A resistor of 50 Ω, a capacitor of (\frac{25}{\pi}) \mu {\F} and an inductor of (\frac{4}{\pi}) {\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 (c) the effective value of current in the circuit.

CBSE Class 12 Physics 2023 Outside Delhi Set 3 Paper

Question : 13 of 13

Marks: +1, -0

A resistor of

50 Ω

, a capacitor of

(\;{25}/{π}) µ {\F}

and an inductor of

(\;{4}/{π}) {\H}

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

V=70

sin \;(100 π t)

. Calculate:
(a) the net reactance of the circuit.
(b) the impedance of the circuit
(c) the effective value of current in the circuit.

Solution:

(a) The net reactance

\;=X_L-X_C

\;=ω_L-\;{1}/{ω c}

\;=100 π × \;{4}/{π}-\;{1}/{100 π × \;{25}/{π} × 10^{-6}}

\;=400-\;{10^6}/{2500}=400-400

\;=0

(b) The impedance of the Circuit

Z\;={√{R^2+ (X_L-X_C) ^2}}

\;={√{50^2+0}}

\;=50 Ω

=\;{70}/{50} {\A}

So, the effective value of current is

I_{\;\text "eff "\;}=\;{70}/{50} × \;{1}/{{√{2}}}=\;{7}/{5 {√{2}}} A

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