For ac circuit shown in figure, R=100 k \Omega and C=100 pF and the phase difference between V_ in and (V_B-V_A) is 90^\circ . The input signal frequency is 10^x rad / sec , where ' x ' is \underline\hspace2cm .

Alternating Current PYQ — JEE Main 2025 · 7 Apr · Shift 1 · Q73 with Solution

View written solutionFree

Correct answer: 5

Let the circuit between nodes $A$ and $B$ be the standard series $RC$ combination driven by $V_{in}$ , and the asked voltage be across the resistor-capacitor branch as $(V_B-V_A)$ . We use the given phase condition.

1. Condition for $90^\circ$ phase difference

In a series $RC$ circuit, the impedance is

Z = R - \frac{j}{\omega C}.

The phase angle of the total impedance is

\phi = -\tan^{-1}\!\left(\frac{1}{\omega RC}\right).

The current leads the input voltage by angle

\tan^{-1}\!\left(\frac{1}{\omega RC}\right).

For the voltage across $R$ and the voltage across $C$ , the phase difference is $90^\circ$ . Hence the given condition implies the operating frequency corresponds to the natural $RC$ angular frequency:

\omega = \frac{1}{RC}.

2. Substitute values

Given:

R = 100\,\text{k}\Omega = 10^5\,\Omega,

C = 100\,\text{pF} = 10^{-10}\,\text{F}.

Therefore,

RC = 10^5 \times 10^{-10} = 10^{-5}\,\text{s}.

So,

\omega = \frac{1}{RC} = \frac{1}{10^{-5}} = 10^5\,\text{rad/s}.

3. Compare with the form $10^x$

Given

\omega = 10^x\,\text{rad/s}.

Thus,

10^x = 10^5 \Rightarrow x=5.

4. Final answer

\boxed{x=5}

The derived answer matches the stored correct answer.

More from Alternating Current

1. Condition for

90^\circ

phase difference

In a series

RC

circuit, the impedance is

Z = R - \frac{j}{\omega C}.

The phase angle of the total impedance is

\phi = -\tan^{-1}\!\left(\frac{1}{\omega RC}\right).

The current leads the input voltage by angle

\tan^{-1}\!\left(\frac{1}{\omega RC}\right).

For the voltage across

R

and the voltage across

C

, the phase difference is

90^\circ

. Hence the given condition implies the operating frequency corresponds to the natural

RC

angular frequency:

\omega = \frac{1}{RC}.

2. Substitute values

Given:

R = 100\,\text{k}\Omega = 10^5\,\Omega,

C = 100\,\text{pF} = 10^{-10}\,\text{F}.

Therefore,

RC = 10^5 \times 10^{-10} = 10^{-5}\,\text{s}.

So,

\omega = \frac{1}{RC} = \frac{1}{10^{-5}} = 10^5\,\text{rad/s}.

Alternating Current question

Alternating Current question

1. Condition for $90^\circ$ phase difference

2. Substitute values

3. Compare with the form $10^x$

4. Final answer

More from Alternating Current

Alternating Current question

Alternating Current question

1. Condition for $90^\circ$ phase difference

2. Substitute values

3. Compare with the form $10^x$

4. Final answer

More from Alternating Current

1. Condition for 90∘90^\circ90∘ phase difference

2. Substitute values

3. Compare with the form 10x10^x10x

4. Final answer

More from Alternating Current

1. Condition for 90∘90^\circ90∘ phase difference

2. Substitute values

3. Compare with the form 10x10^x10x

4. Final answer

More from Alternating Current

1. Condition for $90^\circ$ phase difference

3. Compare with the form $10^x$

1. Condition for $90^\circ$ phase difference

3. Compare with the form $10^x$