JEE Physics Current Electricity MCQs Set 21

Answer: (d) 90 W

Explanation:
1. Series resistance 3R, parallel resistance R/3; \( P=\frac{V^2}{R_{eq}} \).
2. The ratio is \( \frac{3R}{R/3}=9 \), so the power is 9 times as large: 90 W.

Teacher's Note:
1. Power at fixed voltage is inversely proportional to resistance.
2. Compare the two equivalent resistances.

Answer: (b) the 25 W bulb

Explanation:
1. Resistances: \( \frac{220^2}{25}=1936\ \Omega \) and \( \frac{220^2}{100}=484\ \Omega \).
2. Series current \( \frac{440}{2420}=0.18\ \text{A} \).
3. Rated currents: 25 W bulb \( 0.114\ \text{A} \) (exceeded), 100 W bulb \( 0.45\ \text{A} \) (safe).

Teacher's Note:
1. The lower-power bulb has the higher resistance and the smaller rated current.
2. Compare the series current with each rated current.

Answer: (b) 3

Explanation:
1. Each group has \( m=\frac{300}{n} \) cells, so the internal resistance of the battery is \( \frac{m r}{n}=\frac{300\times0.3}{n^2}=\frac{90}{n^2} \).
2. Maximum power when the internal resistance equals the external one: \( \frac{90}{n^2}=10 \), so \( n=3 \).

Teacher's Note:
1. Grouping m in series and n in parallel gives internal resistance mr/n.
2. Equate it to R.

Answer: (d) 49 Ω

Explanation:
1. Drop on internal resistance \( =10-9.8=0.2\ \text{V} \), so \( I=0.2\ \text{A} \).
2. \( R=\frac{9.8}{0.2}=49\ \Omega \).

Teacher's Note:
1. Terminal voltage equals the voltage across the external resistor.
2. Find the current from the internal drop first.

Answer: (a) \( \frac{R_1R_2}{R_1+R_2} \)

Explanation:
1. Seen from R, the two branches act as one source whose internal resistance is the parallel combination \( \frac{R_1R_2}{R_1+R_2} \).
2. Power in R is maximum when R equals the internal resistance of the source (maximum power transfer).

Teacher's Note:
1. Replace the network by its Thevenin equivalent.
2. Equal resistance gives maximum power.

Answer: (c) 90%

Explanation:
1. Efficiency \( \eta=\frac{R}{R+r}=0.6 \) gives \( R=1.5r \).
2. New resistance \( 6R=9r \), so \( \eta=\frac{9r}{9r+r}=90\% \).

Teacher's Note:
1. Efficiency rises as the external resistance rises.
2. Express R in terms of r first.

Answer: (a) 790 Ω

Explanation:
1. Potential gradient \( =\frac{10\ \text{mV}}{40\ \text{cm}} \), so the voltage across 100 cm is 25 mV.
2. Wire current \( I=\frac{25\times10^{-3}}{10}=2.5\ \text{mA} \).
3. \( R+10=\frac{2}{2.5\times10^{-3}}=800 \), so \( R=790\ \Omega \).

Teacher's Note:
1. Find the voltage across the whole wire from the balance point.
2. The same current flows through the series resistance and the wire.

Answer: (c) \( 2\times10^{16} \)

Explanation:
1. \( n=\frac{I}{e}=\frac{3.2\times10^{-3}}{1.6\times10^{-19}}=2\times10^{16}\ \text{s}^{-1} \).

Teacher's Note:
1. Current is charge per second, and each electron carries e.
2. The voltage is not needed for this count.

Answer: (a) 2 kΩ

Explanation:
1. Zero deflection means the voltage across R equals 2 V.
2. Current \( I=\frac{12-2}{10\ \text{k}\Omega}=1\ \text{mA} \), so \( R=\frac{2\ \text{V}}{1\ \text{mA}}=2\ \text{k}\Omega \).

Teacher's Note:
1. At balance no current flows in the galvanometer branch.
2. The same current flows through both series resistors.

Answer: (d) 30 min and \( \frac{20}{3} \) min

Explanation:
1. The heat needed is fixed, so \( t\propto R \) at fixed voltage: \( R_A:R_B=1:2 \).
2. Series: resistance 3 units, time \( 10+20=30\ \text{min} \).
3. Parallel: resistance \( \frac{1\times2}{3}=\frac23 \) units, time \( \frac{2/3}{1}\times10=\frac{20}{3}\ \text{min} \).

Teacher's Note:
1. Time to supply fixed heat at fixed voltage is proportional to resistance.
2. Equivalent resistances give the new times.

Answer: (d) 2 V

Explanation:
1. Let the meter resistances be \( R_A \) and \( R_V \). Initially the current is \( I=\frac{6}{R_A+R_V} \) and \( V=IR_V \).
2. Finally the current is 2I and the voltmeter reads V/2, so the parallel resistance is \( \frac{V/2}{2I}=\frac{R_V}{4} \).
3. Emf equation: \( 6=2I(R_A+\frac{R_V}{4})=I(R_A+R_V) \) gives \( R_A=\frac{R_V}{2} \).
4. Then \( V=6\times\frac{R_V}{1.5R_V}=4\ \text{V} \) initially and 2 V after.

Teacher's Note:
1. Use the same emf for both states.
2. The voltmeter reading is the voltage across the parallel part.

Answer: (a) \( 1.6\times10^{-5}\ \Omega\text{ m} \)

Explanation:
1. Field \( E=\frac{5}{0.1}=50\ \text{V/m} \).
2. From \( v_d=\frac{eE\tau}{m} \) and \( \rho=\frac{m}{ne^2\tau} \): \( \rho=\frac{E}{nev_d}=\frac{50}{8\times10^{28}\times1.6\times10^{-19}\times2.5\times10^{-4}}=1.6\times10^{-5}\ \Omega\text{ m} \).

Teacher's Note:
1. Use J = n e v_d = E / rho.
2. Convert the voltage into a field first.

Answer: (b) 4

Explanation:
1. Let the edges be a, 2a and b. Resistance between a pair of faces is \( R=\frac{\rho L}{A} \).
2. Maximum: length 2a, area ab: \( \frac{2\rho a}{ab}=\frac{2\rho}{b} \). Minimum: length a, area 2ab: \( \frac{\rho}{2b} \).
3. Ratio \( =\frac{2}{1/2}=4 \), independent of b.

Teacher's Note:
1. The unknown third edge cancels out in the ratio.
2. Longest length with smallest area gives the largest resistance.

Answer: (c) \( \frac43J_0 \)

Explanation:
1. \( I=\int_0^R J_0\left(1-\frac rR\right)2\pi r\,dr=2\pi J_0\left(\frac{R^2}{2}-\frac{R^2}{3}\right)=\frac{\pi R^2J_0}{3} \).
2. With \( R^2=\frac4\pi \): \( I=\frac{4}{3}J_0 \).

Teacher's Note:
1. Integrate J over thin rings of area 2 pi r dr.
2. Substitute R squared at the end.

Answer: (a) \( \frac{\rho_0L}{A}\left(1-\frac1e\right) \)

Explanation:
1. \( R=\int_0^L\frac{\rho\,dx}{A}=\frac{\rho_0}{A}\int_0^Le^{-x/L}dx=\frac{\rho_0L}{A}\left(1-e^{-1}\right) \).

Teacher's Note:
1. Slices of the rod are in series, so their resistances add as an integral.
2. The integral of e^(-x/L) is -L e^(-x/L).

Answer: (a) 3 A

Explanation:
1. Voltage across the 60 Ω branch \( =0.75\times60=45\ \text{V} \), which is also across the 15 Ω branch.
2. So \( I_2=\frac{45}{15}=3\ \text{A} \).

Teacher's Note:
1. Branches in parallel share the same voltage.
2. Ammeters here are ideal.

Answer: (b) 40 V

Explanation:
1. The voltmeter is across the 300 Ω resistor: \( \frac{300\times600}{900}=200\ \Omega \).
2. Total resistance \( 200+200+100=500\ \Omega \), current 0.2 A.
3. Voltmeter reading \( 0.2\times200=40\ \text{V} \).

Teacher's Note:
1. A non-ideal voltmeter loads the circuit; include its resistance.
2. Reading is the voltage across the parallel combination.

Answer: (a) an electric field is set up in the wire very quickly, producing a current through every cross-section almost instantly

Explanation:
1. When the switch closes, the field travels along the wire at nearly the speed of light and every free electron starts drifting at once.
2. The bulb receives current as soon as the nearby electrons drift; it does not wait for electrons from the battery to arrive.
3. Random thermal motion has no net direction, so it does not carry the current.

Teacher's Note:
1. Signal speed and drift speed are different things.
2. The field acts on all the electrons in the wire together.

Answer: (d) 5 Ω

Explanation:
1. By symmetry the three edges from the first corner, the six middle edges and the three edges into the last corner form three groups in series: \( \frac R3+\frac R6+\frac R3=\frac{5R}{6} \).
2. With R = 6 Ω: 5 Ω.

Teacher's Note:
1. Points at the same potential by symmetry can be joined.
2. The result 5R/6 is standard for resistors.

Answer: (d) 954 Ω

Explanation:
1. \( R_1=600(1+0.001\times30)=618\ \Omega \); \( R_2=300(1+0.004\times30)=336\ \Omega \).
2. Sum \( =954\ \Omega \).

Teacher's Note:
1. Series resistances simply add after each is corrected for temperature.
2. The temperature rise is 30 K.

Answer: (a) 0.2 A

Explanation:
1. At the junction the currents add to zero: \( \frac{10-V_0}{10}+\frac{6-V_0}{20}+\frac{5-V_0}{30}=0 \).
2. Multiply by 60: \( 6(10-V_0)+3(6-V_0)+2(5-V_0)=0 \), so \( 88=11V_0 \) and \( V_0=8\ \text{V} \).
3. Current in \( R_1 \) \( =\frac{10-8}{10}=0.2\ \text{A} \).

Teacher's Note:
1. Apply Kirchhoff's current law at the junction.
2. Cross-multiply to avoid fractions.

Answer: (b) 2 cal/s

Explanation:
1. The branches have 10 Ω and 5 Ω, so the current in the 10 Ω branch is half that in the 5 Ω resistor: \( I_4=\frac{I_5}{2} \).
2. \( P_5=5I_5^2=10 \), so \( I_5^2=2 \).
3. \( P_4=4\left(\frac{I_5}{2}\right)^2=I_5^2=2\ \text{cal/s} \).

Teacher's Note:
1. Parallel branches share the same voltage, so currents are inversely proportional to resistance.
2. Express both powers in terms of one current.

Answer: (d) \( \frac13\ \Omega \)

Explanation:
1. Let R be the external resistance. Series: \( \frac{3}{R+2r}=1 \). Parallel: \( \frac{1.5}{R+r/2}=0.6 \).
2. So \( R+2r=3 \) and \( R+\frac r2=2.5 \). Subtract: \( 1.5r=0.5 \), \( r=\frac13\ \Omega \).

Teacher's Note:
1. Write the circuit equation for each grouping.
2. Subtracting eliminates R.

Answer: (d) 4

Explanation:
1. Let P have length x and area A. Stretching to length 2x halves the area, so \( R_R=\frac{\rho(2x)}{A/2}=4\frac{\rho x}{A} \).
2. Q has \( R_Q=\frac{\rho(1-x)}{A} \); equal resistances give \( 1-x=4x \), so \( x=0.2\ \text{m} \).
3. Ratio \( \frac{Q}{P}=\frac{0.8}{0.2}=4 \).

Teacher's Note:
1. Stretching keeps volume constant so the area halves when the length doubles.
2. Use R proportional to length divided by area.

Answer: (b) \( \frac{2ER}{2R+r} \)

Explanation:
1. Equivalent battery: emf E, internal resistance \( \frac r2 \).
2. Current \( I=\frac{E}{R+r/2} \), so \( V=IR=\frac{ER}{R+r/2}=\frac{2ER}{2R+r} \).

Teacher's Note:
1. Identical cells in parallel halve the internal resistance.
2. Voltage across R is IR.