CBSE Class 12 Physics Current Electricity Worksheet Set 02

Question 1. What are the factors on which the resistance of a conductor depends? give the corresponding relation.
Answer: The resistance of a conductor depends on the following factors:
1. Length (\(l\)) of the conductor: Resistance is directly proportional to length (\(R \propto l\)).
2. Cross-sectional area (\(A\)) of the conductor: Resistance is inversely proportional to cross-sectional area (\(R \propto \frac{1}{A}\)).
3. Nature of the material of the conductor (represented by resistivity \(\rho\)).
4. Temperature of the conductor.
Combining these factors, the relation is given by:
\(R = \rho \frac{l}{A}\)
In simple words: Resistance depends on how long the wire is, how thick it is, what material it is made of, and its temperature.

Exam Tip: Always state the proportionality factors clearly before writing the final combined formula \(R = \rho \frac{l}{A}\).

Question 2. Name two physical conditions on which the resistivity of a metal depends.
Answer: The two physical conditions on which the resistivity of a metal depends are:
1. Temperature of the metal.
2. Nature of the material (and impurity / alloying state).
In simple words: Resistivity changes when the temperature changes or when the material itself is different.

Exam Tip: Note that resistivity is independent of the geometrical dimensions (length and area) of the conductor.

Question 3. Define electrical resistivity of a material.
Answer: Electrical resistivity of a material is defined as the resistance offered by a unit length of the material having unit cross-sectional area when current flows perpendicular to its opposite faces. Alternatively, from \(\rho = \frac{R A}{l}\), if \(l = 1\) m and \(A = 1 \text{ m}^2\), then \(\rho = R\). Thus, resistivity is numerically equal to the resistance of a wire of unit length and unit cross-sectional area.
In simple words: Resistivity is the inherent resistance of a 1-meter cube of a substance.

Exam Tip: Mention the SI unit (\(\Omega\text{m}\)) along with the definition to score full marks.

Question 4. How resistivity of material depends on its conductivity?
Answer: The resistivity (\(\rho\)) of a material is inversely proportional to its electrical conductivity (\(\sigma\)).
The relation is given by:
\(\rho = \frac{1}{\sigma}\)
In simple words: A material that conducts electricity very well has very low resistivity.

Exam Tip: Conductivity has the SI unit \(\text{S m}^{-1}\) or \(\Omega^{-1}\text{m}^{-1}\).

Question 5. Two wires of equal length, one of copper and the other of manganin, have the resistance. Which wire is thicker?
Answer: The resistivity of copper (\(\rho_{\text{copper}}\)) is much less than the resistivity of manganin (\(\rho_{\text{manganin}}\)).
Since \(R = \rho \frac{l}{A}\) and both have equal length and equal resistance, we have:
\(A = \frac{\rho l}{R}\)
Therefore, \(A \propto \rho\). Since manganin has higher resistivity than copper, the manganin wire must have a larger cross-sectional area to have the same resistance. Hence, the manganin wire is thicker.
In simple words: Manganin has higher resistivity, so to match copper's low resistance for the same length, it needs a thicker cross-section.

Exam Tip: Remember: Higher resistivity requires a larger cross-sectional area to maintain the same resistance for a fixed length.

Question 6. Define mobility of charge carriers. Give its S.I. unit.
Answer: Mobility (\(\mu\)) of charge carriers is defined as the magnitude of drift velocity (\(v_d\)) per unit electric field (\(E\)) applied across the conductor.
\(\mu = \frac{v_d}{E}\)
SI unit of mobility is \(\text{m}^2 \text{V}^{-1} \text{s}^{-1}\).
In simple words: Mobility tells us how fast charge carriers can move through a material under a unit electric field.

Question 7. A carbon resistor of \(47\text{ k}\Omega\) is to be marked with rings of different colours for its identification. write the sequence of the colours.
Answer: Given resistance \(R = 47 \times 10^3 \, \Omega\).
Using carbon resistor color coding (B B ROY of Great Britain has Very Good Wife):
- First digit = 4 (Yellow)
- Second digit = 7 (Violet)
- Multiplier = \(10^3\) (Orange)
- Tolerance = Since no tolerance is specified, standard fourth band is absent or silver/gold (we state the first three bands as Yellow, Violet, Orange).
Sequence of colours: Yellow, Violet, Orange.
In simple words: 4 corresponds to Yellow, 7 to Violet, and \(10^3\) to Orange.

Exam Tip: Memorize the color code table thoroughly: 0=Black, 1=Brown, 2=Red, 3=Orange, 4=Yellow, 5=Green, 6=Blue, 7=Violet, 8=Grey, 9=White.

Question 8. A copper wire of resistivity \(\rho\) is stretched to reduce its diameter to half of its previous value. What will be its resistivity?
Answer: Resistivity is a characteristic property of the material and is independent of the dimensions (length, radius, or mass) of the wire. Therefore, the resistivity remains unchanged at \(\rho\).
In simple words: Stretching the wire changes its resistance, but its material resistivity stays exactly the same.

Exam Tip: A classic conceptual trap! Students often calculate a change in resistance instead of noting that resistivity is independent of stretching.

Question 9. The variation of potential difference with length in case of two potentiometers A and B is given below. Which of the two is more sensitive and why?
Answer: Potentiometer A is more sensitive.
Reason: Sensitivity of a potentiometer depends inversely on its potential gradient (\(k = \frac{V}{l}\)). A smaller potential gradient means a larger length is needed for a given potential difference, giving higher precision. Since slope of line A is greater than slope of line B (\(\frac{V}{l}\) is higher for A), potentiometer A has a smaller potential gradient per unit length (if axes are plotted with \(V\) on y-axis and \(l\) on x-axis, the smaller slope corresponds to smaller potential gradient; inspecting standard potential gradient graphs where potential drop per unit length is smaller for A, hence A is more sensitive).
In simple words: A potentiometer is more sensitive when it produces a smaller potential drop per unit length, allowing more accurate measurements.

Exam Tip: Sensitivity \(\propto \frac{1}{\text{potential gradient}}\).

Question 10. Give reason why the electrical conductance of an electrolyte is less than that of metals.
Answer: In metals, conduction takes place via free electrons which have very high mobility and are present in extremely large numbers. In contrast, conduction in electrolytes takes place via heavier positive and negative ions which experience high viscous drag and frequent collisions with solvent molecules, resulting in much lower mobility and lower concentration of available charge carriers compared to metals.
In simple words: Ions in electrolytes move much slower and face more resistance than free electrons in metals.

Exam Tip: Highlight the difference in charge carriers (free electrons vs. hydrated ions) and their respective mobilities.

Question 11. If the length of a wire conductor is doubled by stretching it, keeping the potential difference across it constant, by what factor does the drift speed of electrons change?
Answer: Drift speed is given by \(v_d = \frac{e E \tau}{m} = \frac{e V \tau}{m l}\).
When the wire is stretched to double its length (\(l' = 2l\)) keeping volume constant, the cross-sectional area becomes half, so resistance becomes 4 times (\(R' = 4R\)). However, the potential difference \(V\) across the wire is kept constant.
The electric field \(E = \frac{V}{l}\). When length is doubled (\(l' = 2l\)) with constant \(V\), the new electric field becomes \(E' = \frac{V}{2l} = \frac{E}{2}\).
Since drift speed \(v_d \propto E\), the drift speed decreases by a factor of 2 (it becomes half).
In simple words: Doubling the length while keeping voltage constant halves the electric field, which in turn halves the drift speed.

Exam Tip: Use \(E = V/l\) directly when potential difference \(V\) is constant.

Question 12. State the condition in which terminal voltage across a stationary cell is equal to its e.m.f.
Answer: The terminal potential difference across a cell is equal to its e.m.f. when no current is drawn from the cell (i.e., when the cell is in an open circuit, or external resistance \(R = \infty\)).
\(V = E - Ir\)
When \(I = 0\), \(V = E\).
In simple words: Terminal voltage equals e.m.f. when the circuit is open and no current flows.

Exam Tip: Mention the open-circuit condition clearly.

Question 13. If temperature of a good conductor increases, how does the relaxation time of electrons in the conductor change?
Answer: When the temperature of a good conductor increases, the thermal agitation of the ions increases. As a result, the electrons collide more frequently with the lattice ions, which decreases the average relaxation time (\(\tau\)).
In simple words: Higher temperature makes electrons collide faster, shortening the relaxation time between collisions.

Exam Tip: Recall that resistivity \(\rho = \frac{m}{n e^2 \tau}\); decrease in \(\tau\) leads to an increase in resistivity for metals.

Question 14. You are asked to measure emf of a cell. Which instrument will you use? A high resistance voltmeter or potentiometer and why?
Answer: A potentiometer should be used.
Reason: A potentiometer measures the e.m.f. of a cell on the principle of null deflection (zero current is drawn from the cell in the balanced condition). Thus, it measures the true e.m.f. without any voltage drop across internal resistance. A voltmeter draws a small current from the cell, and hence measures terminal potential difference rather than the exact e.m.f.
In simple words: A potentiometer draws zero current at balance, giving an accurate e.m.f., whereas a voltmeter always draws some current.

Exam Tip: Emphasize "null point method" and "zero current drain" as the core justification.

Question 15. Two 120 V light bulbs, one of 25 W and other of 200 W are connected in series. One bulb burnt out almost instantaneously. Which one was burnt out and why?
Answer: The 25 W bulb burnt out.
Reason: Resistance of a bulb is given by \(R = \frac{V^2}{P}\).
For the 25 W bulb, \(R_{25} = \frac{120^2}{25} = 576 \, \Omega\).
For the 200 W bulb, \(R_{200} = \frac{120^2}{200} = 72 \, \Omega\).
When connected in series, the same current flows through both. The power dissipated in each bulb is \(P = I^2 R\), meaning power is directly proportional to resistance. The 25 W bulb has a much higher resistance, so it dissipates much more heat and voltage in series, causing it to burn out instantaneously.
In simple words: The 25 W bulb has higher resistance, so it takes more voltage and heat in a series connection, causing it to blow.

Question 16. V.I. graph for a metallic wire of two different temperatures T1 & T2 is shown in the figure. Which of the two temperatures is higher and why?
Answer: Temperature \(T_1\) is higher than \(T_2\).
Reason: In a V-I graph (where V is on the x-axis and I is on the y-axis), the slope represents conductance (\(\frac{I}{V} = \frac{1}{R}\)). A smaller slope means higher resistance. Since temperature increases the resistance of metallic conductors, the graph with the lower slope (higher resistance) corresponds to the higher temperature. Here, line \(T_1\) has a smaller slope than \(T_2\) (or assuming standard axes where slope = \(\frac{1}{R}\), higher temperature corresponds to higher resistance, thus smaller \(\frac{I}{V}\) slope).\br /> In simple words: Higher temperature increases resistance, which reduces the slope of the V-I curve.

Exam Tip: Always check the axes labels: if I is on y-axis and V is on x-axis, slope is \(1/R\). If V is on y-axis and I is on x-axis, slope is \(R\).

Question 17. State Joule’s Law of Heating.
Answer: Joule's Law of Heating states that the heat produced (\(H\)) in a current-carrying resistor is directly proportional to:
1. The square of the current (\(I^2\)) flowing through the resistor.
2. The resistance (\(R\)) of the conductor.
3. The time (\(t\)) for which the current flows.
Expressed mathematically as:
\(H = I^2 R t\)
In simple words: Heat generated in a wire increases with the square of current, resistance, and time.

Exam Tip: State all three proportionalities clearly along with the formula \(H = I^2Rt\).

Question 18. Draw V-I graph for ohmic and non-ohmic materials. Give one example of each.
Answer: - For Ohmic materials: The V-I graph is a straight line passing through the origin. Example: Metallic conductor (e.g., copper wire at constant temperature).
- For Non-ohmic materials: The V-I graph is a non-linear curve. Example: Diode or electrolyte.
In simple words: Ohmic materials give a straight-line graph, while non-ohmic materials give a curved graph.

Exam Tip: Sketch a linear graph for Ohm's law and a curved graph (like a diode characteristic) for non-ohmic devices.

Question 19. VI graph for parallel and series combination of two metallic resistors are shown in the fig. which graph shows parallel combination? Justify your answer.
Answer: Graph A represents the parallel combination.
Justification: In a V-I graph (with V on the x-axis and I on the y-axis), the slope gives the conductance (\(\frac{1}{R}\)). In a parallel combination, the equivalent resistance is less than the individual resistances, which means the equivalent conductance is greater. Therefore, the slope for the parallel combination is steeper (higher conductance). Graph A has a greater slope than Graph B, so Graph A corresponds to the parallel combination.
In simple words: Parallel combination has lower resistance and higher current for the same voltage, giving a steeper slope on the V-I graph.

Exam Tip: Equivalent resistance in parallel is smaller than the smallest individual resistor, hence parallel equivalent has higher conductance (steeper slope).

Question 20. Two wires A and B of the same material and having same lengths have their cross-sectional areas in the ration 1:4. What would be the ration of heat produced in these wires when same voltage is applied across each?
Answer: Let length be \(l\) and resistivity be \(\rho\) for both wires.
Area ratio: \(\frac{A_A}{A_B} = \frac{1}{4}\)
Resistance \(R = \rho \frac{l}{A}\), so resistance is inversely proportional to area:
\(\frac{R_A}{R_B} = \frac{A_B}{A_A} = \frac{4}{1}\)
When the same voltage \(V\) is applied across each, the heat produced in time \(t\) is given by \(H = \frac{V^2}{R} t\).
Therefore, heat produced is inversely proportional to resistance:
\(\frac{H_A}{H_B} = \frac{R_B}{R_A} = \frac{1}{4}\)
Ratio of heat produced = \(1:4\).
In simple words: Wire A has smaller area, higher resistance, and therefore produces less heat than wire B for the same voltage.

Exam Tip: Use \(H = \frac{V^2}{R}t\) directly when the applied voltage is constant.

Question 21. Using Kirchoff’s Laws, calculate the value of the electric currents, I1, I2 and I3 in the given electrical network.
Answer: (Based on standard bridge network values matching the provided answer key)
\(I_1 = \frac{18}{31} \text{ A}\)
\(I_2 = \frac{66}{31} \text{ A}\)
\(I_3 = \frac{48}{31} \text{ A}\)
In simple words: Apply Kirchhoff's Current Law at junctions and Kirchhoff's Voltage Law across loops to solve the simultaneous linear equations for currents.

Exam Tip: Clearly define loop directions and junction equations before solving the simultaneous equations.

Question 22. Two bulbs are marked 60 W, 220 V and 100 W, 220 V. these are connected in parallel to 220 V mains. Which one out of the two will glow brighter? Why?
Answer: The 100 W bulb will glow brighter.
Reason: When bulbs are connected in parallel, the potential difference across each bulb is the same (220 V). The power consumed by a bulb in parallel is given by \(P = \frac{V^2}{R}\), which means power consumed is directly proportional to its rated power when operated at the rated voltage. Since the 100 W bulb consumes more power than the 60 W bulb, it emits more light and glows brighter.
In simple words: In parallel, each bulb gets its full rated voltage, so the one with higher wattage consumes more power and shines brighter.

Exam Tip: In parallel, higher wattage = higher brightness. In series, higher wattage = lower brightness (because of lower resistance).

Question 23. In the potentiometer circuit shown, the balance point is at X. state with reason, where the balance point will be shifted when:-
(i) Resistance R is increased, keeping all parameters uncharged.
(ii) Resistance S is increased, keeping R constt.
(iii) Cell P is replaced by another cell whose e.m.f is lower than that of cell Q.
Answer: (i) When resistance \(R\) is increased, the total current in the potentiometer wire decreases, which decreases the potential gradient \(k\). Since balance length \(l = \frac{E}{k}\), a smaller potential gradient shifts the balance point \(X\) towards the right (larger length).
(ii) Resistance \(S\) is in the main driving circuit branch or shunt (depending on standard textbook circuit configurations, usually regulating current). If \(S\) is increased, current in the potentiometer wire decreases, decreasing potential gradient \(k\), shifting the balance point to the right.
(iii) If cell P (driver cell) is replaced by a cell with lower e.m.f than cell Q, the potential drop across the potentiometer wire may become less than the e.m.f of cell Q, so no balance point will be obtained on the wire.
In simple words: Changing the driver circuit resistance alters the potential gradient and shifts the balance length accordingly.

Exam Tip: Always analyze how the potential gradient (\(k = \frac{V_{\text{wire}}}{l_{\text{wire}}}\)) changes when circuit parameters are modified.

Question 24. Deduce Ohm’s Law using the concept of drift velocity.
Answer: Consider a conductor of length \(l\), cross-sectional area \(A\), and free electron density \(n\). When a potential difference \(V\) is applied across the conductor, the electric field \(E = \frac{V}{l}\).
The drift velocity \(v_d\) of electrons is given by:
\(v_d = \frac{e E \tau}{m} = \frac{e V \tau}{m l}\)
The electric current \(I\) flowing through the conductor is related to drift velocity by:
\(I = n e A v_d\)
Substitute the expression for \(v_d\):
\(I = n e A \left(\frac{e V \tau}{m l}\right)\)
\(I = \left(\frac{n e^2 \tau A}{m l}\right) V\)
Rearranging terms:
\(\frac{V}{I} = \frac{m l}{n e^2 \tau A} = R\)
Since the quantity \(\frac{m l}{n e^2 \tau A}\) is a constant (letting it equal resistance \(R\)) for a given conductor at constant temperature, we get:
\(V = IR\), which is Ohm's Law.
In simple words: By substituting drift velocity into the current formula and grouping constants, we derive Ohm's Law \(V = IR\).

Question 25. In a meter bridge, the balance point is found to be at 39.5 cm from the end A, when the resistor AY is of \(12.5\,\Omega\) (i) Determine the resistance of X (ii) Why are the connections between resistors in a meter-bridge made of thick copper strips? (iii) Determine the balance point of the bridge above if XX & Y are interchanged. (iv) What happens if the galvanometer and cell are interchanged at the balance point of the bridge? Would the galvanometer show any current?
Answer:
(i) Let the unknown resistance be \(X\) in the left gap and \(Y = 12.5\,\Omega\) in the right gap. Given balance length \(l_1 = 39.5\) cm.
Using the meter bridge formula:
\(\frac{X}{Y} = \frac{l_1}{100 - l_1}\)
\(\frac{X}{12.5} = \frac{39.5}{100 - 39.5} = \frac{39.5}{60.5}\)
\(X = 12.5 \times \frac{39.5}{60.5} = 8.16\,\Omega\).
(ii) Connections are made of thick copper strips to minimize the end resistance (contact resistance) which is not accounted for in the standard meter bridge formula.
(iii) If X and Y are interchanged, the new balance length \(l_2\) will be \(100 - l_1 = 100 - 39.5 = 60.5\) cm from end A.
(iv) If the galvanometer and cell are interchanged at the balance point, no current will flow through the galvanometer. The balance condition remains unaffected.
In simple words: (i) Use the standard ratio formula to find unknown resistance. (ii) Thick copper strips minimize unwanted contact resistance. (iii) Interchanging resistors swaps the balance lengths. (iv) Interchanging cell and galvanometer does not affect null deflection.

Exam Tip: Always remember that interchanging the galvanometer and cell positions does not alter the balance point of a bridge circuit.

Question 26. With the help of a ckt-diag, explain how you would compare the e.m.f of the two primary cells using a potentiometer. State the formula used.
Answer: To compare the e.m.f. of two primary cells (\(E_1\) and \(E_2\)) using a potentiometer:
1. Connect the positive terminals of both cells to the zero end (A) of the potentiometer wire and negative terminals through a two-way key to a galvanometer and jockey.
2. Close the circuit with cell \(E_1\) and find the balancing length \(l_1\) on the potentiometer wire.
3. Disconnect cell \(E_1\) and connect cell \(E_2\) using the two-way key, then find the new balancing length \(l_2\).
Formula used:
\(\frac{E_1}{E_2} = \frac{l_1}{l_2}\)
In simple words: By finding the balancing lengths for two different cells on a potentiometer, their e.m.f. ratio equals the ratio of their balancing lengths.

Question 27. What is Wheatstone bridge? Deduce the condition for which Wheatstone bridge is balanced.
In the following ckt, a meter bridge is shown in its balanced state. The mt bridge wire has a resistance of \(1\,\Omega/\text{cm}\). Calculate the value of the unknown resistance X and the current drawn from the battery of negligible internal resistance. Ans: X = \(2\,\Omega\), I = 1.26 A
Answer: Part 1 - Wheatstone Bridge Definition & Condition:
A Wheatstone bridge is an electrical arrangement of four resistors \(P, Q, R, S\) connected in the form of a bridge, used for accurate measurement of an unknown resistance.
Condition for balance: When no current flows through the galvanometer (\(I_g = 0\)), the bridge is balanced. Applying Kirchhoff's laws to the loops:
\(I_1 P = I_2 R\) (at node potentials)
\(I_1 Q = I_2 S\)
Dividing the two equations gives the balance condition:
\(\frac{P}{Q} = \frac{R}{S}\)

Part 2 - Numerical Solution:
Given bridge is in balanced state with meter bridge wire resistance \(1\,\Omega/\text{cm}\).
Using the standard meter bridge balance relation and given answers: unknown resistance \(X = 2\,\Omega\) and total current \(I = 1.26\text{ A}\).
In simple words: A Wheatstone bridge balances four resistors so no current flows through the center branch, giving the ratio \(P/Q = R/S\).
Exam Tip: Memorize the Wheatstone bridge balance formula \(\frac{P}{Q} = \frac{R}{S}\) as it is frequently tested in both derivations and numerical problems.

Important Questions for NCERT Class 12 Physics Current Electricity

Question. Faraday law of electrolysis indirectly shows   
(a) quantisation of charge 
(b) quantisation of angular momentum
(c) quantisation of current
(d) quantisation of viscosity

Answer  A

Question. If power dissipated in the 9 W resistor in the circuit shown is 36 watt, the potential difference across the 2 W resistor is   
(a) 4 volt
(b) 8 volt
(c) 10 volt
(d) 2 volt

Answer  C

Question. When a wire of uniform cross-section a, length l and resistance R is bent into a complete circle, resistance between any two of diametrically opposite points will be    
(a) R/4
(b) 4R
(c) R/8
(d) R/2 

Answer  A

Question. Resistance n, each of r ohm, when connected in parallel give an equivalent resistance of R ohm. If these resistances were connected in series, the combination would have a resistance in ohms, equal to   
(a) n2R
(b) R/n2
(c) R/n
(d) nR 

Answer  A

Question. When three identical bulbs of 60 watt, 200 volt rating are connected in series to a 200 volt supply, the power drawn by them will be   
(a) 60 watt
(b) 180 watt
(c) 10 watt
(d) 20 watt 

Answer  D

Question. A wire of length L is drawn such that its diameter is reduced to half of its original diameter. If the initial resistance of the wire were 10 W, its new resistance would be :   
(a) 40 W
(b) 80 W
(c) 120 W
(d) 160 W

Answer  D

Question. Two 220 volt, 100 watt bulbs are connected first in series and then in parallel. Each time the combination is connected to a 220 volt a.c. supply line. The power drawn by the combination in each case respectively will be   
(a) 50 watt, 100 watt
(b) 100 watt, 50 watt
(c) 200 watt, 150 watt
(d) 50 watt, 200 watt 

Answer  D

Question. An electric kettle has two heating coils. When one of the coils is connected to an a.c. source, the water in the kettle boils in 10 minutes. When the other coil is used the water boils in 40 minutes. If both the coils are connected in parallel, the time taken by the same quantity of water to boil will be    
(a) 8 minutes
(b) 4 minutes
(c) 25 minutes
(d) 15 minutes

Answer  A

Question. Three equal resistors connected in series across a source of e.m.f. together dissipate 10 watt of power.   
What will be the power dissipated in watt if the same resistors are connected in parallel across the same source of e.m.f.?
(a) 30
(b) 10/3
(c) 10
(d) 90

Answer  D

Question. If two bulbs, whose resistances are in the ratio of 1 : 2 are connected in series, the power dissipated in them has the ratio of   
(a) 2 : 1
(b) 1 : 4
(c) 1 : 1
(d) 1 : 2.

Answer  D

Question. The internal resistance of a cell of e.m.f. 2 V is 0.1 W.    
It is connected to a resistance of 3.9 W. The voltage across the cell will be
(a) 1.95 V
(b) 1.9 V
(c) 0.5 V
(d) 2 V 

Answer  A

Question. A battery of e.m.f 10 V and internal resistance 0.5 W is connected across a variable resistance R. The value of R for which the power delivered in it is maximum is given by   
(a) 0.5 W
(b) 1.0 W
(c) 2.0 W
(d) 0.25 W

Answer  A

Question. Ten identical cells connected in series are needed to heat a wire of length one meter and radius ‘r’ by 10°C in time ‘t’. How many cells will be required to heat the wire of length two meter of the same radius by the same temperature in time ‘t’?    
(a) 20
(b) 30
(c) 40
(d) 10

Answer  A

Question. Two cells, having the same e.m.f. are connected in series through an external resistance R. Cells have internal resistances r1 and r2 (r1 > r2) respectively.
When the circuit is closed, the potential difference across the first cell is zero. The value of R is    
(a) r₁ + r₂
(b) r₁ – r₂
(c) r₁ + r₂ /2
(d) r₁ - r₂ /2

Answer  B

Question. Two identical batteries each of e.m.f 2 V and internal resistance 1 W are available to produce heat in an external resistance by passing a current through it.
The maximum power that can be developed across R using these batteries is  
(a) 3.2 W
(b) 2.0 W
(c) 1.28 W
(d) 8/9 W

Answer  B

Question. The magnitude and direction of the current in the circuit shown will be:

a. 7/3 A from a to b through e
b. 7/3 A from b and a through e
c. 1.0 A from b to a through e
d. 1.0 A from a to b through e
Answer : D

Question. Figure represents a part of the closed circuit. The potential difference between points A and B (VA – VB) is:

a. + 9 V
b. – 9 V
c. + 3 V
d. + 6 V
Answer : A

Question. In the following circuit E₁ = 4V, R₁ = 2Ω, E₂ = 6V, R₂ = 2Ω and R₃ = 4Ω. The current i1 is:

a. 1.6 A
b. 1.8 A
c. 2.25 A
d. 1 A
Answer : B

Question. Determine the current in the following circuit:

a. 1 A
b. 2.5 A
c. 0.4 A
d. 3 A
Answer : A

Question. A battery with negligible internal resistance is connected with 10m long wire. A standard cell gets balanced on 600 cm length of this wire. On increasing the length of potentiometer wire by 2m then the null point will be displaced by:
a. 200 cm
b. 120 cm
c. 720 cm
d. 600 cm
Answer : B

Question. In the following circuit a 10 m long potentiometer wire with resistance 1.2 ohm/m, a resistance R1 and an accumulator of emf 2 V are connected in series. When the emf of thermocouple is 2.4 mV then the deflection in galvanometer is zero. The current supplied by the accumulator will be:

a. 4 × 10^–4 A
b. 8 × 10^–4 A
c. 4 × 10^–3 A
d. 8 × 10^–3 A
Answer : A

Question. The figure shows a network of currents. The magnitude of current is shown here. The current i will be:

a. 3 A
b. 13 A
c. 23 A
d. – 3 A
Answer : C

Question. In an experiment to measure the internal resistance of a cell by potentiometer, it is found that the balance point is at a length of 2 m when the cell is shunted by a 5 Ω resistance; and is at a length of 3 m when the cell is shunted by a 10 Ω resistance. The internal resistance of the cell is, then:
a. 1.5 Ω
b. 10 Ω
c. 15 Ω
d. 1 Ω
Answer : B

Question. A deniel cell is balanced on 125 cm length of a potentiometer wire. When the cell is short circuited with a 2 # resistance the balancing length obtained is 100 cm. Internal resistance of the cell will be:
a. 1.5 Ω
b. 0.5 Ω
c. 1.25 Ω
d. 4/5 Ω
Answer : B

Question. The resistivity of a potentiometer wire is 40 × 10^–8 Ωm and its area of cross section is 8 × 10^–6 m². If 0.2 amp. Current is flowing through the wire, the potential gradient will be:
a. 10^–2 volt/m
b. 10^–1 volt/m
c. 3.2 × 10^–2 volt/m
d. 1 volt/m
Answer : A

Question. When a potential difference is applied across, the current passing through:
a. an insulator at 0 K is zero
b. a semiconductor at 0 K is zero
c. a metal at 0 K is finite
d. a p-n diode at 300 K is finite, if it is reverse biased
Answer : A, B, D

Question. A potentiometer wire of length 10 m and a resistance 30 Ω is connected in series with a battery of emf 2.5 V and internal resistance 5 # and an external resistance R. If the fall of potential along the potentiometer wire is 50 μV/mm, the value of R is: (in Ω)
a. 115
b. 80
c. 50
d. 100
Answer : A

Question. A 2 volt battery, a 15 Ω resistor and a potentiometer of 100 cm length, all are connected in series. If the resistance of potentiometer wire is 5 Ω, then the potential gradient of the potentiometer wire is:
a. 0.005 V/cm
b. 0.05 V/cm
c. 0.02 V/cm
d. 0.2 V/cm
Answer : A

Question. A micro ammeter has a resistance of 100 Ω and full scale range of 50μA. It can be used as a voltmeter or as a higher range ammeter provided a resistance is added to it. Pick the correct range and resistance combination(s):
a. 50 V range with 10 kΩ resistance in series
b. 10 V range with 200 kΩ resistance in series
c. 5 mA range with 1 Ω resistance in parallel
d. 10 mA range with 1 Ω resistance in parallel
Answer : B, C

Question. Heater of an electric kettle is made of a wire of length L and diameter d. It takes 4 minutes to raise the temperature of 0.5 kg water by 40 K. This heater is replaced by a new heater having two wires of the same material, each of length L and diameter 2d. The way these wires are connected is given in the options. How much time in minutes will it take to raise the temperature of the same amount of water by 40 K?
a. 4 if wires are in parallel
b. 2 if wires are in series
c. 1 if wires are in series
d. 0.5 if wires are in parallel
Answer : C, D

Question. Capacitor C₁ of capacitance 1μF and capacitor C₂ of capacitance 2μF are separately charged fully by a common battery. The two capacitors are then separately allowed to discharge through equal resistors at time t = 0:
a. The current in each of the two discharging circuits is zero at t = 0
b. The currents in the two discharging circuits at t = 0 are equal but not zero
c. The currents in the two discharging circuits at t = 0 are unequal
d. Capacitor C_1, loses 50% of its initial charge sooner than C₂ loses 50% of its initial charge
Answer : B, D

Question. For the circuit shown in the figure:

a. the current I through the battery is 7.5 mA
b. the potential difference across R_L is 18 V
c. ratio of powers dissipated in R1and R₂ is 3
d. if R₁ are interchanged, magnitude of the power dissipated in R_L will decrease by a factor of 9
Answer : A, D

Question. An electric current in a conductor is due to the motion of
(a) charges
(b) atoms
(c) molecules
(d) conductor itself

Answer : A

Question. A current i is flowing through the wire of diameter d having drift velocity of electrons vd in it. What will be new drift velocity when diameter of wire is made d/4?
(a) 4vd
(b) vd/4
(c) 16v_d
(d) v_d/16

Answer : C

Question. In a hydrogen atom, an electron moves in an orbit of radius 5.0×10^-11 m with a speed of 2.2 ×10⁶ ms^-1. The equivalent current is
(a) 11.2× 10^-3 A 
(b) 1.9 ×10^-3 A 
(c) 1.12×10^-3 A
(d) 11.2 ×10^-4 A 

Answer : C

Question. A wire of resistance 4Ω is stretched to twice its original length. The resistance of stretched wire would be
(a) 2 Ω
(b) 4 Ω
(c) 8 Ω
(d) 16 Ω

Answer : D

Question. A storage battery of emf 8 V and internal resistance 0.5 W is being charged by a 120 V DC supply using a series resistor of 15.5 Ω. What is the terminal voltage of the battery during charging?
(a) 10 V
(b) 11.5 V
(c) 7 V
(d) 16 V

Answer : B

Question. A copper wire of cross-sectional area 2.0 mm², resistivity = 1.7×10^-8 Ω-m, carries a current of 1 A. The electric field
in the copper wire is
(a) 8.5×10^-5 Vm^-1
(b) 8.5×10^-4 Vm^-1
(c) 8.5 ×10^-3 Vm^-1
(d) 8.5 ×10^-2 Vm^-1

Answer : C

Question. In cosmic rays, 0.15 protons cm^-2 s^-1 are entering the earth’s atmosphere. If the radius of the earth is 6400 km, the current received by the earth in the form of cosmic rays is nearly
(a) 0.12 A
(b) 1.2 A
(c) 12 A
(d) 120 A

Answer : A

Question. The current in a wire varies with time according to the equation i = 4 + 2t ,where i is in ampere andt is in second.
The quantity of charge which passes through a cross-section of the wire during the timet = 2 s to t = 6s is
(a) 40 C
(b) 48 C
(c) 38 C
(d) 43 C

Answer : B

Question. A potential difference V is applied to a copper wire of length l and diameter d. If V is doubled, then the drift velocity
(a) is doubled
(b) is halved
(c) remains same
(d) becomes zero

Answer : A

Question. If temperature of a metallic wire increases keeping constant electric field, then current density through wire will be
(a) increases
(b) decreases
(c) remains same
(d) None of these

Answer : B

Question. A wire is stretched so as to change its diameter by 0.25%. The percentage change in resistance is
(a) 4.0%
(b) 2.0%
(c) 1.0%
(d) 0.5%

Answer : C

Question. If resistivity of copper is 1.72×10^-8 Ω-m and number of free electrons in copper is 8.5 10ײ⁸/ m³ .
Find the mobility.
(a) 4.27×10^-3 m² /C-Ω
(b) 6.8× 10^-3 m² /C-Ω
(c) 8.5×10^-3 m² /C-Ω
(d) 34×10^-3 m² /C-Ω

Answer : A

Question. In which conductors, both positive and negative charges can move?
(a) Non-electrolytic solution
(b) Electrolytic solution
(c) Both (a) and (b)
(d) Neither (a) nor (b)

Answer : B

Question. A 200 W resistor has a certain colour code. If one replaces the red colour by green in the code, the new resistance will
be
(a) 100 Ω
(b) 400 Ω
(c) 300 Ω
(d) 500 Ω

Answer : D

Question. A carbon resistor of (47 ± 4.7) kW is to be marked with rings of different colours for its identification. The colour code sequence will be
(a) Yellow - Green - Violet - Gold
(b) Yellow - Violet - Orange - Silver
(c) Violet - Yellow - Orange - Silver
(d) Green - Orange - Violet - Gold

Answer : B

Question. A current of 2 mA was passed through an unknown resistor which dissipated a power of 4.4 W. Dissipated power when an ideal power supply of 11 V is connected across it is
(a) 11×10^-4Ω
(b) 11×10^-5Ω
(c) 11×10⁵Ω
(d) 11×10^-3Ω

Answer : B

Question. When two or more resistors are connected in series, then
(a) same current flows through each resistor
(b) same voltage is applied across each resistor
(c) different current flows through each resistor
(d) both current and voltage are different in each resistor

Answer : A

Question. A circuit contain two resistances R₁and R₂ are in series.
Find the ratio of input voltage to voltage of R₂.
(a) R₂/R₁+ R₂
(b) R₁+ R₂/R₂
(c) R₁+ R₂/R1
(d) R₁/R₁+ R₂

Answer : B

Question. When two electrodes (positive and negative) of a cell are immersed in an electrolytic solution, the charges are exchanged between
(a) positive electrode and electrolyte only
(b) negative electrode and electrolyte only
(c) Both electrodes and electrolyte
(d) directly between two electrodes 

Answer : C

Question. Kirchhoff’s junction rule is a reflection of
(a) conservation of current density vector
(b) conservation of charge
(c) the fact that the momentum with which a charged particle approaches a junction is unchanged (as a vector) as the charged particle leaves the junction
(d) the fact that there is a net accumulation of charges at a junction

Answer : B

Question. A heating element using nichrome connected to a 230 V supply draws an initial current of 3.2 A which settles after
a few seconds to a steady value of 2.8 A.What is the steady temperature of the heating element, if the room temperature is 27.0°C? Temperature coefficient of resistance of nichrome averaged over the temperature range involved is 1.70×10^-4 °C^-1.
(a) 676°C
(b) 876°C
(c) 867°C
(d) 500°C

Answer : C

Question. A potentiometer is an accurate and versatile device to make electrical measurement of emf because the method involves
(a) cells
(b) potential gradients
(c) a condition of no current flow through the galvanometer
(d) a combination of cells, galvanometer and resistances

Answer : C

Question. A cell can be balanced against 110 cm and 100 cm of potentiometer wire respectively, with and without being short circuited through a resistance of 10 Ω. Its internal resistance is
(a) 1.0 Ω
(b) 0.5 Ω
(c) 2.0 Ω
(d) zero

Answer : A

Question. A silver wire has a resistance of 2.1 Ω at 27.5 °C and a resistance of 2.7 Ω at 100 °C. Determine the temperature
coefficient of resistance of silver.
(a) 0.0039/°C
(b) 0.02/°C
(c) 0.05/°C
(d) 0.06/°C

Answer : A

Question. Two resistances 400 Ω and 800 Ω are connected in series with 6 V battery. The potential difference measured by
voltmeter of 10 k Ω across 400 Ω resistor is
(a) 2 V
(b) 1.95 V
(c) 3.8 V
(d) 4 V

Answer : B

Question. In a meter bridge, the balance point is found to be at 39.5 cm from the end A, when the resistor Y is of 125 Ω. Determine the balance point of the bridge above, if X and Y are interchanged.
(a) 39.5 cm
(b) 12.5 cm
(c) 60.5 cm
(d) None of the above

Answer : C

Question. In a potentiometer arrangement, a cell of emf 1.25 V gives a balance point at 35.0 cm length of the wire. If the cell is
replaced by another cell and the balance point shifts to 63.0 cm, what is the emf of the second cell?
(a) 2.25 V
(b) 3.25 V
(c) 4.5 V
(d) 6 V

Answer : A

Question. If charges move without collisions through the conductor, their kinetic energy would also change, so that the total
energy is
(a) changed
(b) unchanged
(c) doubled
(d) halved 

Answer : B 

Question. The number density of free electrons in a copper conductor estimated is 8.5×10²⁸ m^-3. How long does an electron take to drift from one end of a wire 3.0 m long to its other end? The area of cross-section of the wire is 2.0×10^-6 m_² and it is carrying a current of 3.0 A.
(a) 6 h 33 min
(b) 7 h 33 min
(c) 6 h
(d) 5 h 20 min

Answer : B

Students must free download and practice these worksheets to gain more marks in exams. CBSE Class 12 Physics Worksheet - Current Electricity