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  • An AC voltage V V sin100t = 0 is applied to the circuit, the phase difference between current and voltage is found to be <img alt="\frac{\pi}{4}" src="https://entrancecorner.oncodecogs.com/gif.late

An AC voltage V V sin100t = 0 is applied to the circuit, the phase difference between current and voltage is found to be \frac{\pi}{4} , then
                      

Option: 1

\mathrm{R}=100 \Omega, \mathrm{C}=1 \mu \mathrm{F}


Option: 2

\mathrm{R}=1 \mathrm{k} \Omega, \mathrm{C}=10 \mu \mathrm{F}


Option: 3

\mathrm{R}=10 \mathrm{k} \Omega, \mathrm{L}=1 \mathrm{H}


Option: 4

\mathrm{R}=1 \mathrm{k} \Omega, \mathrm{L}=10 \mathrm{H}


Answers (1)

best_answer

In the given graph current is leading the voltage by \mathrm{45^{\circ} }. Therefore circuit is \mathrm{\mathrm{C}-\mathrm{R}. }

\mathrm{\begin{aligned} & \tan \phi=\frac{\mathrm{X}_{\mathrm{C}}}{\mathrm{R}} \text { and } \tan \phi=\frac{\pi}{4} \\ \therefore \quad & \mathrm{X}_{\mathrm{C}}=\mathrm{R} \\ \Rightarrow \quad & \omega \mathrm{CR}=1 \\ \Rightarrow \quad & \mathrm{CR}=\frac{1}{\omega}=\frac{1}{100} \end{aligned} }

When,\mathrm{ \mathrm{R}=1 \mathrm{k} \Omega=10^3 \Omega, then \mathrm{C}=\frac{1}{10}=10^{-5} \mathrm{~F}=10 \mu \mathrm{F} }

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Rishi

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