A lamp consumes only 50% of peak power in an | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$\frac{\pi}{3}$

Vedclass · Answer

(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?

Similar Questions

In a series $LCR$ circuit,the inductive reactance $(X_{L})$ is $10\, \Omega$ and the capacitive reactance $(X_{C})$ is $4\, \Omega$. The resistance $(R)$ in the circuit is $6\, \Omega$. The power factor of the circuit is :

The $r.m.s.$ current in an $ac$ circuit is $2 \ A$. If the wattless current is $\sqrt{3} \ A$,the power factor is:

In an $A.C.$ circuit,the current flowing in an inductance is $I = 5\,\sin\,(100\,t - \pi/2)\,A$ and the potential difference is $V = 200\,\sin\,(100\,t)\,V$. The power consumption is equal to......$W$.

When a voltage $V_{s} = 200 \sqrt{2} \sin(\omega t + 15^{\circ})$ is applied to an $AC$ circuit,the current in the circuit is found to be $I = 2 \sin(\omega t + 45^{\circ})$. The average power consumed in the circuit is:

$A$ coil of inductive reactance $31\,\Omega$ has a resistance of $8\,\Omega$. It is placed in series with a capacitor of capacitive reactance $25\,\Omega$. The combination is connected to an $A$.$C$. source of $110\,V$. The power factor of the circuit is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module