(D) The given potential is $E = E_0 \sin(\omega t)$.The given current is $I = I_0 \sin(\omega t - \frac{\pi}{2})$.The phase difference $\phi$ between

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(D) The given potential is $E = E_0 \sin(\omega t)$.The given current is $I = I_0 \sin(\omega t - \frac{\pi}{2})$.The phase difference $\phi$ between the voltage and the current is $\phi = \omega t - (\omega t - \frac{\pi}{2}) = \frac{\pi}{2}$ or $90^{\circ}$.The average power consumption in an $AC$ circuit is given by the formula $P = E_{rms} I_{rms} \cos \phi$.Since $\phi = 90^{\circ}$,the power factor $\cos \phi = \cos(90^{\circ}) = 0$.Therefore,the power consumption $P = E_{rms} I_{rms} \times 0 = 0$.

Class 12 Physics Alternating Current Power in AC and Power Factor

Easy

If a current $I$ given by $I = I_0 \sin(\omega t - \frac{\pi}{2})$ flows in an $AC$ circuit across which an $AC$ potential of $E = E_0 \sin(\omega t)$ has been applied,then the power consumption $P$ in the circuit will be

A
$P = \frac{E_0 I_0}{\sqrt{2}}$
B
$P = \sqrt{2} E_0 I_0$
C
$P = \frac{E_0 I_0}{2}$
D
$P = 0$

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Alternating Current

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Power in AC and Power Factor

When a voltage $V = V_0 \cos \omega t$ is applied across a resistor of resistance $R$,the average power dissipated per cycle in the resistor is given by

Medium

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$A$ group of electric lamps having a total power rating of $1000 \text{ W}$ is supplied by an $AC$ voltage $E = 200 \sin(310t + 60^\circ)$. Then the $rms$ value of the circuit current is

Medium

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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 ............ $W$.

Easy

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In an $A.C.$ circuit containing $L$, $C$ and $R$ in series, the ratio of apparent power to the true power is ($Z=$ impedance of the circuit and $R=$ resistance).

MediumMHT CET 2026

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In an $AC$ circuit,the current is $i = 5 \sin(100t - \frac{\pi}{2}) \ A$ and the voltage is $e = 200 \sin(100t) \ V$. The power consumption in the circuit is (given $\cos 90^{\circ} = 0$): (in $W$)

MediumMHT CET 2023

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If a current $I$ given by $I = I_0 \sin(\omega t - \frac{\pi}{2})$ flows in an $AC$ circuit across which an $AC$ potential of $E = E_0 \sin(\omega t)$ has been applied,then the power consumption $P$ in the circuit will be

Similar Questions

When a voltage $V = V_0 \cos \omega t$ is applied across a resistor of resistance $R$,the average power dissipated per cycle in the resistor is given by

$A$ group of electric lamps having a total power rating of $1000 \text{ W}$ is supplied by an $AC$ voltage $E = 200 \sin(310t + 60^\circ)$. Then the $rms$ value of the circuit current is

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 ............ $W$.

In an $A.C.$ circuit containing $L$, $C$ and $R$ in series, the ratio of apparent power to the true power is ($Z=$ impedance of the circuit and $R=$ resistance).

In an $AC$ circuit,the current is $i = 5 \sin(100t - \frac{\pi}{2}) \ A$ and the voltage is $e = 200 \sin(100t) \ V$. The power consumption in the circuit is (given $\cos 90^{\circ} = 0$): (in $W$)

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