JEE Main 2022 · 26 July · morning · Single correct · +4 / −1
The total charge on the system of capacitors C₁=1 μ F, C₂=2 μ F, C₃=4 μ F and C₄=3 μ F connected in parallel is :
(Assume a battery of 20 V is connected to the combination)
- (a) 200 μ C
- (b) 200 C
- (c) 10 μ C
- (d) 10 C
Answer and solution
Answer (a) 200 μ C
Solution
Physics sir · 3 steps
In parallel the capacitances just add, so find the total C first and then use Q = CV once.
- Add in parallel. All four capacitors sit across the same 20 V, so their capacitances add directly: C = 1 + 2 + 4 + 3 = 10 μF.
Q = CV = (10 μF) × (20 V) = 200 μC
- Check the parts. Each capacitor takes Qᵢ = Cᵢ V: that is 20, 40, 80 and 60 μC. These add to 200 μC, matching the total.
- So. The total charge is 200 μC, option (a).
The idea to keep. Parallel capacitors share the same voltage, so total charge is (C₁ + C₂ + …)V. Always carry the unit: microfarads times volts gives microcoulombs.
Answer from the NTA final answer key. Solution by PrepWiser.