Back to Course

0% Complete
0/0 Steps

• ## Do you like this content?

Lesson 5, Topic 2
In Progress

# Resistance | Capacitance | Inductance in an A.C Circuit

Lesson Progress
0% Complete

Resistance in an A.C Circuit

The a.c current and voltage flowing through a resistor are in phase, they attain a maximum and minimum values at the same time. The value of current is I = I0Sinωt, the voltage, V = V0Sinωt

from Ohm’s law, V = IR, I = $$\frac{V}{R}$$

If V = V0Sinωt.

R =$$\frac {V_o Sin \omega t}{I_o Sin \omega t}= \frac{Vrms}{Irms}$$

R = $$\frac {V_o }{I_o }= \frac{Vrms}{Irms}$$

Irms =  $$\frac{Vrms}{R}$$

Capacitance in an A.C Circuit

The voltage (V) and current (I) are out of phase, the current lead the voltage by $$\frac {pi}{2 }$$  or 900 or voltage lags the current by $$\frac {pi}{2 }$$

If capacitor oppose flow of current, hence the opposition offered by a capacitor to flow of an a.c is known as capacitive reactance, Xc

Xc = $$\frac {1}{2 \pi f c} \ \frac{1}{\omega c}$$

V = IXc

The unit of Xc is ohms.

Xc = capacitive reactance, c = capacitance of capacitor in Farad

f = frequency of the source in Hertz (Hz).

Inductance in A.C Current

Inductive reactance is the opposition to the flow of an a.c current offered by an inductor.

VL leads IL by $$\frac {\pi}{2 } \scriptsize \; or \; 90^{\circ}$$

XL= if V = VOsinωt, I = IO $$\left( \scriptsize sin\omega t – \normalsize \frac {\pi}{2 } \right)$$

Since an inductor opposes the flow of current, the inductive reactance is

XL, then V =IXL

XL= 2πfL = ωL

The unit of L is henry (H). The unit XL is (Ω).

A reactance is the opposition to the flow of a.c offered by a capacitor or an inductor or both.

Examples

1. An inductor of 0.7H is connected to an A.C Source of 220V. Calculate

(i) Inductive reactance

(ii) current flowing through the circuit

(iii) energy dissipated in the inductor (f =  $$\frac {100}{\pi}$$ Hz)

Solution

(i) XL= 2πfL

= $$\scriptsize 2 \pi \; \times \; \normalsize \frac {100}{\pi} \; \times \; 0.7$$

= 140Ω

(ii) V = I x L

I = $$\frac {V}{XL} = \frac {220}{140}$$

I = 1.57A

(iii) W = $$\frac {1}{2} \scriptsize LI^2$$

=  $$\frac {1}{2} \scriptsize \times 0.7 \times 1.57^2$$

= 0.86J

2. A 2 F capacitor is connected directly across a 150Vrms, 60Hz a.c source. Find (i) r.m.s value of the current (ii) peak value of the current

Solution

XC = $$\frac {1}{2 \pi f c} =\frac {1}{2 \pi \; \times \; 60 \; \times \; 2 \times 10^{-6}f c}$$

XC = 1324.4Ω

V = I x C

Irms= $$\frac {V_{rms}}{x_c} = \frac {150}{1324.4}$$

Irms = 0.11A

Irms =  $$\frac {I_0}{\sqrt{2}}$$

IO = Irms x √2

= 0.11 x 0.707

=0.160A

error: