Question

The power in an AC circuit contains an inductor of 30 mH, a capacitor of 300 μF, a resistor of 70 Ω, and an AC source of 24 V, 60 Hz. Calculate the energy dissipated in the circuit in 1000 s.

(a) 8.22  J

(b) 8.22 × 10^2  J

(c) 8.22 × 10^3  J

(d) 82.2 × 10^3  J

I got this question in an online quiz.

Question is taken from Power in AC Circuit : The Power Factor in portion Alternating Current of Physics – Class 12

Answer 1

Right answer is (c) 8.22 × 10^3  J

To explain: We know that, Pav=Vrms Irms  cos⁡Φ ………1 and cosΦ;=\(\frac {R}{Z}\) …………………2

In LCR, cosΦ=\(\frac {V_R}{V}=\frac {IR}{IZ}\)  and Irms = \( \frac {V_{rms}}{Z}\) ………………….3

 Substituting 2 and 3 in 1

Pav=\(\frac {V_{rms} \big ( \frac {V_rms}{Z} \big )R}{Z}\)

Pav=\(\frac {V_{rms}^2R}{Z^2}\)

Given: Vrms = 24 V; Resistance (R) = 70 Ω; Inductance (I) = 30 mH = 20 × 10^-3 H; Capacitance (C) = 300 μF

XL=2πυL=2π (60)(30 × 10^-3)

XL=11.304 Ω

XC=\(\frac {1}{2\pi υC}=\frac {1}{2\pi (60)(300 \times 10^{-6})}\)

XC=8.846 Ω

For series LCR circuit➔Z=\(\sqrt {R^2+(X_L-X_C)^2}\)

Z=\(\sqrt {70^2+(11.304-8.846)^2}\)

Z=70.04 ≈ 70 Ω

So, the energy used in 1000 seconds is ➔ Pavt=\((\frac {V_{rms}^2R}{Z^2})\)t

Pavt=\(\frac {24^2 \times 70}{70^2}\) × 1000 =  8.22 × 10^3J

Therefore, the energy dissipated in the circuit at 1000 seconds is 8.22 × 10^3J.