The switch in this circuit has been open for a long time, so no current is flowi

Question

The switch in this circuit has been open for a long time, so no current is flowing. The switch S is then closed at t=0.

(a) What is the current through the battery immediately after the switch is closed?

(b) What is the current through the 20 ohm resistor immediately after the switch has been closed?

(c) What is the current through the battery a long time after the switch is closed?

(d) What is the current through the 20ohm resistor a long time after the switch has been closed?

(e) After a long time, the switch S is opened again. When will the current through the inductor reach 1% of the value it had right before the switch was opened?

Can you please show work. Thanks.

The switch in this circuit has been open for a long time, so no current is flowing. The switch S is then closed at t=0. (a) What is the current through the battery immediately after the switch is closed? (b) What is the current through the 20 ohm resistor immediately after the switch has been closed? (c) What is the current through the battery a long time after the switch is closed? (d) What is the current through the 20ohm resistor a long time after the switch has been closed? (e) After a long time, the switch S is opened again. When will the current through the inductor reach 1% of the value it had right before the switch was opened?

Explanation / Answer

a)I=v/(R+r)

=30/(20+10)

=1 A

b)current = 1 A

c)after a long time, there will be no current through 20 ohm.so,

I=v/r

=30/10

=3 A

d)current =0

e)current through the inductor=3A

so i=io(e^(-rt/l)

or t=2.3*10^-3 s

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