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Kelvi

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An inductor $20$ mH, a capacitor $100 \mu F$ and a resistor $50 \Omega$ are connected in series across a source of emf, $V = 10 \sin 314 t.$ The power loss in the circuit is

hard

Alternating Current

2018

physics

$2.74$ W

$0.43$ W

$0.79$ W

$1.13$ W

Explanation

To solve this problem, we need to calculate the power loss in an RLC series circuit. $\\$ Given: $\\$ • Inductance $L = 20$ mH $= 20 \times 10^{-3}$

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H

\\

• Capacitance

C = 100 \mu F = 100 \times 10^{-6}

F

\\

• Resistance

R = 50 \Omega \\

• Source voltage

V = 10 \sin 314t \\

The angular frequency

\omega

is given by the source:

\\ \omega = 314

rad/s

\\

The impedance

Z

of the RLC circuit is 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:

\\ X_L = \omega L = 314 \times 20 \times 10^{-3} = 6.28 \, \Omega \\ X_C = \frac{1}{\omega C} = \frac{1}{314 \times 100 \times 10^{-6}} = 31.85 \, \Omega \\

Substitute

X_L

and

X_C

into the impedance formula:

\\ Z = \sqrt{50^2 + (6.28 - 31.85)^2} \\ Z = \sqrt{50^2 + (-25.57)^2} \\ Z = \sqrt{2500 + 653.92} \\ Z = \sqrt{3153.92} \\ Z \approx 56.15 \, \Omega \\

The RMS voltage

V_{rms}

is given by:

\\ V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{10}{\sqrt{2}} = 7.07 \,

V

\\

The RMS current

I_{rms}

is:

\\ I_{rms} = \frac{V_{rms}}{Z} = \frac{7.07}{56.15} \\ I_{rms} \approx 0.126 \,

A

\\

The power loss (average power) in the resistor is given by:

\\ P = I_{rms}^2 \times R \\ P = (0.126)^2 \times 50 \\ P = 0.015876 \times 50 \\ P \approx 0.79 \,

W

\\

Therefore, the power loss in the circuit is

0.79 \,

W

. \\

This corresponds to Option 3.