In series LCR circuit, R = 10\ and impedance is 30\ . An rms voltage 210\ V is applied across the circuit. The true power consumed in A.C. circuit in watt is

Correct answer: C. 490 Given R = 10\ , Z = 30\ , and V_ rms = 210\ V . The rms current in the circuit is given by: I_ rms = V_ rms Z Substituting the given values: I_ rms = (210)/(30) = 7\ A The true power consumed in an A.C. circuit is given by: P = I_ rms^2 R Substituting the values of I_ rms and R : P = (7)^2 × 10 = 49 × 10 = 490\ W Answer: 490

MHT CET 202621 AprilEvening Shift

Exam: MHT CET
Year: 2026
Date: 21 April
Shift: Evening Shift
Subject: Physics
Chapter: Alternating Current

MHT CET 2026 · 21 April · Evening Shift · Physics · Alternating Current

In series LCR circuit, R = 10\ and impedance is 30\ . An rms...

In series LCR circuit,

R = 10\ \Omega

and impedance is

30\ \Omega

. An rms voltage

210\ \text{V}

is applied across the circuit. The true power consumed in A.C. circuit in watt is

Options

A $49$
B $220$
C $490$
D $1470$

Correct answer

C. $490$

Step-by-step solution

Given

R = 10\ \Omega

Z = 30\ \Omega

, and

V_{\text{rms}} = 210\ \text{V}

.

The rms current in the circuit is given by:

I_{\text{rms}} = \dfrac{V_{\text{rms}}}{Z}

Substituting the given values:

I_{\text{rms}} = \dfrac{210}{30} = 7\ \text{A}

The true power consumed in an A.C. circuit is given by:

P = I_{\text{rms}^2 R

Substituting the values of

I_{\text{rms}}

and

R

P = (7)^2 \times 10 = 49 \times 10 = 490\ \text{W}

Answer:

490

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