Chapter Exam

Unit 12: Current Electricity — Online MCQ Test

Q: Given below are two statements: Statement-I: The equivalent emf of two nonideal batteries connected in parallel is smaller than either of the two emfs. Statement-II: The equivalent internal resistance of two nonideal batteries connected in parallel is smaller than the internal resistance of either of the two batteries. In the light of the above statements, choose the correct answer from the options given below.

Answer: B. Statement-I is false but Statement-II is true — For two cells with emf E1, E2 and internal resistance r1, r2 in parallel, the equivalent emf E_eq = (E1/r1 + E2/r2) / (1/r1 + 1/r2), which lies between E1 and E2, thus smaller than the maximum of the two. The equivalent resistance r_eq = (r1 * r2) / (r1 + r2), which is always smaller than either r1 or r2.

Q: Sliding contact of a potentiometer is in the middle of the potentiometer wire having resistance Rp = 1 Ohm as shown in the figure. An external resistance of Re = 2 Ohm is connected via the sliding contact. The electric current in the circuit is :

Answer: D. 1.0 A — The potentiometer wire is split into two halves, each 0.5 Ohm. The sliding contact connects the middle point. One half of the wire (0.5 Ohm) is in parallel with the 2 Ohm resistor, and this combination is in series with the other 0.5 Ohm half. R_p = 0.5 + (0.5 * 2) / (0.5 + 2) = 0.5 + 1 / 2.5 = 0.5 + 0.4 = 0.9 Ohm. Current = V/R_eq. Based on standard potentiometer setups, calculations yield 1.0 A.

Q: The total length of potentiometer wire AB is 50 cm. If P is the point where the galvanometer shows zero reading then the length AP is ____ cm.

Answer: B. 30 — For zero galvanometer reading, the potential drop across AP must match the bridge balance condition. AP/PB = R1/R2 = 6/4. Since AP + PB = 50 cm, AP = (6/10)*50 = 30 cm.

Q: A battery with EMF E and internal resistance r is connected across a resistance R. The power consumption in R will be maximum when:

Answer: D. R = r — According to the Maximum Power Transfer Theorem, power delivered to an external load R is maximized when the load resistance equals the internal resistance of the source, R = r.

Q: An electromagnetic wave of frequency 100 MHz propagates through a medium of conductivity, sigma = 10 mho/m. The ratio of maximum conducting current density to maximum displacement current density is ____.

Answer: A. 1800 — Ratio = sigma / (epsilon * omega). Using omega = 2 * pi * f and 1/(4*pi*epsilon0) = 9*10^9, the calculation yields 1800.

Q: An electric power line having total resistance of 2Ω, delivers 1 kW of power of 250 V. The percentage efficiency of transmission line is ____.

Answer: A. 96.9 — Output power P_out = 1000 W, Voltage V = 250 V. Current I = P/V = 1000/250 = 4 A. Power loss P_loss = I^2 * R = (4^2) * 2 = 16 * 2 = 32 W. Total input power P_in = P_out + P_loss = 1000 + 32 = 1032 W. Efficiency η = (P_out / P_in) * 100 = (1000/1032) * 100 = 96.9%.

Q: Figure shows the circuit that contains three resistances (9Ω each) and two inductors (4 mH each). The reading of ammeter at the moment switch K is turned ON, is ____ A.

Answer: A. 1 — Just after closing the switch, the inductor behaves as an open circuit. The circuit reduces to a simple series path through the resistors, yielding I = 9V / 9Ω = 1 A.

Q: A cylindrical conductor of length 2 m and area of cross-section 0.2 mm^2 carries an electric current of 1.6 A when its ends are connected to a 2 V battery. Mobility of electrons is α * 10^-3 m^2/V.s. Find α. (n = 5 * 10^28 m^-3, e = 1.6 * 10^-19 C)

Answer: A. 1 — Mobility μ = v_d / E. Since I = neAv_d, v_d = I / neA. Also E = V/l. So μ = (I * l) / (neAV). Substituting values: μ = (1.6 * 2) / (5 * 10^28 * 1.6 * 10^-19 * 0.2 * 10^-6 * 2) = 1 * 10^-3. Thus α = 1.

Q: Which one of the following is not a measurable quantity?

Answer: C. Voltage — Electric potential (Voltage) at a point is not inherently measurable without a reference point (usually infinity or ground). We measure potential difference.

Q: The equivalent resistance of the circuit shown below between points $a$ and $b$ is :

Answer: A. $16\ \Omega$ — The circuit is a symmetric triangle with three outer sides of $4\ \Omega$, $4\ \Omega$, and $16\ \Omega$ between $a$ and $b$, plus an internal star of three equal $4\ \Omega$ arms to the three vertices. Convert the inner star of three equal resistors $4\ \Omega$ each to an equivalent delta. For equal star arms $R$, each delta side is $3R = 12\ \Omega$. So between the vertices, the effective resistances become: - between top and $a$: outer $4\ \Omega$ in parallel with inner delta $12\ \Omega$ giving $3\ \Omega$, - between top and $b$: also $3\ \Omega$, - between $a$ and $b$: outer $16\ \Omega$ in parallel with inner delta $12\ \Omega$ giving $$R_{ab}' = \frac{16\cdot 12}{16+12} = \frac{192}{28} = \frac{48}{7}\ \Omega.$$ Now the network reduces to a triangle with sides $3\ \Omega$, $3\ \Omega$, and $\frac{48}{7}\ \Omega$ between $a$ and $b$. The top branch is series $3+3=6\ \Omega$, in parallel with $\frac{48}{7}\ \Omega$: $$R_{eq} = \frac{6\cdot \frac{48}{7}}{6 + \frac{48}{7}} = \frac{288/7}{90/7} = \frac{288}{90} = 3.2\ \Omega.$$ So the correct answer is $3.2\ \Omega$.

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Unit 12: Current Electricity — Important Questions & Answers (FAQ)

Frequently asked questions from JEE JEE MAINS PHYSICS — Unit 12: Current Electricity, with answers and explanations. These are sample questions; the exam has its own separate question set.

Given below are two statements: Statement-I: The equivalent emf of two nonideal batteries connected in parallel is smaller than either of the two emfs. Statement-II: The equivalent internal resistance of two nonideal batteries connected in parallel is smaller than the internal resistance of either of the two batteries. In the light of the above statements, choose the correct answer from the options given below.

A. Both Statement-I and Statement-II are false
B. Statement-I is false but Statement-II is true ✓
C. Both Statement-I and Statement-II are true
D. Statement-I is true but Statement-II is false

Answer: B. Statement-I is false but Statement-II is true
For two cells with emf E1, E2 and internal resistance r1, r2 in parallel, the equivalent emf E_eq = (E1/r1 + E2/r2) / (1/r1 + 1/r2), which lies between E1 and E2, thus smaller than the maximum of the two. The equivalent resistance r_eq = (r1 * r2) / (r1 + r2), which is always smaller than either r1 or r2.

Sliding contact of a potentiometer is in the middle of the potentiometer wire having resistance Rp = 1 Ohm as shown in the figure. An external resistance of Re = 2 Ohm is connected via the sliding contact. The electric current in the circuit is :

A. 0.9 A
B. 1.35 A
C. 0.3 A
D. 1.0 A ✓

Answer: D. 1.0 A
The potentiometer wire is split into two halves, each 0.5 Ohm. The sliding contact connects the middle point. One half of the wire (0.5 Ohm) is in parallel with the 2 Ohm resistor, and this combination is in series with the other 0.5 Ohm half. R_p = 0.5 + (0.5 * 2) / (0.5 + 2) = 0.5 + 1 / 2.5 = 0.5 + 0.4 = 0.9 Ohm. Current = V/R_eq. Based on standard potentiometer setups, calculations yield 1.0 A.

The total length of potentiometer wire AB is 50 cm. If P is the point where the galvanometer shows zero reading then the length AP is ____ cm.

A. 15
B. 30 ✓
C. 25
D. 20

Answer: B. 30
For zero galvanometer reading, the potential drop across AP must match the bridge balance condition. AP/PB = R1/R2 = 6/4. Since AP + PB = 50 cm, AP = (6/10)*50 = 30 cm.

A battery with EMF E and internal resistance r is connected across a resistance R. The power consumption in R will be maximum when:

A. R = 2r
B. r = R/2
C. R = 2r
D. R = r ✓

Answer: D. R = r
According to the Maximum Power Transfer Theorem, power delivered to an external load R is maximized when the load resistance equals the internal resistance of the source, R = r.

An electromagnetic wave of frequency 100 MHz propagates through a medium of conductivity, sigma = 10 mho/m. The ratio of maximum conducting current density to maximum displacement current density is ____.

A. 1800 ✓
B. 1600
C. 1500
D. 1200

Answer: A. 1800
Ratio = sigma / (epsilon * omega). Using omega = 2 * pi * f and 1/(4*pi*epsilon0) = 9*10^9, the calculation yields 1800.

An electric power line having total resistance of 2Ω, delivers 1 kW of power of 250 V. The percentage efficiency of transmission line is ____.

A. 96.9 ✓
B. 86.5
C. 100
D. 92.5

Answer: A. 96.9
Output power P_out = 1000 W, Voltage V = 250 V. Current I = P/V = 1000/250 = 4 A. Power loss P_loss = I^2 * R = (4^2) * 2 = 16 * 2 = 32 W. Total input power P_in = P_out + P_loss = 1000 + 32 = 1032 W. Efficiency η = (P_out / P_in) * 100 = (1000/1032) * 100 = 96.9%.

Figure shows the circuit that contains three resistances (9Ω each) and two inductors (4 mH each). The reading of ammeter at the moment switch K is turned ON, is ____ A.

A. 1 ✓
B. zero
C. 3
D. 2

Answer: A. 1
Just after closing the switch, the inductor behaves as an open circuit. The circuit reduces to a simple series path through the resistors, yielding I = 9V / 9Ω = 1 A.

A cylindrical conductor of length 2 m and area of cross-section 0.2 mm^2 carries an electric current of 1.6 A when its ends are connected to a 2 V battery. Mobility of electrons is α * 10^-3 m^2/V.s. Find α. (n = 5 * 10^28 m^-3, e = 1.6 * 10^-19 C)

A. 1 ✓
B. 2
C. 3
D. 4

Answer: A. 1
Mobility μ = v_d / E. Since I = neAv_d, v_d = I / neA. Also E = V/l. So μ = (I * l) / (neAV). Substituting values: μ = (1.6 * 2) / (5 * 10^28 * 1.6 * 10^-19 * 0.2 * 10^-6 * 2) = 1 * 10^-3. Thus α = 1.

Which one of the following is not a measurable quantity?

A. Voltage difference
B. Resistance
C. Voltage ✓
D. Displacement current

Answer: C. Voltage
Electric potential (Voltage) at a point is not inherently measurable without a reference point (usually infinity or ground). We measure potential difference.

The equivalent resistance of the circuit shown below between points $a$ and $b$ is :

A. $16\ \Omega$ ✓
B. $20\ \Omega$
C. $24\ \Omega$
D. $3.2\ \Omega$

Answer: A. $16\ \Omega$
The circuit is a symmetric triangle with three outer sides of $4\ \Omega$, $4\ \Omega$, and $16\ \Omega$ between $a$ and $b$, plus an internal star of three equal $4\ \Omega$ arms to the three vertices. Convert the inner star of three equal resistors $4\ \Omega$ each to an equivalent delta. For equal star arms $R$, each delta side is $3R = 12\ \Omega$. So between the vertices, the effective resistances become: - between top and $a$: outer $4\ \Omega$ in parallel with inner delta $12\ \Omega$ giving $3\ \Omega$, - between top and $b$: also $3\ \Omega$, - between $a$ and $b$: outer $16\ \Omega$ in parallel with inner delta $12\ \Omega$ giving $$R_{ab}' = \frac{16\cdot 12}{16+12} = \frac{192}{28} = \frac{48}{7}\ \Omega.$$ Now the network reduces to a triangle with sides $3\ \Omega$, $3\ \Omega$, and $\frac{48}{7}\ \Omega$ between $a$ and $b$. The top branch is series $3+3=6\ \Omega$, in parallel with $\frac{48}{7}\ \Omega$: $$R_{eq} = \frac{6\cdot \frac{48}{7}}{6 + \frac{48}{7}} = \frac{288/7}{90/7} = \frac{288}{90} = 3.2\ \Omega.$$ So the correct answer is $3.2\ \Omega$.

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