Consider an electrical circuit containing a two way switch 'S'. Initially S is open and then T₁ is connected to T₂. As the current in R = 6Ω attains a maximum value of steady state level, T₁ is disconnected from T₂ and immediately connected to T₃. Potential drop across r = 3Ω resistor immediately after T₁ is connected to T₃ is __________ V. (Round off to the Nearest Integer)

Solution

Physics sir · 4 steps

An inductor will not let its current jump, so whatever flowed through L now flows through r.

1. Steady state first. With T₁ on T₂, the battery, R and L form one loop. At steady state the current stops changing, so the ideal inductor has no voltage across it and acts like a plain wire.

   I₀ = E/R = (6 V)/(6 Ω) = 1 A

2. Switch to T3. Now L is cut off from the battery and joined in a loop with r alone. The current in an inductor cannot change instantly, so just after switching it is still 1 A, and all of it passes through r.
3. Drop across r. V = I₀ r = 1 A × 3 Ω = 3 V. After this instant the current decays, but the question asks only about the first moment.
4. So. The potential drop across r immediately after switching is 3 V.

The common slip: 0. Treating the current through r as starting from zero, as in a freshly switched-on circuit, gives 0 V. Here the inductor already carries 1 A and keeps pushing it through r.

The idea to keep. At the instant of switching, an inductor carries the same current it had just before. Find that current from the old circuit, then use it in the new one.

Answer from the NTA final answer key. Solution by PrepWiser.