Physics Q44 – JEE-MAIN 2026

Home Physics Alternating CurrentQuestion #6

An a.c. source of angular frequency

ω

is connected across a resistor

R

and a capacitor

C

in series. The current is observed as

I

. Now the frequency of the source is changed to

ω / 4

, (keeping the voltage unchanged) the current is found to be

I / 3

. The ratio of resistance to reactance at frequency

ω

Clue 1 of 3: RC Series Circuit Impedance

In an RC series circuit, the impedance $Z$ is given by $Z = \sqrt{R^{2} + X_{C}^{2}}$ , where $R$ is resistance and $X_{C}$ is capacitive reactance.

🧠 Full Solution Path

Step 1: Formulate Current Equations for Both Frequencies✦ Active

Let the voltage of the source be $V$ . At angular frequency $ω$ , the capacitive reactance is $X_{C} = \frac{1}{ω C}$ . The impedance is $Z_{1} = \sqrt{R^{2} + X_{C}^{2}}$ . The current is $I = \frac{V}{Z_{1}}$ . Thus, $I^{2} = \frac{V^{2}}{R^{2} + X_{C}^{2}}$ (Equation 1).

When the frequency is changed to $ω^{'} = \frac{ω}{4}$ , the new capacitive reactance is $X_{C}^{'} = \frac{1}{ω^{'} C} = \frac{1}{(ω / 4) C} = \frac{4}{ω C} = 4 X_{C}$ . The new impedance is $Z_{2} = \sqrt{R^{2} + (4 X_{C})^{2}}$ . The new current is $I^{'} = \frac{I}{3} = \frac{V}{Z_{2}}$ . Thus, $(\frac{I}{3})^{2} = \frac{V^{2}}{R^{2} + 16 X_{C}^{2}}$ (Equation 2).

Step 2: Solve for the Ratio of Resistance to Reactance○ Expand

Divide Equation 1 by Equation 2:

\frac{I^{2}}{(I / 3)^{2}} = \frac{\frac{V^{2}}{R^{2} + X_{C}^{2}}}{\frac{V^{2}}{R^{2} + 16 X_{C}^{2}}}

9 = \frac{R^{2} + 16 X_{C}^{2}}{R^{2} + X_{C}^{2}}

9 (R^{2} + X_{C}^{2}) = R^{2} + 16 X_{C}^{2}

9 R^{2} + 9 X_{C}^{2} = R^{2} + 16 X_{C}^{2}

8 R^{2} = 7 X_{C}^{2}

💡 Teacher's Secret Hint

Ensure careful algebraic manipulation when cross-multiplying and rearranging terms.

Step 3: Calculate the Final Ratio○ Expand

From the previous step, we have $8 R^{2} = 7 X_{C}^{2}$ . We need to find the ratio of resistance to reactance at frequency $ω$ , which is $\frac{R}{X_{C}}$ .

\frac{R^{2}}{X_{C}^{2}} = \frac{7}{8}

\frac{R}{X_{C}} = \sqrt{\frac{7}{8}}

This matches option 3.

💡 Teacher's Secret Hint

Remember to take the square root to find the ratio of R to X_C, not R-squared to X_C-squared.

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Physics Question 44 – JEE-MAIN 2026

Clue 1 of 3: RC Series Circuit Impedance

🧠 Full Solution Path