Download Example 21-4 Inductor Voltage

Example 21-4 Inductor Voltage
A 0.500-mH inductor is in a dc circuit that carries a current of 0.500 A. If the current increases to 0.900 A in 0.150 ms
due to a fault in the circuit, how large is the voltage that appears across the inductor? From the perspective of a positive
charge moving through the inductor, is there a voltage rise or drop across the inductor?
Set Up
We’ll use Equation 21-16 to calculate the
voltage induced across the inductor. To decide
whether there’s a voltage rise or drop, we’ll use
the idea that the voltage across the inductor
acts to oppose the change in current.
Solve
Find the voltage that appears in the inductor.
Voltage across an inductor:
i
V = -L (21-16)
t
i = 0.500 A
i increasing
The current increases from 0.500 A to 0.900 A in 0.150 ms =
0.150 * 1023 s, so
i
0.900 A - 0.500 A
=
= 2.67 * 103 A>s
t
0.150 * 10 - 3 s
From Equation 21-16, the voltage across the inductor is
i
= - 10.500 * 10-3 H2 12.67 * 103 A>s2
V = -L
t
= -1.33 H # A>s = -1.33 V
The magnitude of the voltage is 1.33 V. The value of V is negative,
which means that there is a voltage drop of 1.33 V across the inductor. Thus, the voltage opposes the flow of current and so opposes the
increase in current (compare Figure 21-5b).
Reflect
The voltage induced in the inductor is comparable to that produced by a household AA or AAA battery. This voltage
is only present during the time when the current is changing, however; when the current is constant, the inductor voltage
is zero.