The values are: battery V=17.4 V, R1 = 11.4 ohms, R2 = 7.10 ohms, R3 = 7.96 ohms

Question

The values are: battery V=17.4 V, R1 = 11.4 ohms, R2 = 7.10 ohms, R3 = 7.96 ohms, and L = 30.0 H. Find the value of i1 immediately after switch S is closed.

Find the value of i2 immediately after switch S is closed.

Find i1 a long time later. Tries 0/10 Find i2 a long time later.

Find i1 immediately after switch S is opened again.

Find i2 immediately after switch S is opened again.

Find i2 a long time later.

The values are: battery V=17.4 V, R1 = 11.4 ohms, R2 = 7.10 ohms, R3 = 7.96 ohms, and L = 30.0 H. Find the value of i1 immediately after switch S is closed. Find the value of i2 immediately after switch S is closed. Find i1 a long time later. Tries 0/10 Find i2 a long time later. Find i1 immediately after switch S is opened again. Find i2 immediately after switch S is opened again. Find i2 a long time later.

Explanation / Answer

immediately after switch is closed, the inductor will be open circuit.

hence i1=i2=V/(R1+R2)=0.9405 A

after a long time,

the inductor will be short circuit.

then R2 and R3 will come in parallel.

net resistance=3.7527 ohms

so i1=V/(R1+3.7527)=1.1483 A

using current division, i2=i1*R3/(R3+R2)=0.6069 A

after switch S is open, the inductor current wont change

i1=0

i2=1.1483-0.6069=0.5414 A
after long time switch is open, all the energy will be dissipated.
so i2=0

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