JEE Main: Find Average Power in RLC Circuit | Easy Trick Explained ⚡

 

❓ Question

An inductor of reactance $X_L = 100~\Omega$, a capacitor of reactance $X_C = 50~\Omega$, and a resistor of resistance $R = 50~\Omega$ are connected in series with an AC source of 10 V, $f = 50~\text{Hz}$.

The average power dissipated by the circuit is $W$.
Find the value of $W$.

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🖼️ Question Image

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✍️ Step-by-Step Solution

Step 1 — Given data

$$R=50\mathrm{\Omega },X_L=100\mathrm{\Omega },X_C=50\mathrm{\Omega },V_{\text{rms}}=10\text{ V}\mathrm{.}$$

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Step 2 — Find net reactance

The total reactance in a series RLC circuit is the difference between inductive and capacitive reactances:

$$X=X_L-X_C=100-50=50\mathrm{\Omega }\mathrm{.}$$

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Step 3 — Find impedance

The total impedance $Z$ is given by

$$Z=\sqrt{R^2+X^2}=\sqrt{{50}^2+{50}^2}=\sqrt{5000}=70.71\mathrm{\Omega }\mathrm{.}$$

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Step 4 — Find current (rms value)

$$I_{\text{rms}}=\frac{V_{\text{rms}}}{Z}=\frac{10}{70.71}=0.1414\text{ A}\mathrm{.}$$

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Step 5 — Find power factor (cos φ)

Power factor for a series RLC circuit:

$$\cos \varphi =\frac{R}{Z}=\frac{50}{70.71}=\mathrm{0.707.}$$

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Step 6 — Average (true) power dissipated

Average power $P_{\text{avg}} = V_{\text{rms}} I_{\text{rms}} \cos\phi$.

Substitute values:

$$P_{\text{avg}}=10\times 0.1414\times 0.707=0.999\approx 1.0\text{ W}\mathrm{.}$$

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🧮 Image Solution

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✅ Final Answer

$$\boxed{{\displaystyle W=1.0\text{ watt}}}$$

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💡 Concept Recap

• Average (real) power in an AC circuit is dissipated only by the resistor (not by pure L or C).

• In an RLC circuit,

  $$P_{\text{avg}}=I_{\text{rms}}^{2}R=V_{\text{rms}}{I}_{\text{rms}}\cos \varphi \mathrm{.}$$

• Inductors and capacitors only store and release energy — they do not dissipate it.

• The power factor $\cos\phi$ measures how effectively the circuit converts electrical power into heat/work.

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