1. When the voltage across a particular inductor is 0.25 V the current through t

Question

1. When the voltage across a particular inductor is 0.25 V the current through the inductor changes at a rate of 0.05 A in 1.0 ms. What is the inductance of the inductor?

2. In the circuit below the switch is initially open and closes at time t = 0

a) What is the initial value of V_l

b) What is the final value of V_l

c) What is the initial value of I

d) What is the final value of I

3. When a particular inductor has a current of 5 mA through it it stores 100 microJ of energy in its magnetic field. What is its inductance?

Explanation / Answer

1.

Voltage across Inductor is

VL=L(dI/dt)

L=VL/(dI/dt) =0.25/(0.05/1*10-3)

L=5*10-3 H or 5 mH

2.

a)

Initially Inductor acts as open circuit ,so

VL=E

b)

After Long time indcutor acts as short circuit ,so

VL=0

c)

So Iniital current through the circuit is

I=0 A

d)

Final Value of current in the circuit is

I=E/R

3.

Energy stored in inductor

U=(1/2)LI2

(100*10-6)=(1/2)*L*(5*10-3)2

L=8 H

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.