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For the ciruit shown in figure E=50V, R1...

For the ciruit shown in figure `E=50V`, `R_1=10Omega, R_2=20Omega, R_3=30Omega` and `L=2.0 mH`. Find the total current

a. Immediately after switch `S` is closed
b. A long time after `S` is closed.
c. Immediaately after S is reopened
d. A long time after `S` is reopened.

Text Solution

Verified by Experts

a. Resistance offered by inductor immediately after switch is closed will be infinite.
Therefore current through `R_3` will be zero and
current through `R_1`= current through `R_2=E/(R_1+R_2)`
`=50/(10+20)=5/3A`
b. After long time of closing the switch, resistance offered by inductor will be zero.
In that case `R_2` and `R_3` are in parallel, and the resultant of these two is then in series with `R_1`. Hence
`R_("net")=R_1+(R_2R_3)/(R_2+R_3)`
`=10+((20)(30))/(20+30)=22Omega`
Current though the battery (for thorugh `R_1)`
`E/(R_("net"))=50/22A`
This current will distribute in `R_2` and `R_3` in invedrse ratio of resistance. Hence
Current through `R_2=(50/22)(R_3/(R_2+R_3))`
`=(50/22)(20/(30+20))=15/11A`
c. Immediately after switch is reopened, the current through `R_1` will become zero.
But current thorgh `R_2` will be eqwual to the steady state current through `R_3`, which is equal to
`(50/22-15/11)A=0.91`
d. A long after S is reopened, current through all resistors will be zero.
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