Consider the circuit in (Figure 1) . Assume that L = 200.0 mH . Part A Just afte
ID: 1538900 • Letter: C
Question
Consider the circuit in (Figure 1) . Assume that L = 200.0 mH .
Part A
Just after the switch is closed, what is the current through the resistor R1 = 19.5 ?
Express your answer in amperes to three significant figures.
Part B
Just after the switch is closed, what is the current through the resistor R2 = 28.0 ?
Express your answer in amperes to three significant figures.
Part C
After the switch has been closed a long time, what is the current through the resistor R2 = 28.0 ?
Express your answer in amperes to three significant figures.
Part D
After the switch has been closed a long time, what is the current through the resistor R1 = 19.5 ?
Express your answer in amperes to three significant figures.
Part E
After S has been closed a long time, it is opened again. Just after it is opened, what is the current through the R1 = 19.5 resistor?
Express your answer in amperes to three significant figures.
t 60.0 V 0000Explanation / Answer
Potential difference across R1 is always 60.0V as long as switch is closed, so a constant current of V/R1 will pass through it .
hence fro part A and D
I = 60/19.5 = 3.08 Amp
B) As R2 is connected in series with inductor, and current through inductor at t=0 or immediatelt after closing switch is zero. To suddenly change change current through inductor, required potential difference is infinite.
Hence curreent through R2 , just after switch is closed is 0.00 amp
C) After long time current through R2 and inductor becomes steady, so potential difference across inductor is zero.
Hence total potential drop of battrey is across R2
so I = V/R2 = 60.0/ 28 = 2.14 amp
E) Just after switch is opened, ther is no current though battery, but current through inductor does not change instantaneously. So current passing though inductor, completes loop through R1 . So current through R1 is same as current through R2 just before openng the switch, that is 2.14 amp