Four capacitors with capacitances C 1 = 1   μF , C 2 = 1 . 5   μF , C 3 = 2 . 5  …

Four capacitors with capacitances C1=1 μF,C2=1.5 μF, C3=2.5 μF and C4=0.5 μF are connected as shown and are connected to a 30 volt source. The potential difference between points a and b is:

  1. 5 V
  2. 9 V
  3. 10 V
  4. 13 V

Solution

Let q be the charge on capacitors C1 and C2

qC1+qC2=30

q1+q1.5=30

q=18 μC

VA-Va=qC1=18 V

Let q' be the charge on capacitors C3 and C4

q'C3+q'C4=30

q'2.5+q'0.5=30

q'=252 μC

VA-Vb=q'C3=5 V

Va-Vb=VA-Vb-VA-Va

Va-Vb=5-18

Va-Vb=-13 V

Asked in: AP EAMCET 2021 (19 Aug Shift 2)

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