An ideal cell of emf 10 V is connected in circuit shown in figure. Each resistance is 2 Ω. The potential difference (in V) across the capacitor when it is fully charged is ______.

Solution

Physics sir · 4 steps

Once the capacitor is full, its branch carries no current, so find the potential at each of its two plates.

1. Steady state. A fully charged capacitor carries no current, so R₅, which is in series with it, carries none either. The right plate of C is therefore at the same potential as the negative terminal of the cell. Call that 0 V, so the left wire of the circuit is at 10 V.
2. Remaining circuit. Current flows only through R₁ and R₂ in series (4 Ω). That pair is in parallel with R₃ (2 Ω), which gives (4 × 2)/(4+2) = 4/3 Ω. This combination is in series with R₄, so the total is 4/3 + 2 = 10/3 Ω.

   I = 10/(10/3) = 3 A, V_(R₄) = 3 × 2 = 6 V

3. Top junction potential. The parallel part takes the remaining 10 - 6 = 4 V. The R₁–R₂ branch therefore carries 4/4 = 1 A, and R₁ drops 1 × 2 = 2 V. The junction joining R₁, R₂ and C is at 10 - 2 = 8 V.
4. So. The left plate is at 8 V and the right plate is at 0 V. The potential difference across the capacitor is 8 V.

The common slip: 6. Taking the capacitor voltage as the drop across R₄ gives 6 V. That uses the lower junction, but the capacitor is attached to the top junction, which is at 8 V.

The idea to keep. In a DC circuit at steady state, remove the capacitor branch, solve the resistor network, then read the potentials at the two nodes where the capacitor connects. A resistor in series with C drops no voltage.

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