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Standard XII

Physics

Capacitor in an AC Circuit

A coil of ind...

Question

A coil of inductance $L = 0.2 H$ and of resistance $R = 62.8 Ω$ is connected to the mains alternating voltage of frequency $50 H z$ . What can be the capacitance of the capacitor connected in series with the coil if the useful power has to remain unchanged?

10 μ C

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50 μ C

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25 μ C

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100 μ C

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Solution

The correct option is C $25 μ C$
Since powers are equal,
$z_{1} = \pm z_{2}$
$\Rightarrow ω^{2} L^{2} + R^{2} = {(ω L - \frac{1}{ω C})}^{2} + R^{2}$
$- \frac{2 L}{C} + \frac{1}{ω^{2} C^{2}} = 0$
$\Rightarrow C = \frac{1}{2 ω^{2} L} = 25 μ C$

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A coil of inductance L=0.2 H and of resistance R=62.8 Ω is connected to the mains alternating voltage of frequency 50 Hz. What can be the capacitance of the capacitor connected in series with the coil if the useful power has to remain unchanged?

The correct option is C 25 μCSince powers are equal, z1=±z2 ⇒ω2L2+R2=(ωL−1ωC)2+R2 −2LC+1ω2C2=0 ⇒C=12ω2L=25 μC

A coil of inductance $L = 0.2 H$ and of resistance $R = 62.8 Ω$ is connected to the mains alternating voltage of frequency $50 H z$ . What can be the capacitance of the capacitor connected in series with the coil if the useful power has to remain unchanged?

The correct option is C $25 μ C$
Since powers are equal,
$z_{1} = \pm z_{2}$
$\Rightarrow ω^{2} L^{2} + R^{2} = {(ω L - \frac{1}{ω C})}^{2} + R^{2}$
$- \frac{2 L}{C} + \frac{1}{ω^{2} C^{2}} = 0$
$\Rightarrow C = \frac{1}{2 ω^{2} L} = 25 μ C$