An inductor with inductance L = 0.200 H and negligible resistance is connected t
ID: 1557640 • Letter: A
Question
An inductor with inductance L = 0.200 H and negligible resistance is connected to a battery, a switch S, and two resistors, R1 = 14.0 and R2 = 14.0 (Figure 1) . The battery has emf 96.0 Vand negligible internal resistance. S is closed at t = 0.
A)
What is the current i1 just after S is closed?
B) What is the current i2 just after S is closed?
C) What is the current i3 just after S is closed?
D) What is the current i2 after S has been closed a long time?
E) What is the current i3 after S has been closed a long time?
F) What is the current i1 after S has been closed a long time?
G) What is the value of t for which i3 has half of the final value that you calculated in part E?
H) When i3 has half of its final value, what is i2?
I) When i3 has half of its final value, what is i1?
R, 0000-1 2 L R 2Explanation / Answer
A) Just after S is closed inductor acsts open ckt.
so, i1 = V_battery/Rnet
= V_battery/R1
= 96/14
= 6.86 A
B) i2 = i1
= 6.86 A
C) i3 = 0 (since the inductor acts as opne ckt)
D) After a long time the inductpr acts as short ckt.
i2 = V_battery/R1
= 96/14
= 6.86 A
E) i3 = V_battery/R2
= 96/14
= 6.86 A
F) i1 = i2 + i3
= 6.86 + 6.86
= 13.7 A