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in the figure, both S1 and S2 are initially open. S2 is then closed and left closed until a constant current is established. The S2 is closed just as S1 is opened, taking the battery out of the circuit. Assume that R=515? and L= 305.0 mH


PART A

What is the initial current in the resistor just after S2 is closed and S1 is opened

PART B

What is the time constant of the circuit

PART C

What is the current in the resistor after a large number of time constants have elapsed
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Added Tue, 04 Aug '15
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When does S1 get closed? The problem, as of 12/24/15, states that S2 is closed, but S1 is initially open. Bill N Thu, 24 Dec '15
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If S1 is initially closed while S2 is open, the current will climb like an "inverse exponential" from zero to a steady state value of V/R = 0.485 A.

The time constant of an RL circuit is given by $L_equiv/R_equiv$. In this circuit, that is simply 0.305/515 = 0.000592 s or 0.592 ms.

Because there is no power source to replenish the energy in the circuit, the resistor will eventually transfer the electrical energy out and the current will drop, exponentially, to zero: $i(t)=i_o exp(-Lt/R)$.

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Added Thu, 24 Dec '15
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when S1 is closed, current i = V/R

i = 250/515

i = 0.485 Amps

-----------------------------------------

time constant T = L/R

T = 0.305/515

T = 0.592 msecs

----------------------------------------

after many long time constants I = 0.367 Io

I = 0.367 *0.458

I = 0.168 Amps
Edit
Added Tue, 04 Aug '15
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Is this a real answer or a placeholder? The current goes to zero after many time constants. Bill N Thu, 24 Dec '15
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