CBSE Class 12 Physics 2023 Outside Delhi Set 3 Paper

Question : 13 of 13

Marks: +1, -0

A resistor of

50 \Omega

, a capacitor of

\left(\frac{25}{\pi}\right)\;\mu\text{F}

and an inductor of

\left(\frac{4}{\pi}\right)\;\text{H}

are connected in series across an ac source whose voltage (in volt) is given by

V=70

\sin(100\pi t)

. Calculate:

(a) the net reactance of the circuit.

(b) the impedance of the circuit

Solution: 👈: Video Solution

(a) The net reactance

=X_L-X_C

=\omega L-\frac{1}{\omega c}

=100\pi \times \frac{4}{\pi} - \frac{1}{100\pi \times \frac{25}{\pi} \times 10^{-6}}

=400-\frac{10^6}{2500}=400-400

=0

(b) The impedance of the Circuit

Z=\sqrt{R^2+(X_L-X_C)^2}

=\sqrt{50^2+0}

=50\Omega

=\frac{70}{50}\text{A}

So, the effective value of current is

I_{\text{eff}}=\frac{70}{50}\times\frac{1}{\sqrt{2}}=\frac{7}{5\sqrt{2}}\text{A}

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