JEE Physics Current Electricity MCQs Set 16

Answer: (b) \( I_3 > I_1 > I_2 \)

Explanation:
1. Only S1: total resistance \( R+6R = 7R \), so \( I_1 = E/7R \).
2. Only S2: \( R+12R = 13R \), so \( I_2 = E/13R \).
3. Both: \( 6R\parallel12R = 4R \), total \( 5R \), so \( I_3 = E/5R \).
4. So \( I_3 > I_1 > I_2 \).

Teacher's Note:
1. Closing both switches gives a parallel combination whose resistance is smaller than either arm, so the current is highest.

Answer: (d) 20 Ω

Explanation:
1. \( P = \dfrac{V^2}{R_{eq}} \Rightarrow R_{eq} = \dfrac{100}{25} = 4\ \Omega \).
2. \( \dfrac{1}{4} = \dfrac{1}{5}+\dfrac{1}{R} \Rightarrow \dfrac1R = \dfrac1{20} \), so \( R = 20\ \Omega \).

Teacher's Note:
1. The 5 Ω resistor alone dissipates 20 W, so R must dissipate the remaining 5 W: R = 100/5 = 20 Ω.

Answer: (a) \( 3.6\times10^5 \) J

Explanation:
1. Energy = power × time \( = 100\times3600 \) J.
2. \( = 3.6\times10^5 \) J.

Teacher's Note:
1. Convert the time to seconds when the power is in watts. 100 W for one hour is 0.1 kWh.

Answer: (d) 16 kW

Explanation:
1. Total fall of potential \( = 8\times100 = 800 \) V and total resistance \( = 0.4\times100 = 40\ \Omega \).
2. \( I = 800/40 = 20 \) A.
3. Power loss \( = VI = 800\times20 = 16000 \) W = 16 kW.

Teacher's Note:
1. Find the total voltage drop and total resistance over the whole distance before applying P = VI.

Answer: (a) \( P = \left(\dfrac{V}{V_0}\right)^2P_0 \)

Explanation:
1. The resistance of the filament is constant: \( R = V_0^2/P_0 \).
2. \( P = V^2/R = \left(\dfrac{V}{V_0}\right)^2P_0 \).

Teacher's Note:
1. At fixed resistance the power goes as the square of the voltage. This ignores the change of resistance with temperature.

Answer: (d) directly proportional to the squares of their marked voltages

Explanation:
1. For the same rated power, \( R = V^2/P \), so \( R \propto V_{marked}^2 \).
2. In series the current is the same, and the power dissipated is \( I^2R \propto R \propto V_{marked}^2 \).

Teacher's Note:
1. In series the bulb with the higher resistance glows brighter. In parallel it would be the one with the lower resistance.

Answer: (d) \( R_2 = R_3 = 4R_1 \)

Explanation:
1. R2 and R3 have the same voltage and the same power, so \( R_2 = R_3 \) and each carries current i. R1 carries 2i.
2. Equal power: \( (2i)^2R_1 = i^2R_2 \Rightarrow R_2 = 4R_1 \).

Teacher's Note:
1. Use the current division: the series resistor carries the total current, and this is twice that of each parallel branch.

Answer: (c) 72 W

Explanation:
1. Power in the 2 Ω resistor is \( I^2\times2 \), which is maximum when the current is maximum, that is, when the total resistance is minimum.
2. The parallel part is least when R = 0 (the 8 Ω is shorted). Then \( I = 12/2 = 6 \) A.
3. \( P = 6^2\times2 = 72 \) W.

Teacher's Note:
1. Maximum power transfer to a fixed resistor means maximum current, not matching of resistances.

Answer: (a) \( P_1 < P_2 < P_3 \)

Explanation:
1. \( R_1 = \dfrac{220^2}{100} = 484\ \Omega \) and \( R_2 = R_3 = \dfrac{220^2}{60} \approx 807\ \Omega \).
2. B3 has the full 220 V, so \( P_3 = 60 \) W.
3. B1 and B2 share the same current in series. Since \( R_2 > R_1 \), \( P_2 > P_1 \). Each gets less than its rating, so \( P_2 < 60 \) W.
4. Hence \( P_1 < P_2 < P_3 \).

Teacher's Note:
1. A higher-wattage bulb has a lower resistance, so in series it gets the smaller share of the power.

Answer: (d) 87.5%

Explanation:
1. Energy supplied during charging \( = 15\times10\times8 = 1200 \) Wh.
2. Energy delivered on discharge \( = 14\times5\times15 = 1050 \) Wh.
3. Efficiency \( = \dfrac{1050}{1200} = 87.5\% \).

Teacher's Note:
1. Watt-hour efficiency compares energy out with energy in. Ampere-hour efficiency would compare only the charge (75 Ah to 75 Ah).

Answer: (a) 26 Ω

Explanation:
1. The bulb current is \( I = P/V = 500/100 = 5 \) A.
2. The resistor must drop \( 230-100 = 130 \) V.
3. \( R = 130/5 = 26\ \Omega \).

Teacher's Note:
1. The bulb must see its rated voltage and current, so the series resistor takes the excess voltage at the same current.

Answer: (b) 3 Ω

Explanation:
1. The meter takes 10 A, so the shunt takes \( 100-10 = 90 \) A, at the same voltage.
2. \( 10\times27 = 90\times S \Rightarrow S = 3\ \Omega \).

Teacher's Note:
1. The shunt and the meter are in parallel, so the voltage across them is the same.

Answer: (d) 2950 Ω

Explanation:
1. Initially \( I = \dfrac{3}{1950+50} = 1.5 \) mA for 30 divisions.
2. For 20 divisions the current is \( \dfrac{20}{30}\times1.5 = 1 \) mA, so the total resistance is \( 3/0.001 = 3000\ \Omega \).
3. Series resistance \( = 3000 - 50 = 2950\ \Omega \).

Teacher's Note:
1. The deflection is proportional to the current. Find the new current first and then the total resistance.

Answer: (b) putting in parallel a resistance of 12 Ω

Explanation:
1. An ammeter needs a shunt in parallel. The shunt takes \( 6-1 = 5 \) A.
2. \( 1\times60 = 5\times S \Rightarrow S = 12\ \Omega \).

Teacher's Note:
1. A shunt (parallel, small resistance) makes an ammeter. A high series resistance makes a voltmeter.

Answer: (b) 1950 Ω

Explanation:
1. Total resistance needed \( = \dfrac{30}{15\times10^{-3}} = 2000\ \Omega \).
2. Series resistance \( = 2000-50 = 1950\ \Omega \).

Teacher's Note:
1. The series resistance is the total required resistance minus the galvanometer coil resistance.

Answer: (a) \( \dfrac{G^2}{S+G} \)

Explanation:
1. The shunted galvanometer has resistance \( \dfrac{GS}{G+S} \).
2. To restore the total to G, the extra series resistance is \( x = G - \dfrac{GS}{G+S} = \dfrac{G^2}{G+S} \).

Teacher's Note:
1. Shunting lowers the resistance of the instrument, so a series resistance is needed to make up the difference.

Answer: (d) 10%

Explanation:
1. At fixed resistance, \( P \propto V^2 \).
2. \( P'/P = (0.95)^2 = 0.9025 \), a decrease of about 9.75%, i.e. about 10%.

Teacher's Note:
1. For small changes, the percentage change in P is about twice that in V, since P depends on V².

Answer: (d) 0.002

Explanation:
1. At full range, the potential difference across the shunt is 50 mV and it carries nearly the whole 25 A.
2. \( S = \dfrac{50\times10^{-3}}{25} = 0.002\ \Omega \).

Teacher's Note:
1. The shunt voltage is the voltage range of the millivoltmeter, and the shunt carries almost all the current.

Answer: (d) 10 Ω

Explanation:
1. Initially \( \dfrac{5}{R} = \dfrac{l_1}{100-l_1} \).
2. After shunting, the right gap has \( R/2 \): \( \dfrac{5}{R/2} = \dfrac{1.5l_1}{100-1.5l_1} \).
3. Since \( \dfrac{10}{R} = 2\times\dfrac{5}{R} \), we get \( \dfrac{2l_1}{100-l_1} = \dfrac{1.5l_1}{100-1.5l_1} \).
4. Cross-multiplying: \( 200-3l_1 = 150-1.5l_1 \), so \( l_1 = 33.3 \) cm.
5. Then \( \dfrac{5}{R} = \dfrac{33.3}{66.7} = \dfrac12 \), so \( R = 10\ \Omega \).

Teacher's Note:
1. Write the balance condition twice, once before and once after shunting, and eliminate R to find the balance length.

Answer: (a) 2 Ω

Explanation:
1. With R infinite the balance length gives the e.m.f.: \( E \propto 60 \). With R = 10 Ω it gives the terminal voltage: \( V \propto 50 \).
2. \( r = R\left(\dfrac{l_1}{l_2}-1\right) = 10\left(\dfrac{60}{50}-1\right) = 2\ \Omega \).

Teacher's Note:
1. Use r = R(E/V − 1) and replace the ratio E/V by the ratio of the balancing lengths.

Answer: (b) \( \dfrac{G}{50} \)

Explanation:
1. The galvanometer carries 2% and the shunt carries 98% of the current, so \( 0.02G = 0.98S \Rightarrow S = G/49 \).
2. Ammeter resistance \( = \dfrac{GS}{G+S} = \dfrac{G\cdot G/49}{G+G/49} = \dfrac{G}{50} \).

Teacher's Note:
1. A quick check: the current splits 2 : 98 = 1 : 49 across the parallel pair, so the equivalent resistance is G/(1 + 49) = G/50.

Answer: (c) \( \dfrac12\dfrac{E_0r}{(r+r_1)} \)

Explanation:
1. The current in the wire is \( I = \dfrac{E_0}{r+r_1} \), so the potential difference across the whole wire is \( \dfrac{E_0r}{r+r_1} \).
2. The balance length is half the wire, so \( E = \dfrac12\cdot\dfrac{E_0r}{r+r_1} \).

Teacher's Note:
1. The potential gradient is uniform along the wire, so the potential drop is proportional to the balance length.