JEE Physics Current Electricity MCQs Set 20

Answer: (a) not defined

Explanation:
1. An ideal cell fixes the voltage between A and B at 10 V and the other cell would fix it at 15 V; a loop of ideal source and resistance 1 Ω would carry 5 A, but the formula for equivalent emf cannot be applied to a zero-resistance branch.
2. With an ideal cell of different emf the arrangement is not consistent as a simple equivalent cell, so no equivalent emf is defined.

Teacher's Note:
1. Parallel combination of cells with an ideal one imposes its own voltage; here it conflicts.
2. Equivalent emf formulas need non-zero internal resistance in all branches.

Answer: (b) 70 mΩ

Explanation:
1. Open circuit voltage is the emf: \( E=1.52\ \text{V} \).
2. Terminal voltage when 1.0 A flows is 1.45 V, so \( r=\frac{1.52-1.45}{1.0}=0.07\ \Omega=70\ \text{m}\Omega \).

Teacher's Note:
1. The lost volts equal I times r.
2. A high-resistance voltmeter draws almost no current.

Answer: (b) 0.01 Ω

Explanation:
1. A shunt is required: \( S=\frac{I_gG}{I-I_g}=\frac{10^{-3}\times100}{10-10^{-3}}\approx0.01\ \Omega \).

Teacher's Note:
1. An ammeter needs a small resistance in parallel.
2. Neglect Ig next to I in the denominator.

Answer: (a) 1 Ω and 4 W

Explanation:
1. Each side has \( 10\times0.1=1\ \Omega \).
2. The bridge is balanced (B and D are at the same potential by symmetry), so the diagonal BD carries no current. The two paths ABC and ADC are each 2 Ω in parallel: 1 Ω.
3. \( P=\frac{V^2}{R}=\frac{4}{1}=4\ \text{W} \).

Teacher's Note:
1. Symmetry about the diagonal AC puts B and D at equal potentials.
2. A resistor between equal potentials carries no current.

Answer: (a) 0.1 V/m

Explanation:
1. \( k=\frac{I\rho}{A}=\frac{0.2\times4\times10^{-7}}{8\times10^{-7}}=0.1\ \text{V/m} \).

Teacher's Note:
1. Potential drop per metre equals current times resistance per metre.
2. Resistance per metre is rho/A.

Answer: (c) II is correct but I is wrong

Explanation:
1. For cells in parallel \( E_{eq}=\frac{E_1r_2+E_2r_1}{r_1+r_2} \) is a weighted average, so it lies between the two emfs, and statement I is wrong.
2. The internal resistances combine as parallel resistors, \( r_{eq}=\frac{r_1r_2}{r_1+r_2} \), which is smaller than either: statement II is correct.

Teacher's Note:
1. The equivalent emf of cells in parallel is a weighted mean, not smaller than both.
2. Internal resistances in parallel behave like ordinary parallel resistors.

Answer: (a) 3 Ω

Explanation:
1. Power in Y is maximum when Y equals the resistance of the rest of the circuit seen by it: \( 2+\frac{6X}{6+X}=4 \).
2. So \( \frac{6X}{6+X}=2 \), giving \( 6X=12+2X \) and \( X=3\ \Omega \).

Teacher's Note:
1. Treat the battery and the X, 6 ohm pair as one source with internal resistance.
2. Apply maximum power transfer.

Answer: (c) +0.4%

Explanation:
1. Volume is constant: \( A\,l=\text{const} \), so \( l\propto\frac{1}{r^2} \).
2. \( R=\frac{\rho l}{A}\propto\frac{1}{r^4} \), so a 0.1% decrease in r gives a 0.4% increase in R.

Teacher's Note:
1. Stretching at constant volume changes both length and area.
2. R is proportional to the inverse fourth power of the radius.

Answer: (b) 3.75 A

Explanation:
1. A₁ carries the sum of the branch currents: \( 3+0.75=3.75\ \text{A} \).

Teacher's Note:
1. The series part of the circuit carries the total current.
2. The 20 ohm resistor does not change the branch currents.

Answer: (c) 5.56 Ω

Explanation:
1. Full-scale current: \( I_g=\frac{k\theta}{NBA}=\frac{10^{-4}\times0.2}{50\times0.02\times2\times10^{-4}}=0.1\ \text{A} \).
2. Shunt: \( S=\frac{I_gG}{I-I_g}=\frac{0.1\times50}{0.9}=5.56\ \Omega \).

Teacher's Note:
1. Balance magnetic torque NBAI with the spring torque k theta.
2. Then use the shunt formula.

Answer: (b) the 2.5 W bulb will fuse

Explanation:
1. Resistances: \( \frac{110^2}{2.5}=4840\ \Omega \) and \( \frac{110^2}{100}=121\ \Omega \).
2. Series current \( =\frac{220}{4961}=0.044\ \text{A} \).
3. Rated currents are \( \frac{2.5}{110}=0.023\ \text{A} \) and \( \frac{100}{110}=0.91\ \text{A} \). Only the 2.5 W bulb is overloaded.

Teacher's Note:
1. In series the same current passes through both, so compare it with each rated current.
2. The low-power bulb has the higher resistance and takes most of the voltage.

Answer: (c) A will increase and V will decrease

Explanation:
1. The extra resistor in parallel with the voltmeter lowers the resistance of that part of the circuit, so the total resistance falls and the ammeter reading A increases.
2. The voltmeter reads the voltage across a smaller parallel resistance; because that resistance falls by a larger fraction than the current rises, V decreases.

Teacher's Note:
1. Parallel resistance is smaller than the smallest branch.
2. The voltmeter is nearly an open circuit, so it takes almost all of the cell voltage; shunting it reduces the reading.

Answer: (d) increase by 0.2%

Explanation:
1. At constant volume \( R\propto l^2 \), so the fractional change is \( 2\times0.1\%=0.2\% \) (an increase).

Teacher's Note:
1. Stretching reduces the area as the length grows.
2. R varies as the square of the length at constant volume.

Answer: (c) may go from the positive terminal to the negative terminal

Explanation:
1. While the battery supplies current, positive charge inside it moves from the negative to the positive terminal.
2. While a battery is being charged, the current inside it flows from the positive to the negative terminal.
3. So both directions are possible depending on the circuit.

Teacher's Note:
1. Check the direction of the external source in each case.
2. The words 'always' make the other options wrong.

Answer: (b) 8 V

Explanation:
1. \( V_0=\frac{10/10+6/20+5/30}{1/10+1/20+1/30}=\frac{1.4667}{0.18333}=8\ \text{V} \).

Teacher's Note:
1. This is the weighted average with weights 1/R.
2. The result lies between the highest and the lowest end potentials.

Answer: (c) \( \frac{\alpha_1+\alpha_2}{2},\ \frac{\alpha_1+\alpha_2}{2} \)

Explanation:
1. Series: \( R_s=R(2+(\alpha_1+\alpha_2)t)=2R\left(1+\frac{\alpha_1+\alpha_2}{2}t\right) \).
2. Parallel (for small t): \( R_p=\frac{R_1R_2}{R_1+R_2}\approx\frac R2\left(1+\frac{\alpha_1+\alpha_2}{2}t\right) \).
3. Both effective coefficients equal \( \frac{\alpha_1+\alpha_2}{2} \).

Teacher's Note:
1. Expand to first order in temperature.
2. For equal resistances the effective coefficient is the mean.

Answer: (c) 3.2 V and 400 Ω

Explanation:
1. Voltage across the 100 Ω and voltmeter pair: \( 3.4-0.04\times(3+2)=3.2\ \text{V} \).
2. Current in 100 Ω: 0.032 A, so the voltmeter takes \( 0.04-0.032=0.008\ \text{A} \).
3. \( R_V=\frac{3.2}{0.008}=400\ \Omega \).

Teacher's Note:
1. The ammeter reads the total current.
2. The voltmeter current is the difference.

Answer: (b) 4 Ω

Explanation:
1. Lengths of wire: AD = 30 cm and DB = 20 cm. Balance: \( \frac{6}{R}=\frac{30}{20} \).
2. \( R=4\ \Omega \).

Teacher's Note:
1. The wire resistance is proportional to length.
2. Match the 6 ohm resistor with the length from A as written.

Answer: (a) \( \frac1{500} \) K⁻¹

Explanation:
1. Increase of resistance per kelvin \( =600\times0.001+300\times0.004=1.8\ \Omega/\text{K} \).
2. \( \alpha=\frac{1.8}{900}=0.002=\frac1{500}\ \text{K}^{-1} \).

Teacher's Note:
1. Effective coefficient is the weighted mean of the coefficients, weighted by the resistances.
2. Divide the total change by the total resistance.

Answer: (c) \( \frac{50}{73} \) A

Explanation:
1. Ammeter resistance \( \frac{480\times20}{500}=19.2\ \Omega \).
2. Total resistance \( 10+19.2=29.2\ \Omega \), so \( I=\frac{20}{29.2}=\frac{50}{73}\ \text{A} \).

Teacher's Note:
1. The ammeter reads the total current entering the coil and shunt together.
2. Add the parallel pair in series with the resistor.

Answer: (c) 6

Explanation:
1. \( I_1=\frac{9-6}{10}=0.3\ \text{A} \) (discharged, r = 10 Ω).
2. \( I_2=\frac{9-6}{1}=3\ \text{A} \) (charged, r = 1 Ω).
3. \( 10I_1+I_2=3+3=6\ \text{A} \).

Teacher's Note:
1. While charging, the net voltage driving the current is the charger voltage minus the battery emf.
2. The internal resistance changes with the state of charge.

Answer: (a) copper decreases and germanium increases

Explanation:
1. For a metal, resistance falls on cooling because lattice vibrations decrease.
2. For a semiconductor, cooling reduces the number of charge carriers sharply, so resistance rises.

Teacher's Note:
1. Metals have a positive temperature coefficient, semiconductors a negative one.
2. Carrier density dominates in semiconductors.

Answer: (d) 12 A

Explanation:
1. Total power \( =15\times40+5\times100+5\times80+1000=2500\ \text{W} \).
2. Current \( =\frac{2500}{220}=11.4\ \text{A} \), so a fuse of at least 12 A is needed.

Teacher's Note:
1. Appliances are in parallel, so the powers add.
2. Choose the next higher standard rating.

Answer: (a) 0.015 Ω

Explanation:
1. Full-scale current \( I_g=30\times20\ \mu\text{A}=600\ \mu\text{A} \).
2. \( S=\frac{I_gG}{I-I_g}=\frac{6\times10^{-4}\times25}{1-6\times10^{-4}}=0.015\ \Omega \).

Teacher's Note:
1. Full scale is the number of divisions times the sensitivity.
2. A shunt is small compared with the galvanometer.

Answer: (d) increases continuously

Explanation:
1. \( P=\frac{E^2R}{(R+r)^2} \) is maximum at R = r = 5 Ω and rises steadily as R approaches r from below.
2. So over 1 Ω to 5 Ω the power increases continuously, reaching its maximum at 5 Ω.

Teacher's Note:
1. Maximum power transfer occurs when the external resistance equals the internal resistance.
2. Compare the range with the peak position.

Answer: (d) \( j_A>j_B \) and \( E_A>E_B \)

Explanation:
1. The same current crosses every section, so \( j=\frac IA \) is larger where the area is smaller: \( j_A>j_B \).
2. In an ohmic material \( E=\rho j \), so \( E_A>E_B \).

Teacher's Note:
1. Current is the same everywhere in a series conductor.
2. Field is proportional to current density for fixed resistivity.