Put a 10 V battery with a 10 ? internal resistance in series with a switch: hook
ID: 2239847 • Letter: P
Question
Put a 10 V battery with a 10 ? internal resistance in series with a switch: hook this up to a 100 ? resistor in parallel with a 2 H inductor. Assume that the inductor itself is made of superconducting wire so has no resistance of its own. Start with the switch closed for a long time so steady currents exist in the circuit.
Find the battery current, the current in the 100 ? resistor, and the current in the inductor.
Find the initial voltage across the inductor when the switch is opened.
What's the current in the inductor as a function of time t from when you open the switch?
Explanation / Answer
a)
inductor acts as short circuit ,so
I=10/10=1 A
I (100) =0 A
IL =10/10 =0.1 A
b)
V=100*1 =100 V
c)
T=L/R =2/10 =0.2 s
IL =1e^(-t/0.2)
IL =e^(-5t)
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