Question:

An AC voltage V=20sin200πt V = 20 \sin 200 \pi t is applied to a series LCR circuit which drives a current I=10sin(200πt+π3) I = 10 \sin \left( 200 \pi t + \frac{\pi}{3} \right) . The average power dissipated is:

Updated On: Nov 19, 2024
  • 21.6W 21.6 \, \text{W}
  • 200W 200 \, \text{W}
  • 173.2W 173.2 \, \text{W}
  • 50W 50 \, \text{W}
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The Correct Option is D

Solution and Explanation

The average power dissipated P \langle P \rangle in an AC circuit with voltage V V and current I I is given by:

P=VrmsIrmscosϕ, \langle P \rangle = V_{\text{rms}} I_{\text{rms}} \cos \phi,

where Vrms V_{\text{rms}} is the root mean square (RMS) value of the voltage, Irms I_{\text{rms}} is the RMS value of the current, and ϕ \phi is the phase difference between the voltage and current.

The given voltage is V=20sin200πt V = 20 \sin 200 \pi t , so the peak voltage V0 V_0 is 20 V. The RMS value of the voltage Vrms V_{\text{rms}} is:

Vrms=V02=202=102V. V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = \frac{20}{\sqrt{2}} = 10\sqrt{2} \, \text{V}.

Similarly, the given current is I=10sin(200πt+π3) I = 10 \sin (200 \pi t + \frac{\pi}{3}) , so the peak current I0 I_0 is 10 A. The RMS value of the current Irms I_{\text{rms}} is:

Irms=I02=102=52A. I_{\text{rms}} = \frac{I_0}{\sqrt{2}} = \frac{10}{\sqrt{2}} = 5\sqrt{2} \, \text{A}.

The current leads the voltage by π3 \frac{\pi}{3} , meaning the phase difference ϕ \phi is:

ϕ=π3=60. \phi = \frac{\pi}{3} = 60^\circ.

Since ϕ=60 \phi = 60^\circ :

cosϕ=cos60=12. \cos \phi = \cos 60^\circ = \frac{1}{2}.

Now, substitute the values of Vrms V_{\text{rms}} , Irms I_{\text{rms}} , and cosϕ \cos \phi into the formula for average power:

P=VrmsIrmscosϕ=(102)(52)12. \langle P \rangle = V_{\text{rms}} I_{\text{rms}} \cos \phi = (10\sqrt{2})(5\sqrt{2}) \cdot \frac{1}{2}.

Simplify the expression:

P=(1052)12=50W. \langle P \rangle = (10 \cdot 5 \cdot 2) \cdot \frac{1}{2} = 50 \, \text{W}.

The average power dissipated in the circuit is:

50W. 50 \, \text{W}.
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