CBSE Class 12 Physics Current Electricity Frequently Asked Questions

Question. When a potential difference is applied between the ends of a conductor:
a. the free electrons start moving
b. the free electrons are accelerated continuously from the lower potential end to higher potential end of the conductor
c. the free electrons acquire a constant drift velocity from the lower potential end to the higher potential end
d. ions starts vibrating with a larger amplitude
Answer : C, D

Question. Which of the following statements, concerning Kirchoff’s laws, is/are correct?
a. In a network of conductors, the algebraic sum of currents meeting at any junction in the circuit is always equal to zero
b. Kirchoff’s first law is an alternative form of the law of conservation of charge
c. In any closed mesh, the algebraic sum of the products of the currents and resistances of the different parts of the mesh is always equal to the algebraic sum of different.m.f.’s acting in the mesh.
d. none of the above statement is correct
Answer : A, B, C

Question. In the given circuit, the AC source has ω =100 rad/s.
Considering the inductor and capacitor to be ideal, the correct choice(s) is:(are)

a. The current through the circuit, I is 0.3 A
b. The current through the circuit, I is 0.3 √2A
c. The voltage across 100# resistor = 10 √2V
d. The voltage across 50# resistor = 10 V
Answer : A, C

Question. When same quantity of electricity is passes through different electrolytes simultaneously, the masses of different element liberated are found to be proportional to their:
a. electrochemical equivalent
b. chemical equivalent
c. atomic masses
d. atomic numbers
Answer : A, B

Question. If three resistances R₁ , R₂ and R₃ are joined in series through a battery, then:
a. equivalent resistance > highest individual resistance
b. current supplied by source = current in each resistance
c. applied potential difference is shared among the three resistance directly in their ratio
d. applied potential difference is shared among the three resistance inversely in their ratio
Answer : A, B, C

Question. Two resistors having equal resistances are joined in series and a current is passed through the combination. Neglect any variation in resistance as the temperature changes. In a given time interval:
a. equal amounts of thermal energy must be produced in the resistors
b. unequal amounts of thermal energy may be produced
c. the temperature must rise equally in the resistors
d. the temperature may rise equally in the resistors
Answer : A, D

Question. Which of the following statements is/are correct?
a. Resistance of the filament of a bulb is inversely proportional to the power of the bulb
b. Resistance of the filament of a bulb is directly proportional to the power of the bulb
c. Higher is the wattage of bulb, higher is the current that can be allowed to pass through a bulb
d. Higher is the wattage of bulb, lower is the current that can be allowed to pass through a bulb
Answer : A, C

Question. If three resistances R₁, R₂ and R₃ are joined in parallel across some potential source, then:
a. equivalent resistance < lowest individual resistance
b. applied potential difference = potential difference across each resistance
c. current supplied by the source is shared among three resistance inverselyin their ration
d. current supplied by the source is shared among three resistances directly in their ratio
Answer : A, B, C

Question. For the resistance network shown in the figure, choose the correct options(s).

a. The current through PQ is zero
b. I₁ = 3A
c. The potential at S is less than that at Q
d. I₂ = 2A
Answer : A, B, D

Assertion and Reason

Note: Read the Assertion (A) and Reason (R) carefully to mark the correct option out of the options given below:
a. If both assertion and reason are true and the reason is the correct explanation of the assertion.
b. If both assertion and reason are true but reason is not the correct explanation of the assertion.
c. If assertion is true but reason is false.
d. If the assertion and reason both are false.
e. If assertion is false but reason is true.

Question. Assertion: Figure shows the current-voltage (I–V) graphs for a given metallic wire at two different tempera-tures T₁ and T₂. It follows from the graphs that T₂ is greater than T₁.
Reason: The resistance of a metallic conductor increases with increase in temperature.
Answer : A

Question. Assertion: In the metre bridge experiment shown in Fig., the balance length AC corresponding to null deflection of the galvanometer is x. If the radius of the wire AB is doubled, the balanced length becomes 4x.
Reason: The resistance of a wire is inversely proportional to the square of its radius.
Answer : D

Question. Assertion: In the potentiometer circuit shown in Fig. E₁ and E₂ are the emfs of cells C₁ and C₂ respectively with E₁ > E₂. Cell C₁ and C₂ respectively with E₁ > E₂. Cell C₁ has negligible internal resistance. For a given resistor R, the balance length is x. If the diameter of the potentiometer wire AB is increased, the balance length x will decrease.
Reason: At the balance point, the potential difference between AD due to cell C₁= E₂, the emf of cell C_2.
Answer : D

Question. Assertion: Bending a wire does not effect electrical resistance.
Reason: Resistance of wire is proportional to resistivity of material.
Answer : B

Question. Assertion: A 60 W bulb is connected in parallel with a room heater and this combination is connected across the mains. If the 60 W bulb is replaced by a 100 W bulb, the heat produced by the heater will remain the same.
Reason: When resistance are connected in parallel across the mains, the heat produced is inversely proportional to the resistance.
Answer : A

Question. Assertion: If an electric field is applied to a metallic conductor, the free electrons experience a force but do not accelerate; they only drift at a constant speed.
Reason: The force exerted by the electric field is completely balanced by the Coulomb force between electrons and protons.
Answer : C

Question. Assertion: Electric field outside the conducting wire which carries a constant current is zero.
Reason: Net charge on conducting wire is zero.
Answer : A

Question. Assertion: In a simple battery circuit the point of lowest potential is positive terminal of the battery
Reason: The current flows towards the point of the higher potential as it flows in such a circuit from the negative to the positive terminal.
Answer : D

Question. Assertion: In the potentiometer circuit shown in above question, the wire AB is not changed but the value of resistor R is decreased. Then the balance length x will decrease.
Reason: At the balance point, the potential difference between A and D due to cell C₁ = emf E₂ of cell C_2.
Answer : A

Question. Assertion: Electric appliances with metallic body have three connections, whereas an electric bulb has a two pin connection.
Reason: Three pin connections reduce heating of connecting wires.
Answer : C

Question. Assertion: The resistivity of a semiconductor increases with temperature.
Reason: The atoms of a semiconductor vibrate with larger amplitude at higher temperatures thereby increasing its resistivity
Answer : D

Question. Assertion: The e.m.f. of the driver cell in the potentiometer experiment should be greater than the e.m.f. of the cell to be determined.
Reason: The fall of potential across the potentiometer wire should not be less than the e.m.f. of the cell to be determined.
Answer : C

Question. Assertion: The temperature coefficient of resistance is positive for metals and negative for p-type semiconductor.
Reason: The effective charge carriers in metals are negatively charged whereas in p-type semiconductor they are positively charged. 
Answer : B

Question. Assertion: The drift velocity of electrons in a metallic wire will decrease, if the temperature of the wire is increased.
Reason: On increasing temperature, conductivity of metallic wire decreases.
Answer : B

Comprehension Based

Paragraph –I

A conducting balloon of radius ‘a’ is charged to a potential V₀ and held at a large height above the earth surface. The large height of the balloon from the earth ensures that charge distribution on the surface of the balloon remains unaffected by the presence of the earth. It is connected to the earth through a resistance R and a valve in the balloon is opened. The gas inside the balloon escapes from the valve and the size of the balloon decreases. The rate of decrease in radius of the balloon is controlled in such a manner that potential of the balloon remains constant. Assume the electric permittivity of the surrounding air equals to that of free space (ε₀) and charge cannot leak to the surrounding air.

Question. Rate at which radius r of the balloon changes with time is best represented by the equation:

Answer : B

Question. How much heat is dissipated in the resistance R till the moment radius of the balloon becomes half?
a. 0.5πε₀ aV₀²
b. πε₀ aV₀²
c. 2πε₀ aV₀²
d. 4πε₀ aV₀²
Answer : C

Paragraph -II

A thermal power plant produces electric power of 600 kW at 4000 V, which is to be transported to a place 20 km away from the power plant for consumers’ usage. It can be transported either directly with a cable of large current carrying capacity or by using a combination of step-up and step-down transformers at the two ends. The drawback of the direct transmission is the large energy dissipation. In the method using transformers, the dissipation is much smaller. In this method, a step-up transformer is used at the plant side so that the current is reduced to a smaller value. At the consumers’ end, a step-down transformer is used to supply power to the consumers at the specified lower voltage. It is reasonable to assume that the
power cable is purely resistive and the transformers are ideal with the power factor unity. All the currents and voltage mentioned are rms values.

Question. If the direct transmission method with a cable of resistance 0.4 Ω km^–1 is used, the power dissipation (in %) during transmission is:
a. 20
b. 30
c. 40
d. 50
Answer : B

Question. In the method using the transformers, assume that the ratio of the number of turns in the primary to that in the secondary in the step-up transformer is 1 : 10. If the power to the consumers has to be supplied at 200 V, the ratio of the number of turns in the primary to that in the secondary in the step-down transformer is:
a. 200 : 1
b. 150 : 1
c. 100 : 1
d. 50 : 1
Answer : A

Integer

Question. The length of a potentiometer wire is 10 cm. A cell of emf E is balanced at a length (10/3)cm cm from the positive end of the wire. If the length of the wire is increased by 5cm, at what distance (in cm) from positive end will the same cell give a balance point?
Answer : 5

Question. Fig., for what value of R (in ohm) will the current in galvanometer G be zero?

Answer : 7

Question. A battery of emf E and internal resistance r is connected to an external resistance R. The maximum power in the external circuit is 9W. The current flowing in the circuit under the condition is 3A. What is the value of E in volt?
Answer : 6

Question. In the circuit, Fig., what should be the value of r in ohm so that power developed in the resistor r will be maximum?
Answer : 2

Question. When two identical batteries of internal resistance 1# each are connected in series across a resistor R, the rate of heat produced in R is J₁. When the same batteries are connected in parallel across R, the rate is J₂. If J₁ = 2.25 J₂, then the value of R in ohm is:
Answer : 4

Question. A wire of length L and 3 identical cells of negligible internal resistances are connected in series. Due to this current, the temperature of the wire is raised by ΔT in time t. A number N of similar cells is now connected in series with a wire of the same material and cross-section but of length 2L. The temperature of wire is raised by same amount ΔT in the same time t. The value of N is:
Answer : 6

Match the Column

Question. Match the statement of Column with those in Column II:

a. A→3, B→2, C→1, D→4
b. A→4, B→1, C→3, D→2
c. A→1, B→2, C→3, D→4
d. A→2, B→1, C→3, D→4
Answer : 

Question. Match the statement of Column with those in Column II:

a. A→4, B→2, C→3, D→1
b. A→3, B→1, C→4, D→2
c. A→2, B→1, C→3, D→4
d. A→2, B→1, C→4, D→3
Answer : 

Question. Match the statement of Column I with those in Column II

a. A→ 1; B → 2,4; C → 2,4; D → 2,3,4
b. A→ 2; B → 3,4; C → 1,3; D → 1,2,3
c. A→ 3; B → 1,3; C → 1,2; D → 2,3,4
d. A→ 4; B → 3,4; C → 2,3; D → 1,2,3
Answer : 

UNIT II
CURRENT ELECTRICITY

1 MARKS

1. Two wires 'A' & 'B' are of the same metal and of the same length. Their areas of cross-section are in the ratio of 2:1. if the same potential difference is applied across each wire in turn, what will be the ratio of the currents flowing in 'A' & 'B'? [1 Mark]

Answer:
1. Resistance is given by R = \(\frac{\rho l}{A}\), where \(\rho\) is resistivity, l is length, and A is area.
2. Since the wires are of the same metal and length, resistance is inversely proportional to area (R \(\propto\) \(\frac{1}{A}\)).
3. The ratio of areas is \(\frac{A_A}{A_B} = \frac{2}{1}\), so the ratio of resistances is \(\frac{R_A}{R_B} = \frac{1}{2}\).
4. According to Ohm's law, current I = \(\frac{V}{R}\), meaning current is inversely proportional to resistance for a constant potential difference V.
5. Therefore, the ratio of currents is \(\frac{I_A}{I_B} = \frac{R_B}{R_A} = \frac{2}{1}\).

Teacher's Note:
a) Use the fundamental resistance formula involving resistivity, length, and cross-sectional area.
b) Remember that current is inversely proportional to resistance when potential difference remains constant.

2. Explain, with the help of a graph, the variation of conductivity with temperature for a metallic conductor. [1 Mark]

Answer:
1. Conductivity (\(\sigma\)) of a metallic conductor decreases non-linearly with an increase in temperature.
2. As temperature increases, thermal agitation of ions increases, which decreases the relaxation time of free electrons, thereby lowering conductivity.

Teacher's Note:
a) Metals possess a positive temperature coefficient of resistivity, implying conductivity drops as temperature rises.
b) Sketch the curve showing a smooth downward trend concave upwards.

3. Draw V-I graph for ohmic and non-ohmic materials. Give one example for each. [1 Mark]

Answer:
1. For ohmic materials, the V-I graph is a straight line passing through the origin (Example: Metallic conductors like copper wire).
2. For non-ohmic materials, the V-I graph is non-linear (Example: Junction diode or electrolyte).

Teacher's Note:
a) Ohmic devices strictly obey Ohm's Law maintaining constant resistance.
b) Non-ohmic devices show voltage-dependent resistance values.

4. Explain how does the resistivity of a conductor depend upon (i) number density 'n' of free electrons, & (ii) relaxation time 't'. [1 Mark]

Answer:
1. Resistivity is given by the relation \(\rho = \frac{m}{n e^2 \tau}\), where m is mass, e is electronic charge, n is number density, and \(\tau\) is relaxation time.
2. (i) Resistivity is inversely proportional to the number density of free electrons (\(\rho \propto \frac{1}{n}\)).
3. (ii) Resistivity is inversely proportional to the relaxation time of free electrons (\(\rho \propto \frac{1}{\tau}\)).

Teacher's Note:
a) Cite the microscopic expression for electrical resistivity derived from drift velocity.
b) Note that relaxation time decreases with temperature rise in metals.

6. A cell of emf (E) and internal resistance (r) is connected across a variable external resistance (R) Plot graphs to show variation of (i) E with R (ii) terminal p.d. of the cell (V) with R. [1 Mark]

Answer:
1. (i) EMF (E) of a cell is constant and independent of the external resistance R, yielding a straight line parallel to the R-axis.
2. (ii) Terminal potential difference is given by V = \(\frac{E R}{R + r}\), which increases non-linearly with R and approaches E as R approaches infinity.

Teacher's Note:
a) Emf is a characteristic property of the source and does not vary with load.
b) Terminal voltage equals EMF only when open circuit conditions apply (R = \(\infty\)).

7. Explain how electron mobility changes from a good conductor
(i) when temperature of the conductor is decreased at constant potential difference
and (ii) applied potential difference is doubled at constant temperature. [1 Mark]

Answer:
1. Mobility is defined as \(\mu = \frac{v_d}{E} = \frac{e\tau}{m}\).
2. (i) When temperature is decreased, thermal vibrations decrease, increasing relaxation time (\(\tau\)), which in turn increases electron mobility.
3. (ii) Mobility is independent of the applied potential difference or electric field, so it remains unchanged when potential difference is doubled.

Teacher's Note:
a) Mobility depends on relaxation time and effective mass of the charge carriers.
b) Changes in applied voltage alter drift velocity proportionally, keeping mobility constant.

8. Write the mathematical relation between mobility and drift velocity of charge carriers in a conductor. Name the mobile charge carriers responsible for conduction of electric current in: (i) an electrolyte, & (ii) an ionised gas. [2 Marks]

Answer:
1. The mathematical relation between mobility (\(\mu\)) and drift velocity (\(v_d\)) is \(\mu = \frac{v_d}{E}\), where \(E\) is the applied electric field.
2. (i) In an electrolyte, the mobile charge carriers are positive and negative ions.
3. (ii) In an ionised gas, the mobile charge carriers are electrons and positive ions.

Teacher's Note:
a) Mobility is defined as the drift velocity per unit electric field.
b) Electrolytes conduct through ionic motion while gases require ionisation to create free electrons and ions.

9. Define drift velocity. Establish a relation between current & drift velocity. [3 Marks]

Answer:
1. Drift velocity is defined as the average velocity with which free electrons get drifted towards the positive terminal under the influence of an applied electric field.
2. Consider a conductor of length \(l\) and cross-sectional area \(A\). The volume of the conductor is \(Al\), and if \(n\) is the number density of electrons, the total number of free electrons in the conductor is \(nAl\).
3. The total charge passing through the conductor is \(Q = neAl\). Since the time taken by electrons to cross the conductor is \(t = \frac{l}{v_d}\), the electric current is given by \(I = \frac{Q}{t} = \frac{neAl}{l / v_d} = neAv_d\).

Teacher's Note:
a) Drift velocity is extremely small, typically of the order of \(10^{-4}\) m/s.
b) The macroscopic current is directly proportional to the microscopic drift velocity.
c) Always define the symbols like \(n, e, A\) clearly when deriving \(I = neAv_d\).

10. Define the term current density of a metallic conductor. Deduce the relation connecting current density ‘J’ & the conductivity ‘s’ of the conductor when an electric field ‘E’ is applied to it. [3 Marks]

Answer:
1. Current density at any point in a conductor is defined as the amount of charge flowing per unit time per unit normal cross-sectional area, given by \(J = \frac{I}{A}\).
2. We know that current is \(I = neAv_d\) and drift velocity is \(v_d = \frac{eE\tau}{m}\), so substituting this gives \(I = neA\left(\frac{eE\tau}{m}\right) = \frac{ne^2\tau A E}{m}\).
3. Since current density \(J = \frac{I}{A} = \frac{ne^2\tau E}{m}\), and conductivity \(\sigma = \frac{ne^2\tau}{m}\), we obtain the vector relation \(J = \sigma E\).

Teacher's Note:
a) Current density is a vector quantity having the same direction as the electric field.
b) This equation is the microscopic form of Ohm's law.
c) Conductivity \(\sigma\) depends on relaxation time \(\tau\) and electron density \(n\).

11. Why do we prefer potentiometer to compare the e.m.f of cells than the voltmeter. Why? [2 Marks]

Answer:
1. A potentiometer draws no current from the cell at the null point, thereby measuring the true open-circuit emf of the cell.
2. A voltmeter draws a small current for its deflection, causing a potential drop across its internal resistance, and hence measures terminal potential difference rather than true emf.

Teacher's Note:
a) Potentiometer works on the null deflection principle.
b) Infinite effective resistance at balance condition makes it superior to any standard voltmeter.

12. State Kirchhoff’s rules of current distribution in an electric network. [2 Marks]

Answer:
1. Kirchhoff's First Rule (Junction Rule): The algebraic sum of currents meeting at any junction in a closed electrical network is zero (\(\sum I = 0\)), which is based on the conservation of charge.
2. Kirchhoff's Second Rule (Loop Rule): The algebraic sum of changes in potential around any closed loop involving resistors and cells is zero (\(\sum IR + \sum E = 0\)), which is based on the conservation of energy.

Teacher's Note:

a) Sign conventions must be strictly followed while applying the loop rule.
b) Junction rule implies that charge cannot accumulate at a node.

13. The variation of potential difference “V’ with length ‘l’ in the case of two potentiometers ‘X’ & ‘Y’ is as shown in figure. Which one of these two will you prefer for comparing ‘emf’s of two cells and why? [2 Marks]

Answer:
1. Potentiometer 'X' will be preferred for comparing the emfs of two cells.
2. This is because the potential gradient (\(k = \frac{V}{l}\)) of potentiometer 'X' is greater than that of 'Y', providing greater sensitivity and accuracy in measurement.

Teacher's Note:
a) Potential gradient is represented by the slope of the V versus l graph.
b) A steeper slope means a larger potential drop per unit length, leading to a more precise null point.

14. Draw a circuit diagram using a metre bridge and write the necessary mathematical relation used to determine the value of an unknown resistance. Why cannot such an arrangement be used for measuring very low resistance? [2 Marks]

Answer:
1. The unknown resistance S is given by the relation \(S = R\left(\frac{100 - l}{l}\right)\), where R is the known resistance in the left gap and l is the balancing length from the left end.
2. Such an arrangement cannot be used for measuring very low resistance because end errors due to contact resistance at the ends of the copper strips become comparable to the small resistance being measured, leading to large percentage errors.

Teacher's Note:
a) The metre bridge is based on the principle of the Wheatstone bridge.
b) Thick copper strips are used to minimize resistance of connecting wires.

15. With the help of a circuit diagram, explain in brief the use of a potentiometer for comparison of emf's of two cells. [2 Marks]

Answer:
1. Connect the positive terminals of both cells to the high potential end A of the potentiometer wire and negative terminals through a two-way key to a galvanometer and jockey.
2. Obtain the balancing length l₁ for the first cell and l₂ for the second cell, then the ratio of their emfs is given by \(\frac{E_1}{E_2} = \frac{l_1}{l_2}\).

Teacher's Note:
a) Potentiometer is preferred over a voltmeter because it draws no current at the null point.
b) The potential gradient along the wire must remain constant.

16. Prove that the current density of a metallic conductor is directly proportional to the drift speed of electrons. [2 Marks]

Answer:
1. The relation between electric current and drift speed is given by I = neAv_d, where n is electron density, e is electronic charge, A is cross-sectional area, and v_d is drift speed.
2. Current density j is defined as j = I / A, substituting I gives j = nev_d, which proves that j is directly proportional to v_d.

Teacher's Note:
a) Current density is a vector quantity directed along the electric field.
b) n and e are constants for a given conductor.

17. A number of identical cells, n, each of emf E, internal resistance r connected in series are charged by a d.c. source of emf E', using a resistor R.
(i) Draw the circuit arrangement. [1 Mark]
(ii) Deduce the expressions for (a) the charging current and (b) the potential difference across the combination of the cells. [2 Marks]

Answer:
1. (i) Circuit arrangement consists of n cells in series opposing or aiding a d.c. source E' with a series resistor R.
2. (ii) (a) The total emf in the charging circuit is (E' - nE) and total resistance is (R + nr), so the charging current is I = \(\frac{E' - nE}{R + nr}\).
3. (ii) (b) The potential difference across the combination of cells during charging is V = E + Ir.

Teacher's Note:
a) During charging, current enters the positive terminal of the battery.
b) The terminal potential difference is greater than the emf during charging.

18. Derive the principle of wheatstone bridge using Kirchhoff's law. [2 Marks]

Answer:
1. Apply Kirchhoff's current law at junctions B and D in the bridge network to find current relations.
2. Apply Kirchhoff's voltage law to closed loops ABDA and BCDB, and substitute conditions for null deflection (I_g = 0) to get \(\frac{P}{Q} = \frac{R}{S}\).

Teacher's Note:
a) Wheatstone bridge is used for accurate measurement of unknown resistance.
b) The bridge is most sensitive when all four arms have comparable resistances.

19. State Kirchhoff's rules of current distribution in an electrical network. Using these rules determine the value of the current I₁ in the electric circuit given below. [2 Marks]

[Figure: Network showing junctions, resistors of 20 ohms, 20 ohms, and sources of 40V, 80V with branches carrying currents I₁, I₂, I₃]

Answer:
1. Kirchhoff's First Rule (Junction Rule): The algebraic sum of currents meeting at any junction in a closed circuit is zero.
2. Kirchhoff's Second Rule (Loop Rule): The algebraic sum of changes in potential around any closed loop is zero.
3. Applying Kirchhoff's rules to the given network loops, the current I₁ is evaluated to be 1.0 A.

Teacher's Note:
a) Junction rule is based on conservation of charge.
b) Loop rule is based on conservation of energy.

20. Write the mathematical relation for the resistivity of material in terms of relaxation time, number density and mass and charge of charge carriers in it. Explain, using this relation, why the resistivity of a metal increases and that of semi-conductor decreases with rise in temperature. [2 Marks]

Answer:
1. The mathematical relation for resistivity is \(\rho = \frac{m}{ne^2\tau}\), where m is mass, n is number density, e is charge, and \(\tau\) is relaxation time.
2. In metals, with rise in temperature, thermal agitation increases, decreasing relaxation time \(\tau\), which increases resistivity.
3. In semiconductors, with rise in temperature, number density n increases much more rapidly than the decrease in \(\tau\), causing resistivity to decrease.

Teacher's Note:
a) Relaxation time is the average time between successive collisions of electrons.
b) Semiconductors have a negative temperature coefficient of resistivity.

21. Calculate the value of the resistance R in the circuit shown in the figure so that the current in the circuit is 0.2 A. What would be the potential difference between points A and B? [2 Marks]

[Figure: Complex resistor network with 15 ohms, 30 ohms, 5 ohms, 10 ohms resistors, 8V and 3V DC sources, and unknown resistor R carrying 0.2 A]

Answer:
1. Using total circuit current of 0.2 A, apply Ohm's law across the total equivalent resistance of the network including sources.
2. Solving the loop equations yields the unknown resistance R = 25 ohms.
3. The potential difference between points A and B calculated from the branch currents is 2.0 V.

Teacher's Note:
a) Simplify parallel and series combinations of resistors before applying Kirchhoff's laws.
b) Pay careful attention to the polarities of the voltage sources.