In the circuit shown in the figure (Figure 1) , the battery and the inductor hav
ID: 1336615 • Letter: I
Question
In the circuit shown in the figure (Figure 1) , the battery and the inductor have no appreciable internal resistance and there is no current in the circuit. Assume that R = 15.5 and L = 12.0 mH.
A) Immediately after the switch is closed, find the reading of the ammeter A .
B) Immediately after the switch is closed, find the readings of the voltmeters (V1 and V2) .
C) Find the reading of the ammeter A after the switch has been closed for a very long time.
D) Find the readings of the voltmeters (V1 and V2) after the switch has been closed for a very long time.
E) Which answers in parts A, B, C and D would change if the inductance were 2.40×102 mH instead?
a. All the answers would change.
b. The answers in part A and B would change.
c. The answers in part C and D would change.
d. None of the answers would change.
Please explain every step clearly! Thank you!!
Explanation / Answer
A) immediately after switch is closed inductor acts as open circuit ,so
I=0 A
B) the voltmeter V1 shows
V1=0V
the induced current produces same voltage that battery have, but in reverese direction. thats why it acts as open circuit after close the switch. so, the voltage shown in voltmeter 2 is
V2=25 V
C)
after long time inductor acts as short circuit
so ,I=V/R
=25/15.5
=1.61 A
D)
the voltage readings are altered here. that is,
V1=25 V and V2=0
the resistance of the inductor is zero here. thats why the potential drop accross it is zero.
(E)
none of the part will change.. because, all the formulas used there does not contain inductor term.