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Question

Two capacitors of 2μF and 3μF are charged to 150V and 120V respectively. The plates of capacitor are connected as shown in the figure. An uncharged capacitor of capacity 1.5μF falls to the free end of the wire. Then :
214977_dee8363af38647bcb09e3f3595b5d9f7.png
  1. charge of 1.5μF capacitor is 180μC
  2. charge on 2μF capacitor is 120μC
  3. positive charge flows through A from left to right
  4. positive charge flows through A from right to left

A
charge on 2μF capacitor is 120μC
B
positive charge flows through A from right to left
C
charge of 1.5μF capacitor is 180μC
D
positive charge flows through A from left to right
Solution
Verified by Toppr

Initially charge on 2μF=2×150=300μC and charge on 3μF=3×120=360μC

When the uncharged capacitor of capacity 1.5μF falls to the free end of the wire, it will make a closed loop circuit. Let qμC charge flows in the closed loop in clockwise direction. Then final charges on different capacitors are as shown in figure.

Applying Kirchhoff's law for the loop,

300q2q1.5+360q3=0

multiplying by 30 both sides, 450015q20q+360010q=0

or 8100=45qq=180μC

Thus, positive charge flow through A from left to right.

charge on 2μF becomes =300180=120μC

272365_214977_ans_a8b4fa3ae32d4cacae01d36b20fb7c44.png

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