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Average Power in RLC Series Circuit Calculation

Learn how to calculate average power in an RLC series circuit using resistance and reactance values. This method uses impedance and power factor for..

 

❓ Question

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

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


đź–Ľ️ Question Image

An inductor of reactance 100 Ω, a capacitor of reactance 50 Ω, and a resistor of resistance 50 Ω are connected in series with an AC source of 10 V, 50 Hz. Average power dissipated by the circuit is W.


✍️ Short Explanation

This problem is based on:

👉 Series LCR circuit
👉 AC impedance
👉 Average power in AC.

Main idea:

Average power in AC circuit:

P=VIcosϕ\boxed{ P=VI\cos\phi }

or directly:

P=I2R\boxed{ P=I^2R }

where current is:

I=VZI=\frac{V}{Z}

and impedance:

Z=R2+(XLXC)2\boxed{ Z=\sqrt{R^2+(X_L-X_C)^2} }

đź”· Step 1 — Calculate Net Reactance đź’Ż

Given:

XL=100ΩX_L=100\Omega
XC=50ΩX_C=50\Omega

Net reactance:

X=XLXCX=X_L-X_C
=10050=100-50
=50Ω=50\Omega

đź”· Step 2 — Find Impedance

Z=R2+X2Z=\sqrt{R^2+X^2}

Substitute:

=502+502= \sqrt{50^2+50^2}
=502 Î©= 50\sqrt2\ \Omega

đź”· Step 3 — Find Current

I=VZI=\frac{V}{Z}
=10502= \frac{10}{50\sqrt2}
=152 A= \frac1{5\sqrt2}\text{ A}

đź”· Step 4 — Calculate Average Power

Use:

P=I2RP=I^2R
=(152)2×50= \left(\frac1{5\sqrt2}\right)^2 \times50
=150×50= \frac1{50}\times50
=1 W=1\text{ W}

đź”· Step 5 — JEE Trap Alert 🚨

❌ Reactances directly add kar dena

❌ Average power formula me total impedance use kar lena instead of resistance

❌ RMS values confuse kar dena

Remember:

Only resistor dissipates average power.


✅ Final Answer

1 W\boxed{ 1\text{ W} }


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