Question

The equation of A.C. voltage is $E = 220\sin \left( {\omega t + \pi /6} \right)$ and the A.C. current is $I = 10\sin \left( {\omega t - \pi /6} \right)$ . The average power dissipated is:
(A) $150{\text{W}}$
(B) $550{\text{W}}$
(C) $250{\text{W}}$
(D) $50{\text{W}}$

Answer 1

Hint : We have to find out the value of the RMS current and the RMS voltage from the given equations of the AC voltage and the AC current. We also have to find out the value of the phase difference between the AC voltage and the AC current. Then we need to put these values in the formula for the power in an AC circuit to get the final answer.

Formula used: The formula used to solve this question is given by
$P = {V_{rms}}{I_{rms}}\cos \varphi$ , here $P$ is the power dissipated in an AC circuit, ${V_{rms}}$ is the root mean square value of the voltage, ${I_{rms}}$ is the root mean square current, and $\theta$ is the phase difference between the AC voltage and current.

Complete step by step answer
The equation of the A.C. voltage given is
$E = 220\sin \left( {\omega t + \pi /6} \right)$
So the amplitude of the voltage is ${E_0} = 220{\text{V}}$ .
Therefore the RMS value of the voltage becomes
${V_{rms}} = \dfrac{{{E_0}}}{{\sqrt 2 }}$
$\Rightarrow {V_{rms}} = \dfrac{{220}}{{\sqrt 2 }}{\text{V}}$ …………...(1)
Also, the equation of the A.C. current given is
$I = 10\sin \left( {\omega t - \pi /6} \right)$
So the amplitude of the current is ${I_0} = 10{\text{A}}$ .
Therefore the RMS value of current is
${I_{rms}} = \dfrac{{{I_0}}}{{\sqrt 2 }}$
$\Rightarrow {I_{rms}} = \dfrac{{10}}{{\sqrt 2 }}{\text{A}}$ …………...(2)
The phase of the voltage, from the equation given, is
${\varphi _1} = \omega t + \pi /6$ …………...(3)
Also, the phase of the current is
${\varphi _2} = \omega t - \pi /6$ …………...(4)
So the phase difference between the voltage and the current is
$\varphi = {\varphi _1} - {\varphi _2}$
From (3) and (4)
$\varphi = \omega t + \pi /6 - \left( {\omega t - \pi /6} \right)$
$\Rightarrow \varphi = \pi /3$ …………...(5)
Now, we know that the power dissipated in an AC circuit is given by
$P = {V_{rms}}{I_{rms}}\cos \varphi$
Putting (1), (2) and (5) we get
$P = \dfrac{{220}}{{\sqrt 2 }} \times \dfrac{{10}}{{\sqrt 2 }} \times \cos \left( {\pi /3} \right)$
On solving we finally get
$P = 550{\text{W}}$
Thus, the power dissipated is equal to $550{\text{W}}$ .
Hence, the correct answer is option B.

Note
The RMS value of the voltage in the AC circuit is that value of the constant DC voltage which produces the same power as the DC voltage produces. So we calculate the RMS values for calculating the power. Also, do not forget the phase difference term which appears in the formula for the power.