Two capacitors C₁ and C₂ are connected in parallel to a battery. Charge-time graph is shown below for the two capacitors. The energy stored with them are U₁ and U₂, respectively. Which of the given statements is true?

Solution

Physics sir · 4 steps

Ignore how fast each curve climbs; read the final charge, because both capacitors end at the same voltage.

1. Same voltage. In parallel, both capacitors are connected across the same battery. Once charging is over, each has the battery voltage V across it.
2. Read the graph. The steady value each curve levels off at is the final charge. C₁ rises faster at first, but C₂ levels off higher, so Q₂ > Q₁.

   C = Q/V, U = 1/2QV

3. Compare both. With V the same, more charge means more capacitance, so C₂ > C₁. Energy U = 1/2QV also grows with Q, so U₂ > U₁.
4. So. C₂ > C₁ and U₂ > U₁, which is option (a).

The idea to keep. For capacitors in parallel, V is common, so both C and U go up with the final charge. On a q–t graph, the plateau decides this, not the early slope.

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