(D) The average power dissipated in an $AC$ circuit is given by the formula: $P_{avg} = V_{rms} I_{rms} \cos \phi$. Here, $V_{rms} = \frac{V_0}{\sqrt{

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(D) The average power dissipated in an $AC$ circuit is given by the formula: $P_{avg} = V_{rms} I_{rms} \cos \phi$. Here, $V_{rms} = \frac{V_0}{\sqrt{2}}$ and $I_{rms} = \frac{I_0}{\sqrt{2}}$. The phase difference $\phi$ between the voltage and current is $\frac{\pi}{3}$ (or $60^{\circ}$). Substituting these values into the formula: $P_{avg} = \left(\frac{V_0}{\sqrt{2}}\right) \left(\frac{I_0}{\sqrt{2}}\right) \cos 60^{\circ}$. $P_{avg} = \frac{V_0 I_0}{2} \times 0.5$. $P_{avg} = \frac{V_0 I_0}{2} \times \frac{1}{2} = \frac{V_0 I_0}{4}$.

Class 12 Physics Alternating Current Power in AC and Power Factor

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In a series $LCR$ circuit, the alternating e.m.f. and current are given by the equations $V = V_0 \sin(\omega t)$ and $I = I_0 \sin(\omega t + \frac{\pi}{3})$ respectively. The average power dissipated in the circuit over one cycle of a.c. is $(\cos 60^{\circ} = 0.5)$.

MHT CET 2026 All MHT CET

A
zero
B
$\frac{V_0 I_0}{2}$
C
$\frac{\sqrt{3}}{2} V_0 I_0$
D
$\frac{V_0 I_0}{4}$

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Alternating Current

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Power in AC and Power Factor

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For an $AC$ circuit,the potential difference and current are given by $V = 10 \sqrt{2} \sin \omega t$ (in $V$) and $I = 2 \sqrt{2} \cos \omega t$ (in $A$) respectively. The power dissipated in the instrument is ............ $W$.

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$A$ 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:

MediumAIPMT 2006

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In an $ac$ circuit,the current is given by $i = 5 \sin \left( 100 t - \frac{\pi}{2} \right)$ and the $ac$ potential is $V = 200 \sin (100 t) \text{ volts}$. Then the power consumption is $....... \text{ watts}$.

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The potential differences across the resistance,capacitance,and inductance are $80 \, V$,$40 \, V$,and $100 \, V$ respectively in an $L-C-R$ circuit. The power factor of this circuit is:

MediumNEET 2016

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In an $AC$ circuit,$I = 5 \sin \left( 100t - \frac{\pi}{2} \right)$ and $V = 200 \sin 100t$. The power consumed is

Medium

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In a series $LCR$ circuit, the alternating e.m.f. and current are given by the equations $V = V_0 \sin(\omega t)$ and $I = I_0 \sin(\omega t + \frac{\pi}{3})$ respectively. The average power dissipated in the circuit over one cycle of a.c. is $(\cos 60^{\circ} = 0.5)$.

Similar Questions

For an $AC$ circuit,the potential difference and current are given by $V = 10 \sqrt{2} \sin \omega t$ (in $V$) and $I = 2 \sqrt{2} \cos \omega t$ (in $A$) respectively. The power dissipated in the instrument is ............ $W$.

$A$ 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:

In an $ac$ circuit,the current is given by $i = 5 \sin \left( 100 t - \frac{\pi}{2} \right)$ and the $ac$ potential is $V = 200 \sin (100 t) \text{ volts}$. Then the power consumption is $....... \text{ watts}$.

The potential differences across the resistance,capacitance,and inductance are $80 \, V$,$40 \, V$,and $100 \, V$ respectively in an $L-C-R$ circuit. The power factor of this circuit is:

In an $AC$ circuit,$I = 5 \sin \left( 100t - \frac{\pi}{2} \right)$ and $V = 200 \sin 100t$. The power consumed is

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