Physics - EMI/AC Question with Solution | TestHub

Given:

For an LCR series circuit, the applied voltage is

$$v=v_0\sin \omega t$$

At \( t = 0 \), the current phasor leads the voltage phasor by

$$\varphi =\frac{\pi }{3}$$

Hence, the circuit is capacitive.

Current is:

$$i=i_0\sin (\omega t+\varphi )=i_0\sin \left(\omega t+\frac{\pi }{3}\right)$$

Instantaneous power supplied by the source:

$$p(t)=vi=v_0{i}_0\sin \omega t\sin \left(\omega t+\frac{\pi }{3}\right)$$

Using the identity

$$\sin A\sin B=\frac{1}{2}\left[\cos (A-B)-\cos (A+B)\right]$$

$$p(t)=\frac{v_0{i}_0}{2}\left[\cos \frac{\pi }{3}-\cos \left(2\omega t+\frac{\pi }{3}\right)\right]$$

Option (A):

At

$$t=\frac{\pi }{2\omega }\Rightarrow \omega t=\frac{\pi }{2}$$

$$p=\frac{v_0{i}_0}{2}\left[\frac{1}{2}-\cos \left(\pi +\frac{\pi }{3}\right)\right]=\frac{v_0{i}_0}{2}\left[\frac{1}{2}-\cos \frac{4\pi }{3}\right]$$

$$\cos \frac{4\pi }{3}=-\frac{1}{2}\Rightarrow p>0$$

$$Option(A)isfalse.$$

Option (B):

Average power in an LCR circuit:

$$P_{avg}=\frac{v_0{i}_0}{2}\cos \varphi =\frac{v_0{i}_0}{2}\cos \frac{\pi }{3}=\frac{v_0{i}_0}{4}>0$$

$$Option(B)istrue.$$

Option (C):

At

$$t=\frac{5\pi }{6\omega }\Rightarrow \omega t=\frac{5\pi }{6}$$

$$p=\frac{v_0{i}_0}{2}\left[\frac{1}{2}-\cos \left(\frac{5\pi }{3}+\frac{\pi }{3}\right)\right]=\frac{v_0{i}_0}{2}\left[\frac{1}{2}-\cos 2\pi \right]<0$$

$$Option(C)istrue.$$

Option (D):

Since the circuit is capacitive, if ω is increased slightly,

the circuit moves toward resonance and the phase difference decreases.

$$Option(D)istrue.$$

Correct options: (B), (C), (D)

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