Question

A resistance of 12 ohms an Inductance of 0.15 Henry and a capacitance of 130 f are connected in series across a supply of 200 volts 50 He. Calculate the 1.impedenace 2.the current 3.powerfactor 4.phase angle between voltage and current and 5.powerconsumed

Answer 1

Explanation & Calculations

$\qquad\qquad \begin{aligned} \small X_L&=\small 2\pi f L\\ \small X_C&=\small \frac{1}{2\pi f C}\\ \small X_R&=\small R\\ \small Z&=\small R+j(X_C-X_L) \end{aligned}$

1.

$\qquad\qquad \begin{aligned} \small |Z|&=\small \sqrt{R^2+(X_C-X_L)^2}\qquad\cdots(1) \end{aligned}$

2.

$\qquad\qquad \begin{aligned} \small V&=\small IZ\\ \small I&=\small \frac{V}{Z}=\frac{200}{\sqrt{R^2+(X_C-X_L)^2}}\qquad\cdots(2) \end{aligned}$

3.

$\qquad\qquad \begin{aligned} \small IZ\cos\phi&=\small IR\\ \small \cos\phi&=\small \frac{R}{Z}\qquad\cdots(3) \end{aligned}$

4.

$\qquad\qquad \begin{aligned} \small \cos\phi&=\small \frac{R}{Z}\\ \small \phi&=\small \cos^{-1}\Big(\frac{R}{Z}\Big)\qquad\cdots(4) \end{aligned}$

5.

• Power consumed is the active power.

$\qquad\qquad \begin{aligned} \small P&=\small I^2R\qquad\cdots(5) \end{aligned}$