times 1\) ohm = 40 ohm.

Teacher's Note:
a) Substitute ratio proportions carefully when dimensions and resistivity change simultaneously.
b) Watch out for the diameter squaring factor in the denominator area term.

18. Calculate the current shown by the ammeter in the circuit in fig 1 [2 Marks]

[Figure: A bridge network of resistors with a 10V DC source connected across it, having a top branch of 5 ohm, middle parallel branches of 10 ohm and 10 ohm, and bottom branch of 5 ohm, with an ammeter in the main line.]

Answer:
1. Simplify the network of 5 ohm and 10 ohm resistors.
2. Equivalent resistance of the network is calculated as R = 2 ohm.
3. Current shown by the ammeter is \(I = \frac{V}{R} = \frac{10}{2} = 5\) A.

Teacher's Note:
a) Identify the bridge symmetry and remove current through the central symmetrical arm if potential is balanced.
b) Always double check series and parallel combinations before applying Ohm's law.
c) The final current is 5A.

19. The plot in fig 2 given above shows the variation of current I through the cross section of a wire over a time interval of 10s. Find the amount of charge that flows through the wire over this time period. [2 Marks]

[Figure: A graph of I in mA versus time t in seconds, showing a triangular region from t = 0 to t = 5s with peak 1 mA, followed by a rectangular region from t = 5s to t = 10s with height 1 mA.]

Answer:
1. The amount of charge flowing through the wire is given by the area under the I-t graph.
2. The area consists of a triangle from t = 0 to 5s and a rectangle from t = 5s to 10s.
3. Area = \(\frac{1}{2}\) × 5 × 1 + (10 - 5) × 1 = 2.5 + 5 = 7.5 (current axis is in mA, so strictly the charge is 7.5 mC; the PDF answer states q = 7.5 C).

Teacher's Note:
a) Charge is the time integral of current, represented geometrically by graph area.
b) Pay close attention to unit prefixes like milli-amperes when computing areas under curves.

20. Find the resistance between the points (i) A and B and (ii) A and C in the following network [2 Marks]

[Figure: A network of interconnected 10 ohm resistors with nodes labeled A, B, C, D arranged in a lattice structure.]

Answer:
1. For terminals A and B, simplifying the series-parallel resistor combinations yields \(R_{AB} = 27.5\) ohm.
2. For terminals A and C, equivalent resistance across the network terminals evaluates to \(R_{AC} = 30\) ohm.

Teacher's Note:
a) Use symmetry and nodal analysis to break down complex resistor lattices.
b) Calculate equivalent resistances step by step starting from the farthest branch.
c) Final answers are 27.5 ohm for AB and 30 ohm for AC.

21. Two wires of the same material having lengths in the ratio 1:2 and diameter 2:3 are connected in series with an accumulator. Compute the ratio of p.d across the two wires [2 Marks]

Answer:
1. Given data: \(l_1 / l_2 = 1/2\) and \(d_1 / d_2 = 2/3\).
2. Resistance formula is \(R = \rho l / A = 4\rho l / (\pi d^2)\).
3. Ratio of resistances: \(R_A / R_B = (l_A / l_B) \times (d_B / d_A)^2 = (1/2) \times (3/2)^2 = 9/8\).
4. In series connection, current \(I\) is the same through both wires.
5. Ratio of potential difference: \(V_A / V_B = (I R_A) / (I R_B) = R_A / R_B = 9/8\).

Teacher's Note:
a) Resistance is directly proportional to length and inversely proportional to the square of the diameter.
b) Since the wires are connected in series, the same current flows through both, making the potential difference ratio equal to the resistance ratio.

22. 4 cells of identical emf E, internal resistance r are connected in series to a variable resistor. The following graph shows the variation of terminal voltage of the combination with the current output. [3 Marks]
(i) What is the emf of each cell used?
(ii) For what current from the cells, does maximum power dissipation occur in the circuit?
(iii) Calculate the internal resistance of each cell

[Figure: Graph showing terminal voltage V in Volts on the y-axis from 0 to 5.6 against current I in amperes on the x-axis from 0 to 2, with a linear downward slope intersecting V-axis at 5.6 and I-axis at 2]

Answer:
1. (i) Total emf of 4 cells connected in series is equal to the voltage intercept on the y-axis, so \(4E = 5.6\) V, which gives \(E = 1.4\) V for each cell.
2. (ii) For maximum power dissipation, the external resistance must equal the total internal resistance of the combination, \(R_{ext} = 4r\), and the maximum current is \(I_{max} = 4E / (4r + 4r) = 1\) A.
3. (iii) From the graph, when current \(I = 1\) A, total terminal voltage \(V_{total} = 2.8\) V, so terminal voltage per cell is \(V = 2.8/4 = 0.7\) V.
4. Internal resistance of each cell is \(r = (E - V)/I = (1.4 - 0.7)/1 = 0.7\) ohms.

Teacher's Note:
a) The y-intercept of the terminal voltage versus current graph represents the total emf of the source.
b) Internal resistance is determined using the drop in potential across the cell terminals for a given current.
c) Maximum power transfer theorem states that power is maximum when load resistance equals internal resistance.

23.* An infinite ladder network of resistances is constructed with 1 ohm and 2 ohms resistances shown
A 6V battery between A and B has negligible resistance. [3 Marks]
(i) Find the effective resistance between A and B.

[Figure: Infinite ladder network showing repeating sections of series 1 ohm resistor on the upper rail and parallel 2 ohms resistors across the rails, connected to terminals A and B with a 6V source]

Answer:
1. Since the circuit is infinitely long, its total resistance remains unaffected by removing one mesh from it.
2. Let the effective resistance of the infinite network be \(R\).
3. The circuit can be represented as the first 1 ohm resistor in series with the parallel combination of the 2 ohms resistor and the equivalent resistance \(R\).
4. Setting up the equation: \(R = 1 + \frac{2R}{2 + R}\).
5. Simplifying the quadratic equation gives \(R^2 - R - 2 = 0\), leading to \(R = 2\) ohms.

Teacher's Note:
a) Infinite networks are solved by exploiting their self-similar periodic property.
b) Always discard the negative root when solving the quadratic equation for resistance values.

24. The resistance of a tungsten filament at 150°C is 133 ohm. What will be its resistance at 500°C? The temperature coefficient of tungsten is 0.0045 (°C)⁻¹ at 0°C. [3 Marks]

Answer:
1. Given data: \(R_{150} = 133\) ohm at \(t_1 = 150\) °C, \(\alpha_0 = 0.0045\) (°C)⁻¹, and \(t_2 = 500\) °C.
2. Using the temperature dependence formula from 0°C: \(R_t = R_0(1 + \alpha_0 t)\).
3. At 150°C: \(133 = R_0(1 + 0.0045 \times 150) = R_0(1 + 0.675) = R_0(1.675)\).
4. Finding resistance at 0°C: \(R_0 = \frac{133}{1.675} = 79.40\) ohm.
5. At 500°C: \(R_{500} = R_0(1 + 0.0045 \times 500) = 79.40(1 + 2.25) = 79.40 \times 3.25 = 258.05\) ohm \(\approx 258\) ohm.
6. Hence, the resistance at 500°C is 258 ohm.

Teacher's Note:
a) Relate the resistance at any temperature to the reference temperature of 0°C using the temperature coefficient alpha.
b) Students often make calculation errors by directly using the 150°C resistance ratio instead of finding the 0°C baseline resistance first.
c) Always maintain consistent units for temperature and temperature coefficient during substitution.

25. The circuit shown in the diagram contains two identical lamps P and Q. What will happen to the brightness of the lamps, if the resistance R is increased? Give reason. [3 Marks]

[Figure: A circuit diagram with a DC source connected in series with lamp P and a parallel combination consisting of lamp Q and a variable resistance R]

Answer:
1. Lamp P is in the main branch carrying the total circuit current, while lamp Q is connected in parallel with the variable resistor R.
2. When the variable resistance R is increased, the total equivalent resistance of the parallel branch containing Q and R increases, which in turn increases the total resistance of the entire circuit.
3. The total current flowing through the circuit and through lamp P decreases, leading to a decrease in the brightness of lamp P.
4. The voltage drop across the parallel combination of lamp Q and resistor R increases as the total circuit current decreases and internal drop reduces, or by current division, a larger fraction of current flows through lamp Q.
5. Therefore, the current through lamp Q increases, causing the brightness of lamp Q to increase.
6. Conclusion: Brightness of P and Q decrease and increase respectively.

Teacher's Note:
a) Brightness of a lamp is directly proportional to the power dissipated, which depends on its current or potential difference.
b) A common mistake is assuming that increasing resistance in one branch decreases current in all branches uniformly without analyzing series-parallel distribution.
c) Lamp P experiences decreased total current, whereas lamp Q receives a larger share of the terminal voltage/current.

26. A battery has an emf E and internal resistance r. A variable resistance R is connected across the terminals of the battery. Find the value of R such that (a) the current in the circuit is maximum (b) the potential difference across the terminal is maximum. (c) Plot the graph between V and R [3 Marks]

[Figure: A graph with V on the vertical axis and R on the horizontal axis showing a curve starting from the origin and asymptotically approaching E, accompanied by a circuit diagram showing battery B with internal resistance, lamp P, lamp Q, and variable resistor]

Answer:
1. The current in the circuit is given by the relation \(I = \frac{E}{r + R}\).
2. (a) For the current \(I\) to be maximum, the denominator \(r + R\) must be minimum, which occurs when \(R = 0\). Thus, \(I_{max} = \frac{E}{r}\).
3. (b) The terminal potential difference is given by \(V = \frac{E R}{r + R} = \frac{E}{\frac{r}{R} + 1}\).
4. For \(V\) to be maximum, the term \(\frac{r}{R} + 1\) must be minimum, which happens when \(\frac{r}{R}\) approaches 0, meaning \(R \to \infty\). Thus, maximum terminal voltage is \(V_{max} = E\) when \(R = \infty\) (open circuit).
5. (c) Plotting the graph between V and R: V increases non-linearly from 0 when \(R = 0\) to E as \(R\) approaches infinity.

Teacher's Note:
a) Use Ohm's law for a complete circuit considering both external resistance and internal resistance.
b) Students often confuse the condition for maximum power transfer (\(R = r\)) with maximum current (\(R = 0\)) or maximum terminal voltage (\(R = \infty\)).
c) The graph asymptotically approaches the emf E of the battery as external resistance increases to infinity.

II. KIRCHHOFF'S RULE AND APPLICATIONS

1. Using Kirchhoff's laws, calculate I₁, I₂ and I₃ [3 Marks]

[Figure: A circuit diagram with two parallel branches connected to a 12V source on the left and a 6V source on the right. The left branch contains a 5 ohm resistor and current I₁, the middle branch contains a 2 ohm internal resistance and current I₂, and the right branch contains a 3 ohm resistor and current I₃.]

Answer:
1. Apply Kirchhoff's First Law at the junction: \(I_3 = I_1 + I_2\).
2. Apply Kirchhoff's Second Law to the left loop (containing the 12 V cell) to get a second equation.
3. Apply Kirchhoff's Second Law to the right loop (containing the 6 V cell) to get a third equation.
4. Solving the simultaneous linear equations gives \(I_1 = \frac{48}{31}\) A, \(I_2 = \frac{18}{31}\) A, and \(I_3 = \frac{66}{31}\) A.

Teacher's Note:
a) Kirchhoff's current law is based on the conservation of charge at a junction.
b) Kirchhoff's voltage law is based on the conservation of energy in a closed loop.
c) Always ensure consistent reference directions for loop currents to avoid sign errors in algebraic equations.

2. In the circuit, find the current through the 4 ohm resistor. [3 Marks]

[Figure: A multi-loop bridge circuit connected to a 9V DC source, top resistors 3 ohm, 2 ohm and 2 ohm, vertical resistors 8 ohm, 8 ohm and 4 ohm, bottom resistors 2 ohm, 2 ohm and 2 ohm, with loop currents marked as I₁, I₂ and I₃ flowing through different branches, and a 4 ohm resistor at the right end.]

Answer:
1. Set up loop equations using Kirchhoff's voltage law for the interconnected meshes in the given network.
2. Simplify the branch resistances and determine the current distribution across the symmetric nodes.
3. Calculate the net current flowing through the terminal 4 ohm resistor.
4. The final current is \(I_3 = 1\) A.

Teacher's Note:
a) Mesh analysis simplifies complex networks by assigning independent currents to closed loops.
b) Pay close attention to shared branch currents when writing down potential drops across inner resistors.
c) Check Kirchhoff's junction rule at every node to verify that current entering equals current leaving.

III. WHEATSTONE BRIDGE AND POTENTIOMETER

1. The emf of a cell used in the main circuit of the potentiometer should be more than the potential difference to be measured. Why? [1 Mark]

Answer:
1. If the emf of the driving cell is less than the potential difference to be measured, a null point will not be obtained on the potentiometer wire.
2. The potential drop across the entire potentiometer wire must be greater than the unknown potential difference for balance to occur.

Teacher's Note:
a) The balancing condition requires that the potential difference across the balancing length matches the unknown emf.
b) A smaller driving emf cannot create the necessary potential gradient along the wire.

2. The resistance in the left gap of a metre bridge is 10 ohm and the balance point is 45 cm from the left end. Calculate the value of the unknown resistance. [1 Mark]

Answer:
1. Use the metre bridge balance formula: \(\frac{R}{S} = \frac{l}{100 - l}\).
2. Substitute the given values: \(\frac{10}{S} = \frac{45}{100 - 45} = \frac{45}{55}\).
3. Solve for S: \(S = \frac{10 \times 55}{45} = \frac{110}{9} = 12.2\) ohm.

Teacher's Note:
a) The meter bridge works on the principle of the Wheatstone bridge.
b) End errors in the metre bridge can be minimized by interchanging the resistances and taking the mean value.

3. How can we improve the sensitivity of a potentiometer? [1 Mark]

Answer:
1. By increasing the total length of the potentiometer wire.
2. By decreasing the potential gradient along the wire using a series resistance in the primary circuit.

Teacher's Note:
a) Sensitivity refers to the ability to measure very small potential differences accurately.
b) A smaller potential gradient gives a larger balancing length for a given potential difference.

4. Why is potentiometer preferred over a voltmeter? [1 Mark]

Answer:
1. A potentiometer draws no current from the circuit at the null point, measuring the true emf.
2. A practical voltmeter draws a small current and measures terminal potential difference rather than open-circuit emf.

Teacher's Note:
a) The potentiometer acts as an ideal voltmeter with infinite resistance at balance.
b) This eliminates errors caused by the internal resistance of the cell being measured.

5. Write the principle of [2 Marks]
(i) A meter bridge.
(ii) A potentiometer.

Answer:
1. (i) Meter bridge: It works on the principle of Wheatstone bridge, where at null deflection, the ratio of resistances in the two arms equals the ratio of the corresponding wire lengths: \(\frac{P}{Q} = \frac{R}{S}\).
2. (ii) Potentiometer: It works on the principle that the potential drop across any portion of the wire of uniform cross-section is directly proportional to the length of that portion when a constant current flows through it (\(V \propto l\)).

Teacher's Note:
a) Both instruments rely on zero-deflection detection methods for high precision.
b) Uniform cross-section and homogeneous material of the bridge wire are essential for accuracy.

6. How does the balancing point of a Wheatstone bridge get affected when [2 Marks]
i) Position of cell and Galvanometer are interchanged?
ii) Position of the known and unknown resistances is interchanged?

Answer:
1. i) The balancing point of the Wheatstone bridge remains completely unaffected when the positions of the cell and galvanometer are interchanged.
2. ii) The balancing point shifts to a new position, interchanging the ratio arms, so that if the original balance length was l, the new balance length becomes \(100 - l\).

Teacher's Note:
a) This property follows directly from reciprocity theorems applied to bridge networks.
b) Interchanging known and unknown resistances swaps the arm ratios, altering the physical balance location on a metre bridge wire.

7. Explain with a neat circuit diagram, how will you compare emf of two cells using a potentiometer?

[Figure: Circuit diagram of a potentiometer setup to compare emf of two cells having positive terminals connected to the high potential end of the potentiometer wire, connected through a two-way key, a galvanometer, and a jockey to the potentiometer wire.]

Answer:
1. Setup a primary circuit with a driver cell of emf E, a rheostat Rh, and a plug key K connected across a long uniform potentiometer wire AB.
2. Connect the positive terminals of both cells (emf E₁ and E₂) to the high potential terminal A, and their negative terminals to two independent ways of a two-way key.
3. Connect the common terminal of the two-way key through a galvanometer (G) and a jockey (J) to slide along the wire AB.
4. Close key K and plug the key between the positive terminal and cell E₁, then obtain the null point at length l₁.
5. Repeat the procedure for cell E₂ to obtain the null point at length l₂.
6. The ratio of emfs is given by the relation E₁ / E₂ = l₁ / l₂.

Teacher's Note:
a) The potentiometer works on the principle that potential drop across any portion of the wire is directly proportional to its length.
b) Ensure the emf of the driver cell is greater than the emfs of both experimental cells.
c) A sensitive galvanometer is required to accurately locate the exact zero-deflection null point.

8. With the help of a circuit diagram, describe the method of finding the internal resistance of the Primary Cell using a potentiometer. [3 Marks]

[Figure: Circuit diagram showing potentiometer wire AB, a driver cell with rheostat, a primary cell in the secondary circuit with a resistance box R connected across it via a plug key K, and a galvanometer with jockey.]

Answer:
1. Connect the primary cell of emf E and internal resistance r in the secondary circuit with a resistance box R connected across it through a plug key K.
2. First, keep key K open and find the balancing length l₁ on the potentiometer wire, which corresponds to the emf E of the cell.
3. Close the plug key K so that a known resistance R is connected in parallel with the cell, and find the new balancing length l₂ corresponding to the terminal potential difference V.
4. Use the internal resistance formula r = R \(\left(\frac{l_1}{l_2} - 1\right)\) to calculate the internal resistance of the cell.

Teacher's Note:
a) The balancing length l₁ without resistance gives the open-circuit emf.
b) The balancing length l₂ with shunt resistance R gives the closed-circuit potential difference.
c) Always make sure the positive terminals of all circuits are connected to the same end A of the wire.

9. With the help of a neat circuit diagram describe the method to determine the potential difference across the conductor using a potentiometer. [3 Marks]

[Figure: Circuit diagram showing a potentiometer connected to a driver cell, and a secondary circuit with a conductor carrying current through a rheostat, with two potential probe points connected across a galvanometer and jockey.]

Answer:
1. Set up the main potentiometer circuit using a driver cell, rheostat, and key connected across the potentiometer wire AB.
2. Connect the conductor in series with a battery, a rheostat, and a plug key so that a steady current flows through it.
3. Connect the two ends across which potential difference is to be measured to the terminals of the secondary circuit containing a galvanometer and a jockey.
4. Slide the jockey along the potentiometer wire AB to find the null point where the galvanometer shows zero deflection.
5. If the balancing length is l, the potential difference V is calculated as V = k l, where k is the potential gradient of the wire.

Teacher's Note:
a) A potentiometer measures potential difference accurately because no current is drawn from the source at the balance point.
b) The potential gradient k must be determined precisely using a standard cell beforehand.
c) Check that all connections are tight to avoid contact resistance errors.

10. Calculate the current drawn from the battery in the given network.
(A diagram showing a network of resistors 2 Ω, 1 Ω, 5 Ω, 4 Ω and 2 Ω connected to a 4 V battery.)

Answer:
1. Identify the series and parallel groups of resistors in the network.
2. Reduce the 1 Ω, 5 Ω, 4 Ω, 2 Ω and 2 Ω resistors step by step using series and parallel rules.
3. The equivalent resistance of the whole network works out to 2 Ω.
4. Calculate equivalent resistance and apply Ohm's law to find the total current drawn from the 4V battery as I = 2A.

Teacher's Note:
a) Bridge symmetry simplifies complex networks by eliminating redundant branches.
b) Always verify the ratio of opposite arms before removing the central resistor.

11. Find the value of X and current drawn from the battery of emf 6V of negligible internal resistance if current through 10 ohm is zero [3 Marks]
(A diagram showing a bridge circuit with resistors 2 Ω, 10 Ω, 4 Ω, 3 Ω, 2.4 Ω and X connected across a 6 V battery.)

Answer:
1. Since the current through the 10 ohm resistor is zero, the bridge is balanced.
2. Apply the balanced bridge condition: \(\frac{2}{4} = \frac{3}{X}\).
3. Substituting values gives X = 6 ohms.
4. The 2 Ω and 4 Ω resistors in series (6 Ω) are in parallel with the 3 Ω and 6 Ω resistors in series (9 Ω), giving 3.6 Ω; adding 2.4 Ω gives R = 6 Ω, so I = 6/6 = 1 A.

Teacher's Note:
a) Zero current in the galvanometer branch indicates equal potential at both bridge terminals.
b) Simplify the parallel and series combinations step by step to find total equivalent resistance.
c) The internal resistance is negligible, so terminal voltage equals the emf of 6V.

12. Find the value of the unknown resistance X and the current drawn by the circuit from the battery if no current flows through the galvanometer. Assume the resistance per unit length of the wire is 0.01 ohm cm\(^{-1}\). [3 Marks]
(A diagram showing a meter bridge setup with lengths 120 cm and 80 cm, unknown resistance X in series with a 2 Ω resistor, a galvanometer joining their junction to the wire at the balance point, and a 4 V battery.)

Answer:
1. Identify the balancing lengths from the diagram as l₁ = 120 cm and l₂ = 80 cm.
2. Apply the meter bridge balance condition: \(\frac{X}{R}\) = \(\frac{l_1}{l_2}\).
3. Substitute R = 2 ohms, l₁ = 120 cm, and l₂ = 80 cm to get X = 3 ohms.
4. Calculate the total resistance of the circuit and find the current drawn from the 4V battery.

Teacher's Note:
a) Meter bridge ratio is directly proportional to the lengths of the wire segments.
b) Ensure units of length for both segments are consistent.
c) The galvanometer shows no deflection at the exact null point.

13. In the circuit shown, AB is a resistance wire of uniform cross section in which a potential gradient of \(0.01\text{ V cm}^{-1}\) exists.
(a) If the galvanometer G shows zero deflection, what is the emf \(E_1\) of the cell used? [1.5 Marks]
(b) If the internal resistance of the driver cell increases on some account, how will it affect the balance point in the experiment? [1.5 Marks]

[Figure: A potentiometer circuit showing a driver cell connected across a resistance wire AB, with a galvanometer G and a cell \(E_1\) connected in the secondary circuit, showing a balancing length of 180 cm from end A.]

Answer:
1. Potential gradient \(k = 0.01\text{ V cm}^{-1}\) and balancing length \(l = 180\text{ cm}\).
2. (a) Emf of the cell \(E_1 = k \times l = 0.01\text{ V cm}^{-1} \times 180\text{ cm} = 1.8\text{ V}\).
3. (b) If the internal resistance of the driver cell increases, the current in the main circuit decreases, which decreases the potential gradient across the wire AB.
4. Since \(E = k l\) and \(E\) is constant, a decrease in \(k\) requires a larger balancing length \(l\), so the balance point will shift towards end B.

Teacher's Note:
a) The potential gradient represents the potential drop per unit length of the potentiometer wire.
b) Students must remember that current reduction in the primary circuit lowers the potential gradient directly.
c) Always relate the shift of the balance point inversely to the change in the potential gradient \(k\).

14.* In a potentiometer circuit, a battery of negligible internal resistance is set up as shown to develop a constant potential gradient along the wire AB. Two cells of emfs \(E_1\) and \(E_2\) are connected in series as shown in the combination (1) and (2). The balance points are obtained respectively at \(400\text{ cm}\) and \(240\text{ cm}\) from the point A. Find: [3 Marks]
(i) \(E_1 / E_2\)
(ii) balancing length for the cell \(E_1\) only.

[Figure: Potentiometer circuit diagram showing two cells connected in series aiding and opposing each other, with wire AB and galvanometer G.]

Answer:
1. When cells assist each other, the effective emf is \(E_1 + E_2\) and the balancing length is \(l_1 = 400\text{ cm}\), so \(E_1 + E_2 \propto 400\).
2. When cells oppose each other, the effective emf is \(E_1 - E_2\) and the balancing length is \(l_2 = 240\text{ cm}\), so \(E_1 - E_2 \propto 240\).
3. Taking the ratio gives \((\frac{E_1 + E_2}{E_1 - E_2}) = \frac{400}{240} = \frac{5}{3}\), which simplifies to \(3E_1 + 3E_2 = 5E_1 - 5E_2\), yielding \(2E_1 = 8E_2\), or \(E_1 / E_2 = 4\).
4. Letting \(E_1 = 4x\) and \(E_2 = x\), we get \(E_1 + E_2 = 5x \propto 400\text{ cm}\), and for \(E_1\) alone, \(E_1 = 4x \propto l_1\), giving \(l_1 = \frac{4}{5} \times 400 = 320\text{ cm}\).

Teacher's Note:
a) Series aiding adds the emfs, while series opposing subtracts them in the potentiometer loop.
b) A common error is mixing up the numerator and denominator when solving the linear ratio equation.
c) The potential gradient \(k\) remains constant throughout the wire for both measurement setups.

15.* A potentiometer wire of length \(100\text{ cm}\) having a resistance of \(10\,\Omega\) is connected in series with a resistance and cell of emf \(2\text{ V}\) of negligible internal resistance. A source emf of \(10\text{ mV}\) is balanced against a length of \(40\text{ cm}\) of potentiometer wire. What is the value of the external resistance? [3 Marks]

[Figure: Potentiometer wire of length \(100\text{ cm}\) connected in series with cell of \(2\text{ V}\) and external resistor R, balancing a cell of \(10\text{ mV}\) at \(40\text{ cm}\).]

Answer:
1. The total resistance of the potentiometer wire is \(10\,\Omega\), and its length is \(100\text{ cm}\).
2. The resistance of \(40\text{ cm}\) of the wire is \(R_{40} = (\frac{40}{100}) \times 10 = 4\,\Omega\).
3. The current in the primary circuit is \(I = \frac{2}{R + 10}\), so the potential drop across \(40\text{ cm}\) of the wire is \(V_{40} = I \times 4 = (\frac{2}{R + 10}) \times 4\).
4. Equating this potential drop to the balanced emf of \(10\text{ mV}\) (\(10 \times 10^{-3}\text{ V}\)), we get \((\frac{8}{R + 10}) = 0.01\), which gives \(R + 10 = 800\), resulting in \(R = 790\,\Omega\).

Teacher's Note:
a) Current in the primary circuit depends on the total resistance of the driver circuit including the wire and resistor R.
b) Always convert millivolts to volts carefully before equating potential drops.
c) The resistance of the wire segment is proportional to its length since the wire has uniform cross section.

16.* In the potentiometer circuit shown, the balance point is at X. State with reason where the balance point will be shifted when [3 Marks]
(i) Resistance R is increased, keeping all parameters unchanged.
(ii) Resistance S is increased keeping R constant.
(iii) Cell P is replaced by another cell whose emf is lower than that of that cell Q.

[Figure: Potentiometer circuit diagram showing a driver cell E connected across wire AB in series with a resistance R and a cell P, with a galvanometer G connected to a sliding contact X and a cell Q in the secondary circuit.]

Answer:
1. (i) As resistance R is increased, the current in the potentiometer wire decreases, which decreases the potential gradient \(\frac{V}{L}\). Hence, the balance point X will shift towards B to balance the same potential difference.
2. (ii) Resistance S is part of the auxiliary circuit containing cell Q and does not affect the main circuit current or potential gradient across the wire AB. Hence, there is no effect on the balance point.
3. (iii) If the driver cell P is replaced by another cell whose emf is lower than that of cell Q, the potential drop across the potentiometer wire AB will become less than the emf of cell Q. Therefore, the balance point is not found on the wire.

Teacher's Note:
a) The potential gradient \(k = \frac{V}{L}\) depends directly on the current flowing through the potentiometer wire.
b) A common mistake is assuming changes in the secondary circuit affect the potential gradient on the primary wire AB.
c) The balance condition requires that the emf of the test cell must be less than the total potential drop across the potentiometer wire.

17.* A potentiometer wire has a length L and resistance \(R_0\). It is connected to a battery and a resistance combination as shown. Obtain an expression for the potential difference per unit length of the potentiometer wire. What is the maximum emf of a ‘test cell’ for which one can get a balance point on this potentiometer wire? What precautions should one take while connecting this test cell to the circuit? [3 Marks]

[Figure: Circuit diagram showing a battery E connected in series with a resistance R, a switch S, and a potentiometer wire AB of length L and resistance \(R_0\), along with a parallel resistor combination.]

Answer:
1. Total resistance of the circuit is given by \(R_{total} = R_0 + \frac{RS}{R + S}\).
2. Current in the main circuit is \(I = \frac{E}{R_0 + \frac{RS}{R + S}}\).
3. Total potential difference across the potentiometer wire AB is \(V = I R_0 = \frac{E R_0}{R_0 + \frac{RS}{R + S}}\).
4. Therefore, the potential difference per unit length is \(\frac{V}{L} = \frac{E R_0}{L \left(R_0 + \frac{RS}{R + S}\right)}\).
5. The maximum emf of a test cell must be less than the total potential difference V across the wire AB.
6. Precaution: The positive terminal of the test cell and the positive terminal of the driver battery must be connected to the same terminal A of the potentiometer wire.

Teacher's Note:
a) Circuit analysis requires combining series and parallel resistances to find the total current.
b) Students must carefully derive the potential drop across only the potentiometer wire, not the entire circuit.
c) Always verify that the potential gradient formula has dimensions of potential divided by length.

18. The variation of potential difference V with length l in case of two potentiometers X and Y as shown. Which one of these will you prefer for comparing emfs of two cells and why? [3 Marks]

[Figure: Graph showing linear variation of potential difference V versus length l with two straight lines labeled X and Y, where line Y has a smaller slope than line X.]

Answer:
1. The potential gradient \(k = \frac{V}{l}\) is represented by the slope of the V-l graph.
2. Potentiometer Y has a smaller slope (lower potential gradient) compared to potentiometer X.
3. A smaller potential gradient provides a higher sensitivity, allowing small potential differences to be measured more accurately.
4. Therefore, potentiometer Y is preferred for comparing the emfs of two cells.

Teacher's Note:
a) Sensitivity of a potentiometer is inversely proportional to its potential gradient.
b) A smaller gradient means a larger balancing length for a given emf, reducing percentage measurement errors.
c) Always relate the slope of the graphical representation directly to the sensitivity parameter in measurement instruments.

Ans : The potentiometer Y is preferred, as it has low potential gradient (V/l) [1 Mark]

[Figure: Graph showing two lines for potentiometers X and Y on V versus l axes, where Y has a smaller slope representing a lower potential gradient]

Answer:
1. Potential gradient \(k = \frac{V}{l}\).
2. A potentiometer with a smaller potential gradient is more sensitive.
3. Therefore, potentiometer Y is preferred due to its lower potential gradient.

Teacher's Note:
a) Sensitivity of a potentiometer is inversely proportional to its potential gradient.
b) A lower potential gradient allows measurement of smaller potential differences.

19. Two cells of emfs e₁ and e₂ (e₁ > e₂) are connected as shown in figure. When a potentiometer is connected between A and B, the balancing length of the potentiometer wire is 300 cm. On connecting the same potentiometer between A and C, the balancing length is 100 cm. Calculate the ratio of e₁ and e₂. [3 Marks]

[Figure: Circuit diagram showing cells of emf e₁ and e₂ connected in opposition between terminals A, B, and C with balancing lengths]

Answer:
1. When connected between A and B, balancing length l₁ = 300 cm, so e₁ = kl₁ = 300k.
2. When connected between A and C, the cells are in opposing polarities, balancing length l₂ = 100 cm, so e₁ - e₂ = kl₂ = 100k.
3. Dividing the two equations, \(\frac{e_1}{e_1 - e_2} = \frac{300}{100} = 3\).
4. Solving the equation, e₁ = 3(e₁ - e₂) which gives 2e₁ = 3e₂.
5. Hence, the ratio \(\frac{e_1}{e_2} = \frac{3}{2}\).

Teacher's Note:
a) The balancing length is directly proportional to the emf of the cell connected in the secondary circuit.
b) When two cells are connected in opposition, the effective emf is the difference of their individual emfs.
c) Always check the polarity of the cells before writing the effective emf expression.

(VALUE BASED QUESTION)

Q1. That night Vaikunth was preparing for his physics exam. Suddenly the light in his room went off and he could not continue his studies. His cousin brother Vasu who had come to visit him was quick to react. Vasu using the torch (an android application) installed in his mobile phone found that the fuse had blown out. He checked the wiring and located a short circuit. He rectified it and put a fuse wire. The light came to life again. Vaikunth had a sigh of relief, He thanked Vasu and continued his studies. [4 Marks]
(1) What are the values projected by Vaikunth and Vasu? (Any two)
(2) Why did Vasu have to check the wiring?
(3) What is an electric fuse? What characteristics you would prefer for a fuse wire?

Answer:
1. Vaikunth displayed the value of acknowledging the help from others with gratitude.
2. Vasu displayed awareness of technology, helping tendency, and practical knowledge of the subject.
3. Vasu had to check the wiring to locate the root cause of the problem, which was a short circuit that caused the fuse to blow out.
4. An electric fuse is a safety device used in electrical circuits which melts and breaks the circuit when current exceeds the safe limit, possessing a low melting point and high resistivity.

Teacher's Note:
a) Value-based questions assess both scientific understanding and ethical or social attributes.
b) An ideal fuse wire must have a low melting point so it melts quickly on overheating, and high resistivity so it heats up rapidly.
c) Short circuits occur when live and neutral wires come in direct contact, causing a sudden surge of heavy current.
d) Always verify safety device characteristics in household circuits to prevent electrical hazards.

Q2 Kumaran wanted to pay electricity bill that day. He realized that the consumption shown by the meter was unbelievably low. He thought that the meter must have been faulty. He wanted to check the meter. But unfortunately he did not have any idea as to how to do this. There came his friend Subhash to help him. He told Kumaran to run only the electric heater rated 1kW in his house for some time keeping other appliances switched off. He also calculated the power consumed in kilowatt hour and compared the value with the meter. Kumaran was happy and thanked Subash for his timely help and the knowledge. [4 Marks]

(1) What are the values displayed by the friends? [1 Mark]

Answer:
1. Honesty
2. Sharing of knowledge
3. Willingness to help

Teacher's Note:
a) This is a value-based question testing moral attributes in daily life.
b) Students must identify positive social traits depicted in the scenario.

(2) Express kWh in joules. Find the resistance of the heater. [3 Marks]

Answer:
1. 1 kWh = \(1 \times 10^3 \text{ W} \times 3600 \text{ s} = 3.6 \times 10^6 \text{ J}\).
2. Power of the heater, P = 1 kW = 1000 W, assuming standard voltage V = 220 V.
3. Resistance, R = \(\frac{V^2}{P}\) = \(\frac{220^2}{1000}\) = \(\frac{48400}{1000}\) = 48.40 ohms.

Teacher's Note:
a) Conversion from kilowatt-hour to joules uses fundamental SI unit relationships.
b) The formula R = V²/P relates resistance, power, and voltage.
c) Always state the standard operating voltage clearly when it is not explicitly provided in appliance rating problems.

Please click the link below to download CBSE Class 12 Physics Current Electricity chapter notes and important questions.