(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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The power factor of the circuit as shown in the figure is:

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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

The power factor of the circuit as shown in the figure is:

$A$ sinusoidal alternating current of peak value $I_0$ passes through a heater of resistance $R$. What is the mean power output of the heater?

In an $A$.$C$. circuit,the potential difference $V$ and current $I$ are given respectively by $V = 100 \sin(100t) \text{ V}$ and $I = 100 \sin(100t + \frac{\pi}{3}) \text{ mA}$. The power dissipated in the circuit will be (Given: $\cos \frac{\pi}{3} = \frac{1}{2}$)

Can the instantaneous power output of an $AC$ source ever be negative? Can the average power output be negative?

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 $(\cos 60^{\circ} = 0.5)$.

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