JEE Physics Heating Effect of Current MCQs Set 01

Answer: (c) a high specific resistance and a low melting point

Explanation:
1. A fuse must heat up quickly (high resistivity) and melt at a low temperature to break the circuit.

Teacher's Note:
1. Joule heating is \( I^2R \), so a high resistivity helps it heat; low melting point lets it blow.

Answer: (c) 2 : 1

Explanation:
1. \( P = V^2/R \) with the same V, so \( P_1 : P_2 = R_2 : R_1 = 2 : 1 \).

Teacher's Note:
1. In parallel the same voltage acts, so \( P = V^2/R \) is larger for the smaller resistance.

Answer: (a) 2 : 1

Explanation:
1. Half the length gives half the resistance. At the same voltage \( H\propto 1/R \), so the heat is doubled: 2 : 1.

Teacher's Note:
1. At constant voltage a smaller resistance gives more heat.

Answer: (b) Resistance of the tungsten lamp increases but that of carbon diminishes when hot

Explanation:
1. Metals such as tungsten have a positive temperature coefficient of resistance; carbon has a negative one.

Teacher's Note:
1. Carbon is a semiconductor-like material, so its resistance falls on heating.

Answer: (c) free electron theory

Explanation:
1. Drifting free electrons collide with lattice ions and transfer energy to them, which appears as heat.

Teacher's Note:
1. Collisions of electrons with ions produce Joule heating.

Answer: (a) 1 : 2

Explanation:
1. Same current: \( P = I^2R \), so \( P_1:P_2 = R_1:R_2 = 1:2 \).

Teacher's Note:
1. In series the same current flows, so \( P = I^2R \) is larger for the larger resistance.

Answer: (c) the wire is divided into four parts and all four are connected to the battery in parallel

Explanation:
1. Heat per second \( = V^2/R_{eq} \), largest for the smallest \( R_{eq} \).

Teacher's Note:
1. Four equal parts in parallel give \( R/16 \), smaller than \( R \), \( R/4 \) and \( R/2 \).

Answer: (a) the 100 W bulb has the thicker filament

Explanation:
1. \( P = V^2/R \): 100 W means smaller R. For the same length, smaller R means a larger cross-section, i.e. a thicker filament.

Teacher's Note:
1. Write the power in terms of the quantity that is common to the components (current in series, voltage in parallel).

Answer: (d) 90 W

Explanation:
1. Series: \( P_s = \dfrac{V^2}{3R} = 10 \). Parallel: \( P_p = \dfrac{V^2}{R/3} = \dfrac{3V^2}{R} = 9\times10 = 90 \) W.

Teacher's Note:
1. Parallel resistance is 9 times smaller than series resistance for three equal resistors.

Answer: (d) 150

Explanation:
1. Total power \( = 10\times50 = 500 \) W = 0.5 kW; time \( = 10\times30 = 300 \) h.

Teacher's Note:
1. Energy \( = 0.5\times300 = 150 \) kWh.

Answer: (c) Both bulbs will fuse

Explanation:
1. Each bulb is rated for 220 V but receives 440 V, twice its rating, so the power becomes 4 times the rated value.

Teacher's Note:
1. Both exceed their ratings and fuse.

Answer: (c) 80 W

Explanation:
1. Each lamp works at its rated voltage in parallel, so the powers add: 80 W.

Teacher's Note:
1. Write the power in terms of the quantity that is common to the components (current in series, voltage in parallel).

Answer: (a) In series the fine wire liberates more energy, while in parallel the thick wire liberates more

Explanation:
1. The fine wire has the larger resistance.

Teacher's Note:
1. In series \( H = I^2Rt \) is larger for the fine wire; in parallel \( H = V^2t/R \) is larger for the thick wire.

Answer: (d) 25 W

Explanation:
1. \( P\propto V^2 \): \( P' = 100\times\left(\dfrac{110}{220}\right)^2 = 25 \) W.

Teacher's Note:
1. Power varies as the square of voltage for a fixed resistance.

Answer: (a) 25 W bulb

Explanation:
1. \( R = V^2/P \): \( R_{25} = 1936\ \Omega \), \( R_{100} = 484\ \Omega \).

Teacher's Note:
1. Same current, so the larger resistance (25 W bulb) dissipates more power and glows brighter.

Answer: (a) decreases

Explanation:
1. Adding \( R_2 \) in series increases the total resistance, so the current falls.

Teacher's Note:
1. \( P_1 = I^2R_1 \) decreases.

Answer: (c) less than that of the fan

Explanation:
1. \( R = V^2/P \): the fan \( = 484\ \Omega \), the heater \( = 48.4\ \Omega \).

Teacher's Note:
1. Write the power in terms of the quantity that is common to the components (current in series, voltage in parallel).

Answer: (d) 1 : 4

Explanation:
1. If each wire has resistance R: series \( \dfrac{V^2}{2R} \), parallel \( \dfrac{2V^2}{R} \).

Teacher's Note:
1. Ratio \( = \dfrac{1/2}{2} = 1:4 \).

Answer: (c) 500 W

Explanation:
1. Each \( R = \dfrac{220^2}{1000} = 48.4\ \Omega \); series \( R = 96.8\ \Omega \).

Teacher's Note:
1. \( P = \dfrac{220^2}{96.8} = 500 \) W.

Answer: (b) 100 W bulb

Explanation:
1. Both work at rated voltage, so each delivers its rated power; the 100 W bulb is brighter.

Teacher's Note:
1. In parallel the same voltage acts, so \( P = V^2/R \) is larger for the smaller resistance.

Answer: (b) 25 W bulb

Explanation:
1. \( R_{25} = 1936 \), \( R_{100} = 484\ \Omega \); current \( = \dfrac{440}{2420} = 0.18 \) A.

Teacher's Note:
1. The 25 W bulb's rated current is 0.114 A, which is exceeded, while the 100 W bulb's rated current is 0.455 A. So the 25 W bulb fuses.

Answer: (b) 2%

Explanation:
1. \( P = I^2R \Rightarrow \dfrac{\Delta P}{P} = 2\dfrac{\Delta I}{I} = 2\% \).

Teacher's Note:
1. Write the power in terms of the quantity that is common to the components (current in series, voltage in parallel).

Answer: (b) 2.0 W

Explanation:
1. Series: \( I = \dfrac{4}{2.5} = 1.6 \) A, \( P = 1.28 \) W. Parallel: \( E = 2 \) V, \( r = 0.5\ \Omega \), \( I = \dfrac{2}{1} = 2 \) A, \( P = 4\times0.5 = 2 \) W.

Teacher's Note:
1. The larger value, 2.0 W, is the maximum.

Answer: (a) will be doubled

Explanation:
1. \( R = \dfrac{\rho L}{\pi r^2} \): length ×2, area ×4, so \( R \) becomes \( R/2 \).

Teacher's Note:
1. \( P = V^2/R \) is doubled.

Answer: (d) 55 \( \Omega \)

Explanation:
1. \( P = V^2/R \) is largest for the smallest resistance, 55 \( \Omega \).

Teacher's Note:
1. Write the power in terms of the quantity that is common to the components (current in series, voltage in parallel).

Answer: (c) Heat produced in a conductor is proportional to the charge

Explanation:
1. \( H = I^2Rt \) is proportional to R, \( I^2 \) and t. In terms of charge \( q \), \( H = q^2R/t \), not proportional to q.

Teacher's Note:
1. Be careful with statements that hold only for fixed other quantities.