Question

In the figure shown switch $S_1$ remains connected and switch $S_2$ remains open for a long time. Now $S_2$ is also closed. Assuming $\varepsilon = 10V$ and $L = 1H$. Find the magnitude of rate of change of current (in A/s) in inductor just after the switch $S_2$ is closed.

Answer 1

10 A/s

Answer 2

Solution:

1. Before closing S₂:

   – With S₁ closed and S₂ open for a long time, the inductor is in DC steady state so it behaves like a short.

   – The inductor current becomes

   $$I_L(0^-)=\frac{\varepsilon}{R}=\frac{10}{R}\,.$$

2. Immediately after S₂ is closed (t = 0⁺):

   – Now node B is connected to two branches:

   • Left branch: 10 V source with resistor R, current = $(10-V_B)/R$.
   • Right branch: 30 V source (3ε) with resistor 2R, current = $(30-V_B)/(2R)$.

   – Applying KCL at node B (with the inductor current $I_L=10/R$ unchanged due to current continuity):

   $$\frac{10-V_B}{R}+\frac{30-V_B}{2R}=\frac{10}{R}\,.$$

   – Multiply through by $2R$:

   $$2(10-V_B)+(30-V_B)=20\,.$$

   This simplifies to:

   $$20-2V_B+30-V_B=20\quad\Rightarrow\quad50-3V_B=20\,.$$

   Thus,

   $$3V_B=30\quad\Rightarrow\quad V_B=10\,V\,.$$

3. Rate of Change of the Inductor Current:

   – The voltage across the inductor is $V_L= V_B = 10\,V$.

   – Using the inductor relation

   $$V_L=L\frac{dI}{dt}\,,$$

   with $L=1\,H$, we have

   $$\frac{dI}{dt}=\frac{10}{1}=10\,A/s\,.$$

Core Explanation (Minimal):

• Before S₂ closes, $I_L=10/R$.
• At t = 0⁺, applying KCL at node B gives $V_B=10\,V$.
• Thus, by $V_L = L\frac{dI}{dt}$ → $\frac{dI}{dt}=10\,A/s$.