Question

Query in physics. Refer to attachment. Thank you in advance​

Answer 1

It is true

Explanation:

This is indicated this it is correct

Answer 2

Explanation:

Resistance offered by inductor immediately after switch is closed will be infinite.

Therefore current through R3 will be zero and

current through R1= current through R2=ER1+R2

=5010+20=53A

b. After long time of closing the switch, resistance offered by inductor will be zero.

In that case R2 and R3 are in parallel, and the resultant of these two is then in series with R1. Hence

Rnet=R1+R2R3R2+R3

=10+(20)(30)20+30=22Ω

Current though the battery (for thorugh R1)

ERnet=5022A

This current will distribute in R2 and R3 in invedrse ratio of resistance. Hence

Current through R2=(5022)(R3R2+R3)

=(5022)(2030+20)=1511A

c. Immediately after switch is reopened, the current through R1 will become zero.

But current thorgh R2 will be eqwual to the steady state current through R3, which is equal to

(5022−1511)A=0.91

d. A long after S is reopened, current through all resistors will be zero.

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