Question

A lamp consumes only $50\% $of peak power in an A.C. circuit. What is the phase difference between the applied voltage and the circuit current?
A. $\dfrac{\pi }{6}$
B. $\dfrac{\pi }{3}$
C. $\dfrac{\pi }{4}$
D. $\dfrac{\pi }{2}$

Answer 1

Hint: Concept of power in an A.C. circuit.
\[{I_{rms}} = \dfrac{{{I_0}}}{{\sqrt 2 }},\;\;{V_{rms}} = \dfrac{{{V_0}}}{{\sqrt 2 }}\]
$P = V{}_{rms}\;{I_{rms}}\;\cos \phi $

Complete step by step answer:
Power in an A.C. circuit – The rate at which electric energy is consumed in an electric circuit is called its power.
For an A.C. circuit, we generally define instantaneous power which is the product of instantaneous voltage and instantaneous current.
Let us suppose in an A.C. circuit, the voltage and the current at any instant is
$
  V = V{}_0\;\sin \;\omega t \\
  I = {I_0}\;\sin \;(\omega t - \phi ) \\
 $
Where ${V_0}$ and ${I_0}$are peak values of voltage and current and $\phi $ is the phase difference between voltage and current.
So, instantaneous power will be given by
$
  P = VI \\
   = \left( {{V_0}\;\sin \;\omega t} \right)\;\;\;\;\left( {\;{I_0}\;\sin \;(\omega t - \phi )} \right) \\
   = \left( {{V_0}\;\sin \;\omega t} \right)\;\;\;\;\left( {\;\sin \;(\omega t - \phi )} \right) \\
   = \dfrac{{{V_0}{I_0}}}{2}\;[2\;\sin \;\omega t.\;\sin \;(\omega t - \phi )] \\
   = \dfrac{{{V_0}{I_0}}}{2}\;[\cos \phi \; - \;\cos \;(2\omega t - \phi )] \\
 $
$ - (as\;2\sin \;A\;\;\sin \;B\; = \;\cos \;(A - B) - \cos \,\;(A + B))$
Average power dissipated per cycle = Average of $\dfrac{{{V_0}{I_0}}}{2}\;[\cos \phi - \cos \;(2\omega t - \phi )]$
average $[\cos \;(2\omega t - \phi )] = 0$
So,
\[
  {P_{av}} = \dfrac{{{V_0}{I_0}}}{2}\;\cos \phi \\
  {P_{av}}\; = \;{P_{Peak}}\;\cos \phi \\
 \]
Where [PPeak is the peak power and PPeak $ = {V_0}{I_0}$]
So, Pav $ = $PPeak $\cos \phi $ ….(1) ($\phi $ is phase difference)
Now, according to question
\[
  {P_{av}} = 50\% \;of\;{P_{Peak}} \\
  {P_{av}} = \dfrac{{50}}{{100}}\; \times \;{P_{Peak}} \\
 \]
\[{P_{av}} = \dfrac{1}{2}\; \times \;{P_{Peak}}\] ….(2)
From $(1)$ and $(2)$, we have
$
  \dfrac{1}{2}{P_{Peak}} = {P_{Peak}}\,\cos \phi \\
   \Rightarrow \,\cos \phi = \dfrac{1}{2} \\
   \Rightarrow \cos \phi = \cos \dfrac{\pi }{3} \\
   \Rightarrow \phi = \dfrac{\pi }{3} \\
 $
So, the phase difference between applied voltage and the circuit current is $\dfrac{\pi }{3}$.

So, the correct answer is “Option B”.

Note:
 Peak power $ = \dfrac{{{V_0}{I_0}}}{2}$ not $ V_0I_0 $
1. For pure resistive circuit,
$
  \phi = 0 \\
   \Rightarrow \cos \phi = 1 \\
 $
So, ${P_{av}} = \dfrac{{{V_0}{I_0}}}{{\sqrt 2 }}(1) = \dfrac{{{V_0}{l_0}}}{{\sqrt 2 }}$
2. For pure capacitive or pure inductive circuit,
\[
  \phi = {90^ \circ } \\
   \Rightarrow \cos \phi = \cos \;90 = 0 \\
 \]
So, ${P_{av}} = \dfrac{{{V_0}{I_0}}}{{\sqrt 2 }}(0)$

So, the average power consumed in a purely inductive or purely capacitive circuit over a complete cycle is zero.