Question

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

Answer 1

To solve this question, we need to use the concept of power in AC circuit in which power is dependent on RMS values of voltage and current as well as the power factor which is the cosine of the phase difference between voltage and current. We will also use the relation between RMS values and peak values of voltage and current to find our final answer which is the phase difference between the applied voltage and circuit current.

Formulas used:
$P = {V_{rms}}{I_{rms}}\cos \phi$,
where $P$ is power, ${V_{rms}}$ is RMS value of voltage, ${I_{rms}}$ is RMS value of current and $\cos \phi$ is the power factor where $\phi$ is the phase difference between voltage and current
${P_p} = \dfrac{{{V_p}{I_p}}}{2}$,
where ${P_p}$ is the peak power, ${V_p}$ is the peak voltage, ${I_p}$ is the peak current
${V_{rms}} = \dfrac{{{V_p}}}{{\sqrt 2 }}$,
where ${V_{rms}}$ is RMS value of voltage and ${V_p}$ is the peak voltage
${I_{rms}} = \dfrac{{{I_p}}}{{\sqrt 2 }}$,
where ${I_{rms}}$ is RMS value of current and ${I_p}$ is the peak current

Complete step by step answer:
Here, it is given that a lamp consumes only $50\%$ of peak power. Therefore, we can say that
$P = \dfrac{{{P_p}}}{2}$
We know that
$P = {V_{rms}}{I_{rms}}\cos \phi$
Using formulas,
${V_{rms}} = \dfrac{{{V_p}}}{{\sqrt 2 }}$ and ${I_{rms}} = \dfrac{{{I_p}}}{{\sqrt 2 }}$ , and ${P_p} = \dfrac{{{V_p}{I_p}}}{2}$
$P = \dfrac{{{V_p}}}{{\sqrt 2 }} \times \dfrac{{{I_p}}}{{\sqrt 2 }} \times \cos \phi = {P_p}\cos \phi$
But , here it is given that
$P = \dfrac{{{P_p}}}{2}$
Therefore, we get
$\dfrac{{{P_p}}}{2} = {P_p}\cos \phi \\\ \Rightarrow \cos \phi = \dfrac{1}{2} \\\ \Rightarrow \phi = {\cos ^{ - 1}}\left( {\dfrac{1}{2}} \right) \\\ \therefore \phi = \dfrac{\pi }{3} \\\$
Thus, the phase difference between the applied voltage and circuit current when a lamp consumes only $50\%$ of peak power in an AC circuit is $\dfrac{\pi }{3}$ .

Hence, option B is the right answer.

Note: We have seen that power is dependent on RMS values of voltage and current as well as the power factor. There are two important cases: resistive circuit and capacitive circuit. In a resistive circuit, the circuit contains only pure resistance and hence there is no phase difference between voltage and current which means that $\phi = 0 \Rightarrow \cos \phi = 1$. Thus, there is maximum power dissipation in a resistive circuit.In capacitive circuit, the circuit contains only capacitor and hence the phase difference between voltage and current is ${90^0}$ which means that $\phi = \dfrac{\pi }{2} \Rightarrow \cos \phi = 0$. Thus, there is no power dissipation in the capacitive circuit.