Discussion
The switch is in position 1 for long time. It is changed to position 2 at t = 0. The initial condition for current is \(i(0_{-})\) =
Answer: B
The initial current is 1 A, option (B). With the switch in position 1 for a long time, the 10 V battery has been connected to the 10 ohm resistor and 0.5 H inductor for many time constants, so the circuit is in steady state. In d.c. steady state the inductor acts as a short circuit and the current is set by the resistor alone.
\[ i(0^{-}) = \frac{10\ \text{V}}{10\ \Omega} = 1\ \text{A} \]
Because the current in an inductor cannot change at once, this value also holds just after the switch moves to position 2. Option (C), 5 A, would be the final current with the 50 V battery. Option (A), 0 A, and option (D), 25 A, do not follow from either battery.
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