Alternating Current

JEE Main level, numerical answer, medium

60 of 190

Question 60 of 190

For ac circuit shown in figure, R=100kΩ\mathrm{R}=100 \mathrm{k} \Omega and C=100pF\mathrm{C}=100 \mathrm{pF} and the phase difference between Vin \mathrm{V}_{\text {in }} and (VB−VA)\left(V_B-V_A\right) is 90∘90^{\circ}. The input signal frequency is 10xrad/sec10^x \mathrm{rad} / \mathrm{sec}, where ' x ' is ______

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Answer: 5


θ+θ=90∘;θ=45∘\theta+\theta=90^{\circ} ; \theta=45^{\circ}
tan⁡θ=XCR\tan \theta=\frac{\mathrm{X}_{\mathrm{C}}}{\mathrm{R}}
XC=R⇒1 WC=R\mathrm{X}_{\mathrm{C}}=\mathrm{R} \Rightarrow \frac{1}{\mathrm{~W}_{\mathrm{C}}}=\mathrm{R}
W=1RC=1100×103×100×10−12\mathrm{W}=\frac{1}{\mathrm{R}_{\mathrm{C}}}=\frac{1}{100 \times 10^3 \times 100 \times 10^{-12}}
=1012107=105=\frac{10^{12}}{10^7}=10^5

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