CBSE Class 12 Physics 2023 Delhi Set 3 Paper

Question : 13 of 14

Marks: +1, -0

SECTION - C

An alternating current

I=14 \sin(100 \pi t)

A passes through a series combination of a resistor of 30

\Omega

and an inductor of

\frac{2}{5\pi}

H. Taking

\sqrt{2}=1.4

, calculate the

(i) rms value of the voltage drops across the resistor and the inductor, and

(ii) power factor of the circuit.

Solution: 👈: Video Solution

(i)

V_{\text{rms}}

drop across resistor,

R

= I_{\text{rms}} \times R

= \frac{14}{\sqrt{2}} \times 30

= 294 \,\mathrm{V}

V_{\text{rms}} \text{ drop across inductor, } L

= I_{\text{rms}} \times \omega L

= \left(\frac{14}{\sqrt{2}}\right) \times (100\pi) \times \left(\frac{2}{5\pi}\right)

= 392 \,\mathrm{V}

(ii) Power factor,

\frac{R}{Z}

\frac{R}{Z} = \frac{R}{\sqrt{R^2 + X_L^2}}

\frac{R}{Z} = \frac{30}{\sqrt{30^2 + \left(100\pi \times \frac{2}{5\pi}\right)^2}}

= \frac{30}{\sqrt{30^2 + 40^2}}

= \frac{30}{50}

= 0.6

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