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  • When an AC source of emf  is connected across a circuit, the

When an AC source of emf  E=E_{0} \sin(100 \;t) is connected across a circuit, the phase difference between the emf E and the current i in the circuit is observed to be \frac{\pi}{4} , as shown in the diagram.  If the circuit consists possibly only of R-C or R-L or L-C series, find the relationship between the two elements.

Option: 1

R=1k\Omega , C=10 \mu F


Option: 2

R=1k\Omega , C=1 \mu F


Option: 3

R=1k\Omega , L=10 H


Option: 4

R=1k\Omega , L=1 H


Answers (1)

best_answer

\\ \text{Here voltage lags by} \\ \frac{\pi}{4} \\ \text{i.e. circuit is capacitive} \\ \begin{aligned} & \therefore \cos \theta=\frac{R}{\sqrt{R^2+X_c^2}}=\frac{1}{\sqrt{2}} \\ & \omega=100 \\ & 2 R^2=R^2+\frac{1}{100^2 C^2} \\ & 100^2 R^2=\frac{1}{c^2} \\ & 100 R C=1 \\ & \text { if } R=1000 \Omega \\ & C=10 \mu F \end{aligned}

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shivangi.bhatnagar

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