In a series resonant $$LCR$$ circuit, the voltage across $$R$$ is 100 volts and $$R = 1k\Omega $$ with $$C = 2\mu F.$$ The resonant frequency $$\omega $$ is $$200 rad/s.$$ At resonance the voltage across $$L$$ is
A.
$$2.5 \times {10^{ - 2}}V$$
B.
$$40 V$$
C.
$$250 V$$
D.
$$4 \times {10^{ - 3}}V$$
Answer :
$$250 V$$
Solution :
Across resistor, $$I = \frac{V}{R} = \frac{{100}}{{1000}} = 0.1A$$
At resonance, $${X_L} = {X_C} = \frac{1}{{\omega C}} = \frac{1}{{200 \times 2 \times {{10}^{ - 6}}}} = 2500$$
Voltage across $$L$$ is $$I\,{X_L} = 0.1 \times 2500 = 250V$$
Releted MCQ Question on Electrostatics and Magnetism >> Alternating Current
Releted Question 1
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