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: Power in the resistor
Only the resistor dissipates power on average: $P = I_{rms}^2R$.

Step 2: Find the current
$I_{rms} = \dfrac{210}{30} = 7$ A.

Step 3: Compute
$P = 7^2\times10 = 490$ W.

Step 4: Check
Option (D) 600 W and option (B) 420 W do not equal $I^2R = 490$ W.

Step 5: Why the other options fail
Apparent power is $V_{rms}I_{rms} = 210 \times 7 = 1470$ W, which is larger than every listed option. The true power is always less than or equal to this and depends on the power factor $\frac{R}{Z} = \frac{1}{3}$. Any listed value other than 490 W would need a different power factor, such as $\frac{2}{7} \approx 0.29$ for 420 W or about 0.41 for 600 W, and neither equals $\frac{1}{3}$.

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