# LR circuit , deriving $I(t)$ [closed]

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 !

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 $$I_1(0)=0$$.

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 ($$I$$, $$I_1$$, $$I_2$$).