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  • When a 15 V dc source was applied across a choke coil then a current of 5 Amp flows in it. If the same coil is connected to a 15 V, 50 rad/s AC source, a current of 3 A flows in the circuit. f

When a 15 V dc source was applied across a choke coil then a current of 5 Amp flows in it. If the same coil is connected to a 15 V, 50 rad/s AC source, a current of 3 A flows in the circuit. find the power developed in the circuit is

Option: 1

0.18 W


Option: 2

0.21 W


Option: 3

0.24 W


Option: 4

0.27 W


Answers (1)

best_answer

For a coil, \mathrm{Z=\sqrt{R^2+\omega^2 L^2}, \quad I=\frac{V}{Z}=\frac{V}{\sqrt{R^2+\omega^2 L^2}}

For dc source, \mathrm{\omega=0 \quad I=\frac{V}{R} \quad \text { i.e., } R=\frac{15}{5}=3 \Omega \quad \ldots \text { (i) }}

When ac is applied

\mathrm{\begin{aligned} & I=\frac{V}{Z} \quad \text { i.e. } \quad Z=\frac{15}{3.0}=5 \Omega \\ & \therefore R^2+x_L^2=25 \end{aligned}}

\mathrm{\begin{aligned} & x_L^2=25-9=16 \Omega \Rightarrow x L=4 \Omega \\ & L=\frac{4}{40}=0.08 \text { Henry. } \end{aligned}}

Now when the capacitor is connected is series.

\mathrm{\begin{aligned} & \therefore x_C=\frac{1}{\omega C}=\frac{1}{50 \times 2500 \times 10^{-6}}=8 \Omega \\ & Z=\sqrt{R^2+\left(x_L-x_0\right)^2}=\sqrt{3+(4-8)^2}=5 \Omega \\ & \therefore I=\frac{15}{5}=3 A \quad \therefore P_{a v}=v_{\text {moss }} I_{m n s} \cos \phi=\left.\right|_{m a} ^2 R=(3)^2 \times 3=27 \text { W. } \end{aligned}}

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Ajit Kumar Dubey

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