Q

# 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\: V$$50\: Hz$  supply. The resistance of the circuit is negligible. Obtain the current amplitude and rms values.

Views

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:

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

Exams
Articles
Questions