For the AC circuit shown in the figure,R = 10 | vedclass

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(B) Let the input voltage be $V_{	ext{in}}$. The circuit consists of two potential dividers in parallel. For the left branch,the voltage at point $A$

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$5$

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(B) Let the input voltage be $V_{	ext{in}}$. The circuit consists of two potential dividers in parallel. For the left branch,the voltage at point $A$ with respect to the bottom node is $V_A = V_{	ext{in}} \cdot \frac{-jX_C}{R - jX_C}$. For the right branch,the voltage at point $B$ with respect to the bottom node is $V_B = V_{	ext{in}} \cdot \frac{R}{R - jX_C}$. Thus,the potential difference $(V_B - V_A)$ is given by: $V_B - V_A = V_{	ext{in}} \cdot \frac{R + jX_C}{R - jX_C}$. Let $Z = R - jX_C$. Then $V_B - V_A = V_{	ext{in}} \cdot \frac{R + jX_C}{R - jX_C}$. The phase of $(V_B - V_A)$ relative to $V_{	ext{in}}$ is the phase of $\frac{R + jX_C}{R - jX_C}$. Let $	an 	heta = \frac{X_C}{R}$. Then the phase of $(R + jX_C)$ is $	heta$ and the phase of $(R - jX_C)$ is $-	heta$. The phase of the ratio is $	heta - (-	heta) = 2	heta$. Given the phase difference is $90^{\circ}$,we have $2	heta = 90^{\circ}$,so $	heta = 45^{\circ}$. Therefore,$	an 45^{\circ} = \frac{X_C}{R} \implies 1 = \frac{1}{\omega RC} \implies \omega = \frac{1}{RC}$. Substituting the values: $R = 10^5 \ \Omega$,$C = 100 	imes 10^{-12} \ F = 10^{-10} \ F$. $\omega = \frac{1}{10^5 	imes 10^{-10}} = \frac{1}{10^{-5}} = 10^5 \ rad/sec$. Comparing with $10^x$,we get $x = 5$.

For the $AC$ circuit shown in the figure,$R = 100 \ k\Omega$ and $C = 100 \ pF$. The phase difference between $V_{\text{in}}$ and $(V_B - V_A)$ is $90^{\circ}$. The input signal frequency is $10^x \ rad/sec$,where $x$ is . . . . . . .

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