For ac circuit shown in figure, R = 100 kΩ and C = 100 pF and the phase difference between V in and (VB - VA) is 90°. The input signal frequency is 10x rad/sec, where ' x ' is __________. Numerical answer type. Work it out, then open the answer.

Answer: 5. Left branch ( R then C): VA = Vin1 + jω RC. Right branch ( C then R): VB = jω RC Vin1 + jω RC. VB - VA = Vin(jω RC - 1)/(jω RC + 1), magnitude Vin, phase π - 2tan ⁻¹(ω RC). Phase = 90° ⇒ tan ⁻¹(ω RC) = 45° ⇒ ω RC = 1 ω = 1/RC = 110⁵ × 10⁻¹⁰ = 10⁵ rad/s So x = 5.

For ac circuit shown in figure, R = 100 kΩ and C = 100 pF and the… | JEE Main 2025 (7 Apr, Shift 1)

Q19JEE Main 2025 · 7 APR · SHIFT 1Difficult

RL and RC Series Circuits

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

Numerical answer type. Work it out, then open the answer.

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Answer: 5

Left branch ( $R$ then $C$ ): $V_A = \frac{V_{in}}{1 + j\omega RC}$ . Right branch ( $C$ then $R$ ): $V_B = \frac{j\omega RC\,V_{in}}{1 + j\omega RC}$ .
$V_B - V_A = V_{in}\frac{j\omega RC - 1}{j\omega RC + 1}$ , magnitude $V_{in}$ , phase $\pi - 2\tan^{-1}(\omega RC)$ .
Phase $= 90^\circ \Rightarrow \tan^{-1}(\omega RC) = 45^\circ \Rightarrow \omega RC = 1$
$\omega = \frac{1}{RC} = \frac{1}{10^5 \times 10^{-10}} = 10^5$ rad/s
So $x = 5$ .

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