JEE Main 2023 · 31 January · evening · Numerical answer · +4 / −0
For the given circuit, in the steady state, |V_B-V_D|= ________ V.
Answer and solution
Answer 1
Solution
Physics sir · 4 steps
In steady state the capacitors block current, so first find which resistors still carry any.
- Dead branch. The 2 Ω resistor after B ends at a junction that leads only to the two 1 μF capacitors. Once they are charged, no current flows through that 2 Ω. Between B and C, current takes only the 1 Ω path, whose far end is wired straight to C.
- Two branches. The 6 V cell sits across A and C. Branch ABC is 2 + 1 = 3 Ω, so it carries 6/3 = 2 A. Branch ADC is 10 + 2 = 12 Ω, so it carries 6/12 = 0.5 A.
- Drops from A. The drop from A to B is 2 × 2 = 4 V. The drop from A to D is 0.5 × 10 = 5 V.
V_B - V_D = (V_A - 4) - (V_A - 5) = 1 V
- So. |V_B - V_D| = 1 V.
The common slip: 0.5. Putting the 2 Ω in parallel with the 1 Ω between B and C gives an AB drop of 4.5 V and an answer of 0.5. That 2 Ω carries no current in steady state.
The idea to keep. In a DC steady state, a charged capacitor acts as an open switch. Any resistor that leads only to capacitors carries no current, so remove it before you do Kirchhoff.
Answer from the NTA final answer key. Solution by PrepWiser.