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  • A resistor R is connected in series to an alternating voltage source  

A resistor R is connected in series to an alternating voltage source V_s =V_0 \sin wt .  What will be the power dissipated in the resistor when the circuit is closed 

Option: 1

\frac{V_{0}^{2}}{R}


Option: 2

\frac{V_{0}^{2}}{2R}


Option: 3

\frac{V_{0}^{2}}{4R}


Option: 4

0 


Answers (1)

best_answer

As we have learned

Power in AM waves -

The power dissipated in any circuit.

P=\frac{ V^{2}_{rms}}{R}

- wherein

V_{rms} is the root mean square value

R = Resistance

 

 The power dissipated in resistor r = V_{rms}^{2}/ R

we shall have to calculate V_{rms} for a sinusoidal waveform V_{s = V_0 \sin wt

 V_{rms}= \frac{\int_{0}^{T} V_{s}^{2}dt}{\int_{0}^{T}dt}

= \frac{V_{0}^{2}[\frac{t}{2}+\frac{\sin 2wt}{2w}]_{0}^{T}}{T}

\frac{V_{0}^{2}}{2}

Hence power dissipated = = \frac{V_{0}^{2}}{2R}

 

 

 

 

 

 

 

 

Posted by

mansi

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