Question

An alternating emf of 0.2 V is applied across an L-C-R series circuit having R = 4Q, C = 80µF, and L = 200 mH. At resonance the voltage drop across the inductor is

पर्याय

• 10V

• 2.5V

• 1V

• 5V

Answer 1

2.5V

Explanation:

Given,
V_rms = 0.2V
R = 4Ω
C = 80µF = 80 × 10^-6 F
and L = 200 mH = 200 x 10^-3 H

The impedance of the series L-C-R circuit,

`"Z"=sqrt("R"^2+("X"_"L"-"X"_"C")^2`

At resonance, X_L =X_C

⇒ Z + R + 4Ω

∴ Voltage drop across inductor,

(V_rms)_L = l_rms × X_L

= `"V"_"rms"/"Z"xxomega"L"`

`="Vrms"/"R"xx"L"/(sqrt"LC")`   `(because omega=1/sqrt"LC")`

`="Vrms"/"R"xxsqrt("L"/"C")`

`=0.2/4xxsqrt((200xx10^-3)/(80xx10^-6))`

=0.05 × `sqrt2500`

= 0.05 × 50 = 2.5 V