Question

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

Answer 1

Hint:First let us see about power-
Power may be described as the amount of work performed in unit time. The SI unit of power is Watt or joules per second.Average power may be described as the total energy consumed divided by total time taken. It can also be said as the average amount of work performed per unit time.There are actually two formulae of power:
Power = voltage × current × cosine of phase difference
$P = {V_{rms}} \times {I_{rms}} \times \cos \phi $
Also, power = work done/time taken
$P = W/t$
Here we shall use the first formula to find the phase difference.

Complete step by step answer:
Given,
A lamp consumes only $50\% $ of maximum power in an A.C. circuit.
We know that-
Multiplying the rms values of current and voltage, the average ac power is detected. By dividing the rms voltage by the impedance, Ohm’s rule for the rms value is discovered. A phase angle between the source voltage and the current is identified in an ac circuit, which can be determined by measuring the resistance.
Power = voltage × current × cosine of phase difference
$P = {V_{rms}} \times {I_{rms}} \times \cos \phi \\
= \dfrac{{{v_p}}}{{\sqrt 2 }} \times \dfrac{{{I_p}}}{{\sqrt 2 }} \times \cos \phi \\
= {P_{peak}}\cos \phi \\ $
According to question,
$P = {P_{peak}} \times \dfrac{1}{2} = {P_{peak}}\cos \phi \\
\cos \phi = \dfrac{1}{2} \\\phi = \dfrac{\pi }{3} = {60^ \circ } \\ $
Hence, option B is correct.

Additional information:
RMS means root mean square which is a way of describing an AC voltage or current quantity in ways that are technically similar to DC. It is often known as the value of an AC voltage or current equivalent or DC equivalent. The RMS value for a sinusoidal wave is approximately $0.707$ of its peak.

Note:Here we have to equate the two peak powers so we can get the phase difference. Also we have to multiply the cosine of angle to get the peak power.