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Question Statement

Let the switch remain open a long time and then be flipped closed at $t=0$. Find the current $I(t)$ for $t \ge 0$. Note that it is necessary to consider the complexity of the parallel branch construction.

Attempt at Solution

From the law of meshes I know that $$I = I_1 + I_2 $$ My potential difference conservation equation should then be : $$ V_0 - IR_{1} - L\frac{dI_{1}}{dt} - I_{2}R_{2} = 0$$ From here I don't know how to transform this equation such that only $I$ appears in it.

Thank you for help !

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2 Answers 2

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You seem to be misusing the potential difference law. It actually states that the voltage on $R_2$ and $L$ is the same,

$$L\dot I_1=R_2I_2$$

As well as the total voltage in the circuit going through just one of these two components is $V_0$

$$V_0=R_1I+R_2I_2$$

So together with charge conservation $I=I_1+I_2$ you have three equations and three unknowns so you can reduce this to a first order linear ODE. The boundary condition is $I_1(0)=0$.

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Your first equation is correct (due to the current rule).

But your second equation is wrong.

You need to apply the voltage rule twice:
for the left loop (consisting of $V_0$, $R_1$ and $L$),
and for the right loop (consisting of $L$ and $R_2$).

So you have 3 equations for 3 unknowns ($I$, $I_1$, $I_2$).

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