Voltage V and current i in A C circuit are given by V=50 (50 t) volt, i=50 (50 t+(π)/(3)) mA The power dissipated in the circuit is

Correct answer: C. 1.25 W Given V=50 (50 t) V Maximum voltage, V_ 0 =50 ~V i= ( 50 t+ (π)/(3)) mA Maximum current, i_ 0 =50 ~mA=50 × 10^ -3 ~A Power dissipated, P= i_ 0 2 × V_ 0 2 = 50 × 50 × 10^ -3 2 = 2500 × 10^ -3 2 =1.25 ~W

BITSAT 2014

Exam: BITSAT
Year: 2014
Subject: Physics
Chapter: Alternating Current

BITSAT 2014 · Physics · Alternating Current

Voltage V and current i in A C circuit are given by V=50 (50 t)...

Voltage

V

and current

i

A C

circuit are given by

\mathrm{V}=50 \sin (50 \mathrm{t})

volt,

\mathrm{i}=50 \sin \left(50 \mathrm{t}+\frac{\pi}{3}\right) \mathrm{mA}

The power dissipated in the circuit is

Options

A $5.0 \mathrm{ W}$
B $2.5 \mathrm{ W}$
C $1.25 \mathrm{ W}$
Dzero

Correct answer

C. $1.25 \mathrm{ W}$

Step-by-step solution

Given

V=50 \sin (50

V

Maximum voltage,

\mathrm{V}_{0}=50 \mathrm{~V}

\left(

50 t+

\frac{\pi}{3}\right) \mathrm{mA}

Maximum current,

i_{0}=50 \mathrm{~mA}=50 \times 10^{-3} \mathrm{~A}

Power dissipated,

\mathrm{P}=\frac{\mathrm{i}_{0}}{\sqrt{2}} \times \frac{\mathrm{V}_{0}}{\sqrt{2}}=\frac{50 \times 50 \times 10^{-3}}{2}=\frac{2500 \times 10^{-3}}{2}=1.25 \mathrm{~W}

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