Question

In an LCR series circuit, a non-inductive resistor of 150 Ω, a coil of 0.2 H inductance and negligible resistance, and a 30 $\mu F$ capacitor are connected across an ac power source of 220 V, 50 Hz. The power loss across the resistor is _________W (Round off to 2 decimal places).

Answer 1

Correct Answer: 297

1. Identify the Given Values

Resistance ($R$) = $150 \, \Omega$

Inductance ($L$) = $0.2 \, \text{H}$

Capacitance ($C$) = $30 \, \mu\text{F} = 30 \times 10^{-6} \, \text{F}$

RMS Voltage ($V_{\text{rms}}$) = $220 \, \text{V}$

Frequency ($f$) = $50 \, \text{Hz}$

2. Calculate Inductive and Capacitive Reactance

First, determine the angular frequency: $\omega = 2\pi f = 2\pi(50) = 100\pi \, \text{rad/s}$.

Inductive Reactance ($X_L$):

$$X_L = \omega L = 100\pi \times 0.2 = 20\pi \approx 62.83 \, \Omega$$

Capacitive Reactance ($X_C$):

$$X_C = \frac{1}{\omega C} = \frac{1}{100\pi \times 30 \times 10^{-6}} = \frac{1000}{3\pi} \approx 106.10 \, \Omega$$

3. Calculate Impedance ($Z$)

The net reactance is $X = |X_L - X_C| = |62.83 - 106.10| = 43.27 \, \Omega$.

The total impedance of the series LCR circuit is:

$$Z = \sqrt{R^2 + (X_L - X_C)^2}$$

$$Z = \sqrt{150^2 + (43.27)^2} = \sqrt{22500 + 1872.29} = \sqrt{24372.29} \approx 156.12 \, \Omega$$

4. Calculate Power Loss

Power loss in an LCR circuit occurs only across the resistor. We can calculate the power using the formula $P = \frac{V_{\text{rms}}^2 R}{Z^2}$ (derived from $P = I^2R$ and $I = V/Z$).

$$P = \frac{(220)^2 \times 150}{(156.12)^2}$$

$$P = \frac{48400 \times 150}{24373.45}$$

$$P = \frac{7260000}{24373.45} \approx 297.87 \, \text{W}$$

Alternatively, finding the current first:

$$I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{220}{156.12} \approx 1.409 \, \text{A}$$

$$P = I_{\text{rms}}^2 R = (1.409)^2 \times 150 \approx 297.88 \, \text{W}$$

Final Answer:

The power loss across the resistor is 297.88 W.