Question

In series LCR circuit \(R = 10\,\Omega\) and impedance is \(30\,\Omega\). An r.m.s. voltage \(210\) V, is applied across the circuit. The true power consumed in AC circuit is

• A. \(360\) W
• B. \(420\) W
• C. \(490\) W
• D. \(600\) W

Hint:

True power is Vrms squared times R over Z squared.

Answer 1

The Correct Option is C

Step 1: Understand the concept
The true (average) power in an AC circuit is \(P = V_{rms}I_{rms}\cos\phi\), where the power factor is \(\cos\phi = \dfrac{R}{Z}\).

Step 2: Find the current
\(I_{rms} = \dfrac{V_{rms}}{Z} = \dfrac{210}{30} = 7\) A.

Step 3: Find the power factor
\(\cos\phi = \dfrac{10}{30} = \dfrac{1}{3}\).

Step 4: Compute
\[ P = 210\times7\times\frac{1}{3} = 490\ \text{W} \]
This equals \(I_{rms}^2R = 49\times10 = 490\) W. Option (C). The value 1470 W would be the apparent power \(V I\), which is not asked.

Final Answer:
The true power is 490 W. This is option (C). \[ \boxed{\text{(C) }490\ \text{W}} \]