Question

The instantaneous values of alternating current and voltages in a circuit are given as
$i = \dfrac{1}{{\sqrt 2 }}\sin \left( {100\pi t} \right)ampere$
$e = \dfrac{1}{{\sqrt 2 }}\sin \left( {100\pi t + \dfrac{\pi }{3}} \right)volt$
The average power in watts consumed in the circuit is:
(A) $\dfrac{{\sqrt 3 }}{4}$
(B) $\dfrac{1}{2}$
(C) $\dfrac{1}{8}$
(D) $\dfrac{1}{4}$

Answer 1

The root means square value of both voltage and current is their peak value divided by root 2. Average power is given by the product of the root mean square value of the current and voltage and the power factor.
Formula used: In this solution we will be using the following formulae;
${P_{ave}} = {e_{rms}}{i_{rms}}\cos \theta$ where ${P_{ave}}$ is the average power consumed in an ac circuit, $e$ is the voltage and ${e_{rms}}$ signifies the root mean square value of the voltage, and ${i_{rms}}$ similarly signifies the root mean square value of the current. $\theta$ is the phase angle between the voltage and current.
${e_{rms}} = \dfrac{{{e_p}}}{{\sqrt 2 }}$ where ${e_p}$ is the peak value of the voltage.
${i_{rms}} = \dfrac{{{i_p}}}{{\sqrt 2 }}$ where ${i_p}$ is the peak value of the current.

Complete Step-by-Step Solution:
Observing the equation given, it can be found that the peak value of the current is
${i_p} = \dfrac{1}{{\sqrt 2 }}$ and peak value of voltage is
${e_p} = \dfrac{1}{{\sqrt 2 }}$
The root mean square values can be given as
${e_{rms}} = \dfrac{{{e_p}}}{{\sqrt 2 }}$ where ${e_p}$ is the peak value of the voltage.
$\Rightarrow {e_{rms}} = \dfrac{1}{{\sqrt 2 }} \times \dfrac{1}{{\sqrt 2 }} = \dfrac{1}{2}V$
And similarly for current, we have
${i_{rms}} = \dfrac{1}{{\sqrt 2 }} \times \dfrac{1}{{\sqrt 2 }} = \dfrac{1}{2}A$
Now, the average power consumed in an ac circuit can be given as
${P_{ave}} = {e_{rms}}{i_{rms}}\cos \theta$ where ${P_{ave}}$ is the average power consumed in an ac circuit, ${e_{rms}}$ signifies the root mean square value of the voltage, and ${i_{rms}}$ similarly signifies the root mean square value of the current. $\theta$ is the phase angle between the voltage and current.
Hence by inserting known values, we have
${P_{ave}} = \dfrac{1}{2} \times \dfrac{1}{2}\cos \left( {\dfrac{\pi }{3}} \right)$
${P_{ave}} = \dfrac{1}{8}{\text{W}}$

Hence, the correct option is C

Note: For understanding, the term $\cos \theta$ is called the power factor. We can observe that if the angle is $\dfrac{\pi }{4}rad$ or 90 degrees, the average power consumed would be zero. This is exactly the case of a circuit with only an inductor. In application, a purely inductive circuit is used to store electricity for a short while as it ideally does not consume the electricity given to it. Hence, the same electricity can be made to go around it a few times. However, in reality, pure inductive circuit is impossible, and hence the electricity will eventually be consumed.