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← JEE Main 2015 · 11 Apr · Shift 1
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JEE Main 2015 · 11 Apr · Shift 1, Physics question 16

In figure is shown a system of four capacitors connected across a 10 V battery. Charge that will flow from switch S when it is closed is :

Two branches across the 10 V battery: the upper has 2 μF then 3 μF with junction a between them; the lower has 3 μF then 2 μF with junction b; switch S joins a and b.
Choose one option
Answer and worked solution

Official answer: A

Before closing. Upper branch: 2 μF and 3 μF in series carry 12 μC each. Lower branch: 3 μF and 2 μF also carry 12 μC each. The node charges at a and b are both zero.

After closing. a and b are joined, so the two left capacitors (2 + 3 μF) and the two right ones (3 + 2 μF) each form 5 μF, with 5 V across each.

  • At a: the left plate carries −10 μC (2 μF) and the right +15 μC (3 μF), a net +5 μC.
  • At b: −15 μC + 10 μC, a net −5 μC.

So 5 μC passes through S from b to a.

Topics:Series and parallel combinations
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