(C) The power factor of an $LCR$ circuit is given by $\cos \phi = \frac{R}{Z}$.The impedance $Z$ of the series $LCR$ circuit is calculated as:$Z = \sq

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(C) The power factor of an $LCR$ circuit is given by $\cos \phi = \frac{R}{Z}$.The impedance $Z$ of the series $LCR$ circuit is calculated as:$Z = \sqrt{R^2 + (X_L - X_C)^2}$Given values are $R = 6\, \Omega$,$X_L = 10\, \Omega$,and $X_C = 4\, \Omega$.Substituting these values:$Z = \sqrt{6^2 + (10 - 4)^2}$$Z = \sqrt{36 + 6^2}$$Z = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}\, \Omega$Now,calculating the power factor:$\cos \phi = \frac{R}{Z} = \frac{6}{6\sqrt{2}} = \frac{1}{\sqrt{2}}$

Class 12 Physics Alternating Current Power in AC and Power Factor

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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 :

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A
$\frac{1}{2}$
B
$\frac{1}{2\sqrt{2}}$
C
$\frac{1}{\sqrt{2}}$
D
$\frac{\sqrt{3}}{2}$

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Alternating Current

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Power in AC and Power Factor

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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

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In a series $LCR$ circuit,an alternating emf $(v)$ and current $(i)$ are given by the equations $v = v_{0} \sin \omega t$ and $i = i_{0} \sin \left(\omega t + \frac{\pi}{3}\right)$. The average power dissipated in the circuit over a cycle of $AC$ is:

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The power dissipated in a resistor of resistance $R$ when an alternating current with a peak value of $I_p$ flows through it is:

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The current and voltage functions in an $AC$ circuit are:
$I = 100 \sin(100t) \, mA$
$V = 100 \sin(100t + \frac{\pi}{3}) \, V$
The power dissipated in the circuit is ...... $W$.

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Assertion $(A):$ If the frequency of the applied $AC$ is doubled, then the power factor of a series $R-L$ circuit decreases.
Reason $(R):$ Power factor of a series $R-L$ circuit is given by $\cos \theta = \frac{2R}{\sqrt{R^2 + \omega^2 L^2}}$.

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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 :

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

In a series $LCR$ circuit,an alternating emf $(v)$ and current $(i)$ are given by the equations $v = v_{0} \sin \omega t$ and $i = i_{0} \sin \left(\omega t + \frac{\pi}{3}\right)$. The average power dissipated in the circuit over a cycle of $AC$ is:

The power dissipated in a resistor of resistance $R$ when an alternating current with a peak value of $I_p$ flows through it is:

The current and voltage functions in an $AC$ circuit are:$I = 100 \sin(100t) \, mA$$V = 100 \sin(100t + \frac{\pi}{3}) \, V$The power dissipated in the circuit is ...... $W$.

Assertion $(A):$ If the frequency of the applied $AC$ is doubled, then the power factor of a series $R-L$ circuit decreases.Reason $(R):$ Power factor of a series $R-L$ circuit is given by $\cos \theta = \frac{2R}{\sqrt{R^2 + \omega^2 L^2}}$.

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The current and voltage functions in an $AC$ circuit are:
$I = 100 \sin(100t) \, mA$
$V = 100 \sin(100t + \frac{\pi}{3}) \, V$
The power dissipated in the circuit is ...... $W$.

Assertion $(A):$ If the frequency of the applied $AC$ is doubled, then the power factor of a series $R-L$ circuit decreases.
Reason $(R):$ Power factor of a series $R-L$ circuit is given by $\cos \theta = \frac{2R}{\sqrt{R^2 + \omega^2 L^2}}$.