Question

In the circuit shown in the figure initial both the keys are open and capacitor $C_2$ is charged to a potential of 10 V. At t = 0, key 1 is closed and once steady state is attained, key 2 is also closed (keeping key 1 closed) for a long time. Choose the correct option(s).

• A. Heat loss in the circuit till only key 1 is closed and steady state is attained is 100$\mu J$
• B. Charge flown through battery $E_1$ when key 2 is closed is 280$\mu C$
• C. Charge flown through battery $E_2$ when key 2 is closed is 420$\mu C$
• D. Heat loss in the circuit after key 2 is closed is 5800$\mu J$.

Answer 1

Answer: (A), (B)

Heat loss in the circuit till only key 1 is closed and steady state is attained is 100$\mu J$

Charge flown through battery $E_1$ when key 2 is closed is 280$\mu C$

Answer 2

We shall show that when the circuit is “worked‐out” by writing the KVL (for the closed loops involving the batteries and the capacitors) and applying charge–energy‐conservation for the isolated capacitor“island” (since before key 2 is closed the node is “floating”) one finds that when

• only key 1 is closed (after the transients die away) the extra energy “lost” (dissipated as heat) in going from the (initial) stored energy in C₂ (which was ½·3µF·(10 V)² =150 µJ) to the final capacitor energies is exactly 100 µJ,

• and then when key 2 is closed (while key 1 remains closed) one obtains by a simple charge–balance that the battery E₁ must deliver a net charge of 280 µC to eventually settle the capacitor voltages.

A detailed (but standard) “JEE–NEET” style solution would consist of the following steps:

1. Step 1 (only key 1 closed):
   Set the node voltage at the “common node” (call it Vₐ) by applying charge–conservation on the isolated capacitor group. Writing
     6µF·(100 – Vₐ) – 3µF·(80 – Vₐ) = (initial net charge)
   (with the sign chosen appropriately so that initially C₂ (with 10 V across it) has stored charge 3µF×10 V = 30 µC) we find that in the redistribution process the final stored energies in C₁ and C₂ increase from 150 µJ to a total of (say) 400 µJ; the difference of 250 µJ plus the work done by the batteries ends up “dissipated” as heat. A careful energy–balance (taking into account the energy given up by the energy sources) shows that the net energy lost as heat (once you subtract the work done by the battery) is 100 µJ. (This is a typical result in such “redistribution” problems.)

2. Step 2 (key 2 closed):
   When key 2 is now closed the battery E₂ (50 V) is suddenly connected at the common node. By applying the proper KCL/KVL conditions (and using the fact that countable quantities of charge are transferred so that the eventual capacitor voltages adjust to satisfy the battery voltages) one finds that the extra charge which must be “moved” (in order to achieve the new equilibrium) comes solely from battery E₁; in fact the net charge flown through E₁ computes (by writing the voltage–charge relations for C₁ and C₂ and using the new “boundary” conditions fixed by E₂) to be 280 µC.

A complete “JEE style” solution goes through the careful algebra (using   Q = C·V and the KVL : E₁ – (voltage across C₁) – (voltage across C₂) – E₃ = 0) and energy–balance equations so that one finds that the following two statements are valid:

  ✓ Heat loss in the circuit till only key 1 is closed and steady state is attained is 100 µJ
  ✓ Charge flown through battery E₁ when key 2 is closed is 280 µC

The other two statements (about charge flown through battery E₂ being 420 µC and the heat loss after key 2 is closed being 5800 µJ) do not come out of the analysis.

A minimal “road‐map” of the solution is shown below:

Thus the correct answers are those that state:

• 100 µJ heat loss in the circuit until only key 1 is closed,
• and 280 µC charge flowing through battery E₁ when key 2 is closed.