JEE Physics Current Electricity MCQs Set 08

Answer: (c) 2 Ω in series with the parallel combination of 3 Ω and 6 Ω

Explanation:
1. 3 Ω ∥ 6 Ω = (3 × 6)/(3 + 6) = 2 Ω.
2. Adding the 2 Ω resistor in series gives 2 + 2 = 4 Ω.
3. Check the others: 3 + 1.5 = 4.5 Ω, 6 + 1.2 = 7.2 Ω, and 2 ∥ 2 = 1 Ω.

Teacher's Note:
1. Try combinations one by one.
2. 3 Ω and 6 Ω in parallel give exactly 2 Ω.

Answer: (c) \( \sqrt{69}\ \Omega \)

Explanation:
1. Total = 3 + 10(3 + R)/(10 + 3 + R) = R.
2. 3(13 + R) + 10(3 + R) = R(13 + R), so 69 + 13R = 13R + R².
3. R² = 69, so R = √69 Ω.

Teacher's Note:
1. Write the equivalent resistance of the parallel part first.
2. The linear terms in R cancel, leaving a simple square.

Answer: (d) \( n^2R \)

Explanation:
1. Parallel: R = r/n, so r = nR.
2. Series: nr = n(nR) = n²R.

Teacher's Note:
1. Find r from the parallel result first.
2. Series total = n times one resistance.

Answer: (c) 8

Explanation:
1. In a parallel combination the equivalent resistance is smaller than either resistor, so R₁ > 6 Ω.
2. Only 8 Ω qualifies; then R₂ = (6 × 8)/(8 − 6) = 24 Ω.

Teacher's Note:
1. Parallel resistance is less than the smallest branch.
2. Check: 8 ∥ 24 = 192/32 = 6 Ω.

Answer: (a) \( 1\ \Omega \)

Explanation:
1. 3 Ω ∥ 6 Ω = 2 Ω.
2. Total resistance = 6 V/2 A = 3 Ω.
3. R = 3 − 2 = 1 Ω.

Teacher's Note:
1. Find the equivalent of the parallel part first.
2. An ideal ammeter has no resistance.

Answer: (d) \( 24\ \Omega \)

Explanation:
1. P = I²(6) = 6 W, so I = 1 A.
2. Voltage across the parallel part = 12 − 1 × 6 = 6 V, so its resistance is 6 Ω.
3. R × 8/(R + 8) = 6 gives 8R = 6R + 48, so R = 24 Ω.

Teacher's Note:
1. First find the current from the power in the known resistor.
2. Then use the remaining voltage to find the parallel equivalent.

Answer: (b) 2 A

Explanation:
1. 4 = 1(2 + r), so r = 2 Ω.
2. Short-circuit current = E/r = 4/2 = 2 A.

Teacher's Note:
1. Find the internal resistance first.
2. In a short circuit only r limits the current.

Answer: (b) 6 V

Explanation:
1. With R = r, current I = E/2r and terminal voltage V = IR = E/2.
2. E = 2 × 3 = 6 V.

Teacher's Note:
1. Equal resistances share the e.m.f. equally.
2. Terminal voltage = E − Ir.

Answer: (b) 14 V

Explanation:
1. Current inside the cell flows from negative to positive, i.e. the cell is discharging.
2. Terminal voltage = E − Ir = 5 − 2 × 0.5 = 4 V.
3. V₊ = 10 + 4 = 14 V.

Teacher's Note:
1. Inside a discharging cell the terminal voltage is E − Ir.
2. The positive terminal is higher than the negative by the terminal voltage.

Answer: (b) \( 2\ \Omega \)

Explanation:
1. Terminal voltage = 2 × 2 = 4 V = E/2, so E = 8 V.
2. E = I(R + r): 8 = 2(2 + r), so r = 2 Ω.

Teacher's Note:
1. The terminal voltage equals the voltage across the external resistance.
2. Then E = V + Ir.

Answer: (b) 1.5 V

Explanation:
1. E = 0.5(2 + r) and E = 0.25(5 + r).
2. 1 + 0.5r = 1.25 + 0.25r, so r = 1 Ω.
3. E = 0.5 × 3 = 1.5 V.

Teacher's Note:
1. Two equations, two unknowns E and r.
2. Check: 0.25 × 6 = 1.5 V.

Answer: (b) \( 0.5\ \Omega \)

Explanation:
1. 0.9(2 + r) = 0.3(7 + r).
2. 1.8 + 0.9r = 2.1 + 0.3r, so 0.6r = 0.3.
3. r = 0.5 Ω.

Teacher's Note:
1. The e.m.f. and r are the same in both cases.
2. Equate the two expressions for E.

Answer: (a) \( 0.5\ \Omega \)

Explanation:
1. Power in R is P = E²R/(R + r)².
2. It is maximum when R = r (maximum power transfer theorem).
3. R = 0.5 Ω.

Teacher's Note:
1. Differentiate P with respect to R to see that R = r gives the maximum.
2. The maximum power is E²/4r.

Answer: (d) \( P = (V/V_0)^2P_0 \)

Explanation:
1. P = V²/R with R constant.
2. P/P₀ = V²/V₀², so P = (V/V₀)²P₀.

Teacher's Note:
1. R = V₀²/P₀ is fixed for the bulb.
2. Power varies as the square of the voltage.

Answer: (a) 25 W

Explanation:
1. R₂₅ = 200²/25 = 1600 Ω and R₁₀₀ = 200²/100 = 400 Ω.
2. Current in series = 400/2000 = 0.2 A.
3. Rated currents: 25 W bulb 0.125 A and 100 W bulb 0.5 A. The 25 W bulb carries more than its rating (0.2 A > 0.125 A), so it fuses.

Teacher's Note:
1. In series both carry the same current, so compare with each bulb's rated current.
2. The lower-power bulb has the higher resistance and the greater voltage drop.

Answer: (a) 1 A

Explanation:
1. With n cells in series in each of m parallel rows (nm = 48) the current is maximum when nr/m = R.
2. n × 4/m = 12 gives n = 3m, so 3m² = 48, m = 4 and n = 12.
3. I = nE/(R + nr/m) = 24/(12 + 12) = 1 A.

Teacher's Note:
1. Maximum current occurs when the external resistance equals the net internal resistance.
2. Use nm = 48 as a constraint.

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

Explanation:
1. R = V²/P, so for equal P, R ∝ V².
2. In series the same current flows, and power dissipated I²R ∝ R ∝ V².
3. Brightness ∝ (marked voltage)².

Teacher's Note:
1. In series P = I²R, so the higher-resistance bulb is brighter.
2. Same rated power means resistance proportional to rated V squared.

Answer: (d) 3 > 1 > 2 > 4

Explanation:
1. Same area and same V, so P = V²/R ∝ 1/(ρL).
2. ρL (in units of ρL): wire 1 = 1, wire 2 = 1.44, wire 3 = 0.81, wire 4 = 1.5.
3. Smallest R dissipates most: 3 > 1 > 2 > 4.

Teacher's Note:
1. At fixed V, power is inversely proportional to resistance.
2. R ∝ ρL for equal diameters.

Answer: (d) \( R = r/2 \)

Explanation:
1. Two identical cells in parallel have e.m.f. E and internal resistance r/2.
2. Maximum power occurs when R equals the net internal resistance: R = r/2.

Teacher's Note:
1. Parallel identical cells halve the internal resistance but keep the e.m.f.
2. Apply the maximum-power condition R = r_net.

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

Explanation:
1. If R₁ carries current I, then R₂ and R₃ each carry I/2 (equal by symmetry).
2. P₁ = I²R₁ and P₂ = (I/2)²R₂ = I²R₂/4.
3. Equal power gives R₂ = 4R₁, and similarly R₃ = 4R₁.

Teacher's Note:
1. The current divides equally between equal branches.
2. Power = I²R.

Answer: (b) (b) and (c)

Explanation:
1. P = 16R/(R + 1)², which is maximum at R = 1: P_max = 16/4 = 4 W.
2. So any power above 4 W is impossible: 5 W and 8 W cannot occur, while 2 W and 4 W can.

Teacher's Note:
1. Use the maximum power theorem: P_max = E²/4r.
2. Any value below P_max can occur for two different values of R.

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

Explanation:
1. The parallel combination of the voltmeter and the resistor has lower resistance than the voltmeter alone, so the total resistance falls and A increases.
2. The voltmeter reads the voltage across this lower parallel resistance, which is a smaller share of the e.m.f.: V decreases.

Teacher's Note:
1. The voltmeter has a very high resistance, so it takes almost the whole e.m.f. at first.
2. Lower parallel resistance gets a smaller share of the voltage.

Answer: (a) \( 11.11\ \Omega \)

Explanation:
1. S = I_g G/(I − I_g) = (10 × 100)/(100 − 10).
2. S = 1000/90 = 11.11 Ω.

Teacher's Note:
1. The shunt carries I − I_g = 90 mA.
2. Shunt and galvanometer have equal voltage across them.

Answer: (a) \( (n-1)G \)

Explanation:
1. Full-scale current I_g = V/G.
2. For range nV the total resistance must be nG, so the series resistance is nG − G = (n − 1)G.

Teacher's Note:
1. Total resistance of the voltmeter must be n times larger.
2. The extra series resistance is the difference.

Answer: (d) 50 cm

Explanation:
1. Initially l/(100 − l) = 10/30, so l = 25 cm.
2. After interchange l/(100 − l) = 30/10, so l = 75 cm.
3. The shift is 75 − 25 = 50 cm.

Teacher's Note:
1. Use the metre bridge balance condition R/S = l/(100 − l).
2. Interchanging resistances gives the mirror point.