27. Switching Inductor Circuit (Corrected)
Step 1: Initial State (Switch at 2)
Before $t=0$, the switch connects the battery to the inductors. In steady state, ideal inductors act as short circuits (zero resistance).
- The resistor $R$ is shorted out by the path through $L_1$ and $L_2$ (assuming $R$ is in parallel with the inductor branch).
- Current is limited only by internal resistance $r$.
$$ I_0 = \frac{\mathcal{E}}{r} $$
This current $I_0$ flows through both $L_1$ and $L_2$.
Step 2: Switching Analysis ($t > 0$)
At $t=0$, the switch throws to Position 3. This disconnects the battery and connects the input of $L_1$ to the node between $L_1$ and $L_2$.
Insight 1 ($L_1$ is Shorted): The switch places $L_1$ in a closed loop with zero resistance.
$$ V_{L1} = 0 \implies L_1 \frac{di_1}{dt} = 0 \implies i_1(t) = \text{constant} $$
Current in $L_1$ remains $I_0$ forever (in this ideal model).
Insight 2 ($L_2$ Decays): The current $I_0$ in $L_2$ must continue flowing. It flows through $L_2$, down the return wire, up through $R$, and back to $L_2$ via the switch connection.
$$ i_2(t) = I_0 e^{-t/\tau} \quad \text{where } \tau = \frac{L_2}{R} $$
Part (a): Current through the Switch
We apply Kirchhoff’s Current Law (KCL) at the node between $L_1$ and $L_2$ (where the switch connects).
- Current coming from $L_1$ = $I_0$ (constant).
- Current leaving into $L_2$ = $i_2(t)$ (decaying).
- The difference must flow through the switch wire.
$$ I_{\text{switch}} = I_{L1} – I_{L2} $$
$$ I_{\text{switch}}(t) = I_0 – I_0 e^{-\frac{R}{L_2}t} $$
$$ I_{\text{switch}}(t) = \frac{\mathcal{E}}{r} \left( 1 – e^{-\frac{Rt}{L_2}} \right) $$
Part (b): Total Heat Dissipated
Heat is generated by the resistor $R$. The energy source for this heat is the magnetic field of the inductors.
- $L_1$ has constant current $I_0$, so its stored energy ($\frac{1}{2}L_1 I_0^2$) remains trapped and is not dissipated.
- $L_2$ current decays to zero through $R$. Therefore, all energy initially stored in $L_2$ turns into heat.
$$ \text{Total Heat} = U_{\text{initial}}(L_2) = \frac{1}{2} L_2 I_0^2 $$
$$ \text{Total Heat} = \frac{1}{2} L_2 \left( \frac{\mathcal{E}}{r} \right)^2 $$
$$ H = \frac{L_2 \mathcal{E}^2}{2r^2} $$