The switch, S, in the circuit has been open for a long time. At t=0 it suddenly closes. Calculate the current through the inductor, $I_L$, as a function of time. Resistance in the inductor can be neglected.

I am trying to understand what the flow of the current will look like at different $t$s. At $t<0$, the switch is closed and the inductor is at steady state, does the current not flow through the second resistor at this point? And when $t>0$, do you only focus on the right part of the circuit with the inductor and second resistor?

Or can I just ignore the second resistor?

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closed as off-topic by ZeroTheHero, John Rennie, Emilio Pisanty, user191954, Kyle Kanos Nov 14 '18 at 12:34

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At t<0 the switch is open, so no charge can flow. Try naming the amount of charge flowing from the battery through the first resistor I1, the amount of charge flowing through the Inductor I2 and the amount of charge flowing through the second resistor I3. Then we certainly have I1=I2+I3. Try to think of other equations that I1, I2 and I3 must satisfy, and what their values are at t=0.

This should give you a differential equation with the initial value at t=0.

  • $\begingroup$ Thank you for your answer! I got that I1 = I2= 0.05A and then I3= 0 (?). From the differential equation I got the equation I(t) = I x (1-e^(-R/L)t) which then when substituting in the varibles becomes, I(t)=0.05 x (1-e^(2x10^7 t) ). Is this correct? $\endgroup$ – P.ython Nov 8 '18 at 16:19
  • $\begingroup$ Do you understand why d/t I2 *10^-5=I3*200? $\endgroup$ – fibo11235 Nov 10 '18 at 20:57

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