Question

Consider the circuit shown in the figure.

a. What is the voltage across the inductor in the instant just after the switch is closed? Use the following data: V_b = 6.40, V; R = 146.0 ?; L = 8.07×10^-1 H.

b. After the switch is closed for a long time, what is the energy stored in the inductor?

Answer 1

Solution:

Part (a) Given V_b = 6.40V; resistance R = 146.0 Ω and inductance of inductor L = 8.07*10^-1 H = 0.807.

Voltage across the inductor is given by

V_L = L*(di/dt)     where (di/dt) is the rate of change of current with time.

When the switch is closed, then the rate of change of current is maximum thus all of the supply voltage appears across the inductor.

Thus the voltage across the inductor in the instant just after the switch is closed is V_b = 6.40 V

(note that after a long time, rate of change of current through the inductor is zero, di/dt = 0 A/s; thus voltage across the inductor is also zero; inductor acts as a normal wire carrying current).

Thus the answer is 6.40 V

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Part (b) After the switch is closed for long time, the current through the inductor rises to a steady value,

i = V_b/R

i = (6.40 V)/(146.0Ω)

i =4.38356*10^-2 A

When the inductor carries a current i, then the energy stored in the inductor in the form of magnetic field is given by,

U = (1/2)*L*i²

U = (1/2)*(0.807 H)*(4.38356*10^-2 A)²

U = 7.75*10^-4 J

Thus the answer is 7.75*10^-4 J