Question

An impedance coil having a resistance of 15 ohms and an inductance of 0.08 henry is connected in series with a variable capacitor. If a 117 volt 60 cps source is connected to the circuit, what will be the capacitance of the capacitor and the current

(a) when the circuit has a power factor of 0.866 leading and (b) when the voltage drop across the coil is equal to the voltage drop across the capacitor?

Answer 1

Explanations & Calculations

• Consider the sketch attached

• Knowns are

$\qquad\qquad \begin{aligned} \small R_L&= \small 15\Omega\qquad\qquad\qquad X_L=2\pi fL=30.16\Omega\\ \small \cos\theta&=\small 0.866\to\theta =30^0 (for\,part\,a\,only) \end{aligned}$

• All the resistance, inductance & capacitance are connected in series hence the same current flows through each.

a)

• From the phasor diagram for part (a),

$\qquad\qquad \begin{aligned} \small V_R&= \small 117\cos\theta =117\times 0.866=101.32V \end{aligned}$

• Then by $\small V=iR$,

$\qquad\qquad \begin{aligned} \small i&= \small \frac{101.32V}{15\Omega}=\bold{6.75A} \end{aligned}$

• Then the voltage drop across the inductive part is,

$\qquad\qquad \begin{aligned} \small V_L&= \small iX_L=203.58\,V \end{aligned}$

• Finally,

$\qquad\qquad \begin{aligned} \small 117\sin\theta&= \small V_c-V_L\\ \small V_c&= \small 117\sin30+203.58=262.08V\\ \small C&= \small \frac{6.75A}{2\pi\times 60Hz\times 262.08V}\\ &= \small \bold{68.32\,\mu F} \end{aligned}$

b)

• By refering to the phasor diagram for part (b),

$\qquad\qquad \begin{aligned} \small V_{coil}&= \small V_c\\ \small iZ_{coil}&= \small iX_c\\ \small \sqrt{R^2+X_L^2}&=\small X_c=\frac{1}{2\pi f C}\\ \small C&=\small \frac{1}{2\pi \times 60\times \sqrt{15^2+30.16^2}}\\ &= \small \bold{78.75\,\mu F} \end{aligned}$

• Then capacitive reactance is

$\qquad\qquad \begin{aligned} \small X_c&= \small\frac{1}{2\pi f C}\\ &= \small \frac{1}{2\pi\times 60\times 78.75\times10^{-6} }\\ &= \small 33.68 \,\Omega \end{aligned}$

• FInally,

$\qquad\qquad \begin{aligned} \small 117^2 &= \small V_R^2+(V_c-V_L)^2\\ &=\small i^2\big[R^2+(X_c-X_L)^2\big]\\ \small i&= \small \frac{117}{\sqrt{15^2+(33.68-30.16)^2}}\\ &= \small \bold{7.59\,A} \end{aligned}$