Question

An RLC circuit is connected to 60hz, 240volts a.c supply. If R=50ohms, C=100uf and M=0.12h. Calculate the power used in circuit

Answer 1

Explanations & Calculations

• To calculate the needs, the total current should be first calculated, and then by $\small V.I$, consumed power can be calculated.
• Since there are all kinds of units: resistive, inductive & capacitive involved, that product gives the sum of active power & the reactive power collectively known as apparent power.
• Voltage-current relationship for each component (with respect to the applied voltage V in vector notation) is given by,

$\qquad\qquad \begin{aligned} \small V_r&= \small i_rR\qquad V_L=j\omega L\cdot i_L\qquad V_c=\frac{1}{j\omega C}\cdot i_c \end{aligned}$

When series-connected,

$\qquad\qquad \begin{aligned} \small 240V&= \small iR+j\omega L.i+\frac{1}{j\omega C}.i\\ &= \small i\Bigg[R+\Big(\omega L-\frac{1}{\omega C}\Big)j\Bigg]\\ &=\small iZ \end{aligned}$

• What's in the [] bracket is called the total impedance (z) & it's in vector form. Then,

$\qquad\qquad \begin{aligned} \small |i|&=\small\frac{240V}{|Z|}\\ &= \small \frac{240}{\sqrt{R^2+(\omega L-\frac{1}{\omega C})^2}}\\ &= \small \frac{240}{\sqrt{R^2+(2\pi f L-\large\frac{1}{2\pi fC})^2}}\\ &= \small 4.4955A \end{aligned}$

• Then the power consumption is $\small 240V\times4.4955A=1078.92\,\,VA$

When parallel-connected,

$\qquad\qquad \begin{aligned} \small 240V&= \small V_r=V_L=V_c\\ \small 240&= \small i_rR=j\omega Li_L=\frac{1}{j\omega C}i_c\\ \\ \small I_{total}&= \small i_r+i_L+i_c\\ \small I_t&= \small 240\Bigg[\frac{1}{R}+\frac{1}{j\omega L}+j\omega C\Bigg]\\ &= \small 240\Bigg[\frac{1}{R}+\Big(\omega C-\frac{1}{\omega L}\Big)j \Bigg]\\ \small |I_t| &= \small240\times\sqrt{\frac{1}{R^2}+\Big(2\pi fC-\frac{1}{2\pi f L}\Big)^2} \\ &= \small 6.0866\,A \end{aligned}$

• Then the power consumption is $\small 240V\times6.0866A=1460.78\,\,VA$