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(B) The instantaneous power in an $AC$ circuit is given by $P = V_{rms} I_{rms} \cos \phi = \frac{V_0 I_0}{2} \cos \phi$.Since the peak power is defin

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$\frac{\pi}{3}$

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(B) The instantaneous power in an $AC$ circuit is given by $P = V_{rms} I_{rms} \cos \phi = \frac{V_0 I_0}{2} \cos \phi$.Since the peak power is defined as $P_{peak} = V_0 I_0$,the average power is $P = \frac{1}{2} P_{peak} \cos \phi$.Given that the lamp consumes $50\%$ of the peak power,we have $P = 0.5 P_{peak} = \frac{1}{2} P_{peak}$.Equating the two expressions: $\frac{1}{2} P_{peak} = \frac{1}{2} P_{peak} \cos \phi$.This simplifies to $\cos \phi = 1$,which implies $\phi = 0$. However,checking the standard interpretation of power consumption relative to peak power in this context: $P_{avg} = V_{rms} I_{rms} \cos \phi = \frac{1}{2} V_0 I_0 \cos \phi$. If $P_{avg} = \frac{1}{2} P_{peak}$,then $\frac{1}{2} V_0 I_0 \cos \phi = \frac{1}{2} V_0 I_0$,which gives $\cos \phi = 1$. Re-evaluating the problem statement: If the power consumed is $50\%$ of the maximum possible power $(V_{rms} I_{rms})$,then $\cos \phi = 0.5$,which leads to $\phi = \frac{\pi}{3}$.

$A$ lamp consumes only $50\%$ of peak power in an $AC$ circuit. What is the phase difference between the applied voltage and the circuit current?

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