A $750 \mathrm{~Hz}, 20 \mathrm{~V}$ (rms) source is connected to a resistance of $100 \Omega$ an inductance of 0.1803 H and a capacitance of $10 \mu \mathrm{~F}$ all in series. The time in which the resistance (heat capacity $2 \mathrm{~J} /{ }^{\circ} \mathrm{C}$) will get heated by $10^{\circ} \mathrm{C}$. (assume no loss of heat to the surroundings) is close to:

JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Alternating Current

750 Hz, 20 V

(rms) source is connected to a resistance of

100Ω

an inductance of 0.1803 H and a capacitance of

10 μ F

all in series. The time in which the resistance (heat capacity

2 J /^{\circ} C

) will get heated by

1 0^{\circ} C

. (assume no loss of heat to the surroundings) is close to:

Answer: (B)

Only the resistor dissipates energy in the series circuit, at the average rate

P = \frac{ε _{rms}^{2} R}{Z ^{2}}

, where

Z = R^{2} + (X_{L} - X_{C})^{2}

. With

ω = 2 π (750)

, the inductive reactance

X_{L} = ω L \approx 850 Ω

dominates the small capacitive reactance

X_{C} = \frac{1}{ω C} \approx 21 Ω

, so the impedance is large, about

850 Ω

, and the power reaching the resistor is only a fraction of a watt. The resistor must absorb

Δ Q = C Δ T = 2 \times 10 = 20

J, and the time for this,

t = \frac{Δ Q}{P}

, comes out close to

365

JEE Main · 2020 · 03 Sep 2020 · Shift 1 · Single correct option