Physics Q50 – JEE-MAIN 2026

Home Physics Alternating CurrentQuestion #5

An inductor of

10 mH

, capacitor of

0.1 µF

and a resistor of

100 Ω

are connected in series across an a.c power supply

220 V

70 Hz

. The power factor of the given circuit is

0.5

. The difference in the inductive reactance and capacitance reactance is

\sqrt{3} α Ω

. The value of

α

is _______.

Enter Calculated Numerical Value:

Clue 1 of 3: Power Factor Definition

The power factor ( $\cos ϕ$ ) in an AC circuit relates the resistance ( $R$ ) to the total impedance ( $Z$ ).

🧠 Full Solution Path

Step 1: Calculate Impedance from Power Factor✦ Active

The power factor of an AC series RLC circuit is given by $\cos ϕ = \frac{R}{Z}$ , where $R$ is the resistance and $Z$ is the total impedance. Given $R = 100 Ω$ and $\cos ϕ = 0.5$ .

Z = \frac{R}{\cos ϕ} = \frac{100 Ω}{0.5} = 200 Ω

Step 2: Determine Difference in Reactances○ Expand

The impedance $Z$ of a series RLC circuit is also given by $Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$ , where $X_{L}$ is the inductive reactance and $X_{C}$ is the capacitive reactance. We can find the magnitude of the difference in reactances, $| X_{L} - X_{C} |$ .

Z^{2} = R^{2} + (X_{L} - X_{C})^{2}

(X_{L} - X_{C})^{2} = Z^{2} - R^{2}

(X_{L} - X_{C})^{2} = (200 Ω)^{2} - (100 Ω)^{2}

(X_{L} - X_{C})^{2} = 40000 - 10000 = 30000 Ω^{2}

| X_{L} - X_{C} | = \sqrt{30000} = \sqrt{10000 \times 3} = 100 \sqrt{3} Ω

💡 Teacher's Secret Hint

Note that the values of L, C, V, and f are distractors and are not needed for this specific calculation.

Step 3: Find the Value of α○ Expand

We are given that the difference in the inductive reactance and capacitance reactance is $\sqrt{3} α Ω$ . Comparing this with our calculated value:

\sqrt{3} α = 100 \sqrt{3}

α = 100

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Physics Question 50 – JEE-MAIN 2026

Clue 1 of 3: Power Factor Definition

🧠 Full Solution Path