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## Homework Statement

In the following figure, R1= 3.1 Ω, R2= 4.9 Ω, L1= 0.155 H, L2= 0.2 H and the ideal battery has

*ε*= 5.6 V. Just after switch S is closed, at what rate is the current in inductor 1 changing?

## Homework Equations

[itex]{ \varepsilon }_{ induced }\quad =\quad L\frac { di }{ dt } \\ V\quad =\quad IR[/itex]

## The Attempt at a Solution

Don't know where to start. I know that the inductor will induce an emf in the opposite direction of the battery's. Right after the switch is closed, I guess it has two resistors in parallel, so the current can be calculated from Ohm's law. But that doesn't seem right since there will be a "resistance" from the inductors.