JEE Main 2022 · 28 Jun · Shift 1 · Physics Q54

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PhysicsInteger+4 / -1Question 54 of 90

An AC source is connected to an inductance of 100 mH, a capacitance of 100 $\mu$ F and a resistance of 120 $\Omega$ as shown in figure. The time in which the resistance having a thermal capacity 2 J/°C will get heated by 16°C is ______ s.

Series circuit of a 100 mH inductor, a 100 μF capacitor and a 120 Ω resistor across an AC source V = 20√2 sin(100t).

Your answer

Answer and worked solution

Official answer: 15

Impedance. $X_L=10$ Ω, $X_C=100$ Ω: $Z=\sqrt{120^2+90^2}=150$ Ω.

Power. $I_{rms}=\frac{20}{150}$ , so $P=I^2R=\frac{32}{15}$ W.

Time. Heat needed $2\times16=32$ J: $t=15$ s.

Answer: 15

Topics:Power in AC circuits Reactance and impedance

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JEE Main 2022 · 28 Jun · Shift 1, Physics question 54