Introduction to Electric Circuits

Figure 4.10

XL

o q-

4.2 Single-phase a.c. circuits in the steady state 75

I

0 r f

and as

f "---> ~ SO XL---'~ and I---,0

Example 4.5

In the circuit of Fig. 4.8, L = 5 mH, i = 10 sin 27rft and f = 400 Hz. Calculate

(1) the inductive reactance of the circuit, (2) the rms value of the supply

voltage.

Solution

1 The inductive reactance, XL = 2-rrfL = 27r 400 5 • 10 -3 = 12.56 1"),.

2 The rms value of the voltage is given by V = IXL, where I is the rms value

of the current. Now

I = Im/V2 = 10/V/2 = 7.07 A

SO

V = 7.07 • 12.56 = 88.8 V

Purely capacitive circuits

The diagram of Fig. 4.11 shows a pure capacitor C connected to a voltage

source V. We have seen (Chapter 2) that i = dq/dt and that q = CV so that

i = d(Cv)//dt = Cdv/dt if C is constant. Let the source voltage be v = Vm sin wt.

Then

i- Cdv/dt- Cd(Vm sin oJt)/dt = C[VmW COS ~ot]

vt ~_c

Figure 4.11