Electrostatics Question 285
A network of four capacitors of capacity equal to $ C _{1}=C,C _{2}=2C,C _{3}=3C $ and $ C _{4}=4C $ are connected in a battery as shown in the . The ratio of the charges on $ C _{2} $ and $ C _{4} $ is [CBSE PMT 2005]
Options:
CE
B) $ \frac{CER _{1}}{R _{2}-r} $
C) $ \frac{CER _{2}}{R _{2}+r} $
D) $ \frac{CER _{1}}{R _{1}-r} $
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Answer:
Correct Answer: C
Solution:
In steady state current drawn from the battery $ i=\frac{E}{(R _{2}+r)} $ In steady state capacitor is fully charged hence No current will flow through line (2) Hence potential difference across line (1) is $ V=\frac{E}{(R _{2}+r)}\times R _{2} $ , the same potential difference appears across the capacitor, so charge on capacitor $ Q=C\times \frac{ER _{2}}{(R _{2}+r)} $
BETA
