In the given circuit, rms value of current (Irms) through the res… — Solution | JEE Physics

MCQEasyJEE 2023LCR Circuits & Resonance

In the given circuit, rms value of current ( $I_{rms}$ ) through the resistor $R$ is:

$Series AC circuit containing an inductor with $$X_L = 200 \, \Omega$$, a capacitor with $$X_C = 100 \, \Omega$$, a resistor with $$R = 100 \, \Omega$$, and source voltage $$V_{rms} = 200\sqrt{2} \, \text{V}$$.$

A
$2 \, \text{A}$
B
$\frac{1}{2} \, \text{A}$
C
$20 \, \text{A}$
D
$2\sqrt{2} \, \text{A}$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: A series $RLC$ circuit has $X_L = 200 \, \Omega$ , $X_C = 100 \, \Omega$ , $R = 100 \, \Omega$ , and $V_{rms} = 200\sqrt{2} \, \text{V}$ .

Find: The rms current through the resistor $R$ .

For a series $RLC$ circuit, the impedance is

Z = \sqrt{R^2 + (X_L - X_C)^2}

Substituting the values,

Z = \sqrt{100^2 + (200 - 100)^2}

= 100\sqrt{2} \, \Omega

Now,

I_{rms} = \frac{V_{rms}}{Z} = \frac{200\sqrt{2}}{100\sqrt{2}}

= 2 \, \text{A}

Therefore, the rms current through the resistor is $2 \, \text{A}$ . The correct option is A.

The solution labels the option as C, but its own working gives $2 \, \text{A}$ , which matches option A in the listed options.

Common mistakes

Using $X_L + X_C$ instead of $X_L - X_C$ for net reactance. In a series $RLC$ circuit, inductive and capacitive reactances oppose each other. Use $|X_L - X_C|$ before combining with $R$ .
Taking impedance as $R + (X_L - X_C)$ directly. Resistance and reactance combine vectorially, so the correct relation is $Z = \sqrt{R^2 + (X_L - X_C)^2}$ .
Using peak voltage in place of rms voltage. The question already gives $V_{rms}$ , so current must be found from $I_{rms} = \frac{V_{rms}}{Z}$ without any extra $\sqrt{2}$ conversion.

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Answer

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