Introduction to Electric Circuits

8.2 Circuits containing resistance and inductance 175

At t = 0, the rate of change of current is V/L amperes per second and if the

current maintained this rate of growth for a time equal to the time constant of

the circuit it would reach

(V/L) x "r = (V/L) • (L/R)

• V/R amperes
  This is its final steady state value I.

Example 8.1

A d.c. voltage of 200 V is suddenly applied to a circuit consisting of a resistor of

20 lI resistance in series with an inductor having an inductance of 0.5 H.

Determine (1) the time constant of the circuit, (2) the final steady state value of

the current, (3) the value of the current after a time equal to three time

constants.

Solution

1 From Equation (8.4) the time constant is given by

L/R - 0.5/20 - 0.025 s - 25 ms

2 The final (steady state) value of the current is given by

I = V/R- 200/20- 10 A

3 The time equivalent to 3~" is 3 x 25 - 75 ms. From Equation (8.2)

i = I[1 - exp (-Rt/L)].

With I = 10 A, R - 20 11, L - 0.5 H and t - 0.075 s, we obtain

i3~.- 10[1 - exp (-20 x 0.075/0.5)] = 9.5 A

Example 8.2

A resistor having a resistance of 2 11 is connected in series with an inductor of

20 H inductance. A step voltage is suddenly applied to the series combination

and the initial rate of rise of current is 4 A/s. Determine (1) the time constant

of the circuit, (2) the magnitude of the applied voltage.

Solution

The circuit is as shown in Figure 8.1 with R - 2 11 and L - 20 H.

1 From Equation (8.4) the time constant is given by ~"- L/R - 20/2 = 10 s.

2 From Equation (8.7) with t = 0 we see that the initial rate of rise of current

is given by V/L. It follows that V = (initial rate of rise of current)x L

=4x20-80V.