In the LCR series circuit, an alternating e.m.f. $e$ and current $i$ are given by the equation $e = 100\sin(100t)\,V$ and $i = 100\sin\left(100t + \dfrac{\pi}{3}\right)mA$ .
The average power dissipated in the circuit will be
$\{\sin(\pi/3) = 3^{0.5}/2,\ \cos(\pi/3) = 1/2\}$

Question

In the LCR series circuit, an alternating e.m.f.

e

and current

i

are given by the equation

e = 100\sin(100t)\,V

and

i = 100\sin\left(100t + \dfrac{\pi}{3}\right)mA

.
The average power dissipated in the circuit will be

\{\sin(\pi/3) = 3^{0.5}/2,\ \cos(\pi/3) = 1/2\}

Quantrex Academy · Accepted Answer

Correct answer: D. 2.5 W

The given equations for alternating e.m.f. and current are:

e = 100\sin(100t)

V

i = 100\sin \left(100t + \dfrac{\pi}{3}\right) mA<br><br>From the equations, the peak voltage

E_0 = 100

V, the peak current

I_0 = 100

mA

= 0.1

A, and the phase difference

\phi =

\dfrac{\pi}{3}.<br><br>The average power dissipated in an AC circuit is given by:<br><br>

P_{avg} = E_{rms} I_{rms} \cos\phi =

\dfrac{E_0 I_0}{2} \cos\phi<br><br>Substituting the given values:<br><br>

P_{avg} =

\dfrac{100 \times 0.1}{2} \cos \left(\dfrac{\pi}{3}\right)<br><br>

P_{avg} =

\dfrac{10}{2} \times \dfrac{1}{2} =2.5

W

Answer:

2.5

W

In the LCR series circuit, an alternating e.m.f. e and current i...