Question

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

$A)$ Obtain the current amplitude and rms values?

A circuit containing a $80mH$ inductor and a $60\mu F$ capacitor in series is connected to a $230V,50Hz$ supply. The resistance of the circuit is negligible.
$B)$ Obtain the rms values of potential drops across each element.

A circuit containing a $80mH$ inductor and a $60\mu F$ capacitor in series is connected to a $230V,50Hz$ supply. The resistance of the circuit is negligible.
$C)$ What is the average power transferred to the inductor?

A circuit containing a $80mH$ inductor and a $60\mu F$ capacitor in series is connected to a $230V,50Hz$ supply. The resistance of the circuit is negligible.
$D)$ What is the average power transferred to the capacitor?

A circuit containing a $80mH$ inductor and a $60\mu F$ capacitor in series is connected to a $230V,50Hz$ supply. The resistance of the circuit is negligible.
$E)$ What is the total average power absorbed by the circuit? [‘Average’ implies ‘averaged over one cycle’.]

Answer 1

$A)$ Given, $L=80mH=80\times {10}^{-3}H$
$C=60\mu F=60\times {10}^{-6}F$

$R\approx 0$

Impedance of the circuit is

$Z=\sqrt{R^2+(X_L-X_C{)}^2}$

Putting values,

$Z=\left|\right(X_L-X_C\left)\right|$

$\Rightarrow Z=|2\pi \times 50\times 80\times {10}^{-3}-\frac{1}{2\pi \times 50\times 60\times {10}^{-6}}|$

$Z=|25.13-53.05|$

$Z=27.92\Omega$

The RMS current is given by,

$I_{rms}=\frac{V_{rms}}{Z}$

$I_{rms}=\frac{230}{27.92}=8.23A$

And the current amplitude is,

$I_o=\sqrt{2}{I}_{rms}$

$I_o=8.23\sqrt{2}=11.63A$

Final Answer:

$I_0=11.6A,I_{rms}=8.24A$

$B)$ Step 1: Calculate $RMS$ current

Given, $L=80mH=80\times {10}^{-3}H$
$C=60\mu F=60\times {10}^{-6}F,R\approx 0$

Impedance of the circuit is

$Z=\left|\right(X_L-X_C\left)\right|$

$\Rightarrow Z=|2\pi \times 50\times 80\times {10}^{-3}-\frac{1}{2\pi \times 50\times 60\times {10}^{-6}}|$

$Z=|25.13-53.05|=27.92\Omega$

The RMS current is given by,

$I_{rms}=\frac{V_{rms}}{Z}$

$I_{rms}=\frac{230}{27.92}=8.23A$
Step 2: Calculate potential drop across each element

Rms Potential drop across inductance:

$V_{L_{r}ms}=I_{rms}{X}_L$

Where, $X_L=2\pi \times 50\times 80\times {10}^{-3}$

$=25.13\Omega$
$\Rightarrow V_{L_{rms}}=8.23\times 25.13$

$V_{L_{rms}}=206.81V\approx 207V$

Rms Potential drop across capacitance:

$V_{C_{rms}}=I_{rms}{X}_C$
Where,
$X_C=\frac{1}{2\pi \times 50\times 60\times {10}^{-6}}=53.05\Omega$

$\Rightarrow V_{C_{rms}}=8.23\times 53.05$

$V_{C_{rms}}=436.6V\approx 437V$
Final Answer:

$V_{L_{rms}}=207V{V}_{C_{rms}}=437V$

NOTE:$437V–207V=230V$ is equal to the applied rms voltage as should be the case. The voltage across $L$ and $C$ gets subtracted because they are ${180}^{\circ }$ out of phase.

$C)$ Phasor diagram:


Whatever be the current $I$ in $L,$ actual voltage leads current by $\frac{\pi }{2}.$

$\varphi =\frac{\pi }{2}$
$\cos \varphi =0$

Average Power transfer to inductor:

$P=V_{rms}{I}_{rms}\cos \varphi$

$\Rightarrow P=0$

Final Answer : Whatever be the current $I$ in $L,$ actual voltage leads current by $\frac{\pi }{2}.$ Therefore, average power consumed by $L$ is zero.

$D)$
For capacitor, voltage lags by $\frac{\pi }{2}.$

$\varphi =\frac{\pi }{2}$

$\cos \varphi =0$

Average power transfer to capacitor:

$P=V_{rma}{I}_{rma}\cos \varphi$

$\Rightarrow P=0$

Final Answer: For $C,$ voltage lags by $\frac{\pi }{2}.$
Again, average power consumed by $C$ is zero.

$E)$ As the average power transferred to the capacitor and inductor over a complete cycle is zero, therefore average power consumed by the circuit is also zero.

Final answer: Total average power absorbed is zero.