Question

In a series L−C−R circuit, R=200 Ω and the voltage and the frequency of the main supply is 220 V and 50 Hz respectively. On taking out the capacitance from the circuit, the current lags behind the voltage by 30°. On taking out the inductor from the circuit, the current leads the voltage by 30°. The power dissipated in the L−C−R circuit is

A. 305 W
B. 210 W
C. Zero
D. 242 W

Answer 1

The correct answer is (D) 242 W. The power dissipated in the L-C-R circuit is calculated at resonance, where the power factor is 1. Using the formula Pave = Irms * Vrms * cos φ, where ϕ is the phase angle and Irms is the current through the resistor, we find the answer to be 242 W.

Step-by-step explanation:

To calculate the power dissipated in a series L-C-R circuit, we need to identify the phase relationship between voltage and current, as well as the power factor of the circuit.

Since the current lags behind the voltage by 30° upon removing the capacitor and leads by 30° upon removing the inductor, we can infer that the circuit is at resonance when all components are in place.

At resonance, the inductive and capacitive reactances cancel each other, resulting in the current being in phase with the voltage, giving us a power factor of 1 (cos 0° = 1).

Given the mains supply voltage Vrms at 220 V and the resistor R at 200 Ω, the current Irms can be calculated using Ohm's law (Irms = Vrms / R), which gives us Irms = 220 V / 200 Ω = 1.1 A.

Thus, the average power Pave dissipated in the circuit can be calculated using the formula Pave = Irms * Vrms * cos ϕ, where ϕ is the phase angle between the current and the voltage.

Since the circuit is at resonance and the phase angle is 0°, the power factor is 1.

Therefore, Pave = 1.1 A * 220 V * 1 = 242 W.