JEE Physics 2025 Question with Solution

Given: $R = 100 \times 10^3 \, \Omega$, $C = 100 \times 10^{-12} \, \text{F}$, and the phase difference between $V_{in}$ and $(V_B - V_A)$ is $90^\circ$.

Find: The value of $x$ if the input angular frequency is written as $10^x$ rad/sec.

Using the phase condition stated in the solution for the RC circuit:

$$\omega RC = 1$$

Now substitute the given values:

$$\omega \times (100 \times 10^3) \times (100 \times 10^{-12}) = 1$$

So,

$$\omega \times 10^{-5} = 1$$

Hence,

$$\omega = 10^5 \, \text{rad/sec}$$

Therefore, in the form $10^x$ rad/sec, we get $x = 5$.

Given: Two identical RC branches are connected as shown, with midpoint potentials $V_A$ and $V_B$. Also, $R = 100 \times 10^3 \, \Omega$ and $C = 100 \times 10^{-12} \, \text{F}$.

Find: The exponent $x$ in the angular frequency $10^x$ rad/sec.

As stated in the extracted solution, the branch phase angles are equal in magnitude and opposite in direction. Therefore, if the phase difference between $V_{in}$ and $(V_B - V_A)$ is $90^\circ$, each branch contributes $45^\circ$.

For an RC branch at $45^\circ$, the resistive and capacitive reactances are equal:

$$R = X_C$$

Using

$$X_C = \frac{1}{\omega C}$$

we get

$$R = \frac{1}{\omega C}$$

So,

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

Now substitute the values:

$$RC = (100 \times 10^3)(100 \times 10^{-12}) = 10^{-5}$$

Thus,

$$\omega = \frac{1}{10^{-5}} = 10^5 \, \text{rad/sec}$$

Therefore, the required exponent is 5.

Given: $R = 100 \times 10^3 \, \Omega$ and $C = 100 \times 10^{-12} \, \text{F}$.

Find: $x$ in $10^x$ rad/sec.

When the phase condition leads to $\omega RC = 1$, the angular frequency is immediately:

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

Now,

$$RC = 10^5 \times 10^{-10} = 10^{-5}$$

Therefore,

$$\omega = 10^5 \, \text{rad/sec}$$

So the frequency is of the form $10^x$ rad/sec with $x = 5$.

This shortcut works because the required phase condition fixes the circuit exactly at the point where the RC time constant satisfies $\omega \tau = 1$ with $\tau = RC$.