For ac circuit shown in figure, $R=100$ k$\Omega$ and $C=100$ pF and 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

JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Alternating Current

For ac circuit shown in figure,

R = 100

Ω

and

C = 100

pF and the phase difference between

V_{in}

and

(V_{B} - V_{A})

9 0^{\circ}

. The input signal frequency is

1 0^{x}

rad/sec, where 'x' is

Answer: 5

Along the left arm the source drives a resistor

R

followed by a capacitor

C

to ground, giving

V_{A}

; along the right arm it drives a capacitor followed by a resistor, giving

V_{B}

. Writing

x = ω R C

V_{A} = \frac{V _{in}}{1 + j x}, V_{B} = \frac{j x V _{in}}{1 + j x},

so that

V_{B} - V_{A} = \frac{j x - 1}{1 + j x} V_{in} .

The phase of this difference relative to

V_{in}

18 0^{\circ} - 2 tan^{- 1} x

. Equating it to

9 0^{\circ}

gives

tan^{- 1} x = 4 5^{\circ}

, i.e.

x = ω R C = 1

, so

ω = \frac{1}{R C} = \frac{1}{( 100 \times 1 0 ^{3} ) ( 100 \times 1 0 ^{- 12} )} = 1 0^{5} rad/s .

With the input frequency written as

1 0^{x}

rad/s, the value of

x

5

JEE Main · 2025 · 07 Apr 2025 · Shift 1 · Integer answer