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A circuit containing a 80 mH inductor and a 60 μF capacitor in series is connected to a 230 V, 50 Hz supply. The resistance of the circuit is negligible.

Q7.18 (a) A circuit containing a 80mH inductor and a 60\mu F  capacitor in series is connected to a 230\: V50\: Hz  supply. The resistance of the circuit is negligible. Obtain the current amplitude and rms values.

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The inductance of the inductor L=80mH=80*10^3H

The capacitance of the capacitor C=60\mu F

Voltage supply V = 230V

Frequency of voltage supply f=50Hz.

Here, we have

 V=V_{max}sinwt=V_{max}sin2\pi ft

Impedance

 Z=\sqrt{R^2+\left ( wL-\frac{1}{wC} \right )^2}

Z=\sqrt{0^2+\left ( 2\pi*50*80*10^{-3}-\frac{1}{2\pi 50*60*10^{-6}} \right )^2}=8\pi-\frac{1000}{6\pi }

Now,

Current in the circuit will be 

I=\frac{V}{Z}=\frac{V_{max}sinwt}{Z\angle \phi }=I_{max}sin(wt-\phi )

where, 

I_{max}=\frac{V_{max}}{Z}=\frac{\sqrt{2}*230}{8\pi-\frac{1000}{6\pi}}=-11.63A

The negative sign is just a matter of the direction of current.so,

I=11.63sin(wt-\phi )

here

 tan\phi=\frac{wL-\frac{1}{wC}}{R}

But, since the value of R is zero(since our circuit have only L and C)

\phi=90^0

Hence

 I=11.63sin(wt-\frac{\pi}{2} )

Now, 

RMS value of this current:

 I_{rms}=\frac{I_{max}}{\sqrt{2}}=\frac{11.63}{\sqrt{2}}=8.22A.

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