SATHEE: Electrostatics Question 285

Electrostatics Question 285

Question: A network of four capacitors of capacity equal to $C_{1} = C, C_{2} = 2 C, C_{3} = 3 C$ and $C_{4} = 4 C$ are conducted in a battery as shown in the . The ratio of the charges on $C_{2}$ and $C_{4}$ is [CBSE PMT 2005]

Options:

A) CE

B) $\frac{C E R_{1}}{R_{2} - r}$

C) $\frac{C E R_{2}}{R_{2} + r}$

D) $\frac{C E R_{1}}{R_{1} - r}$

Show Answer

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 $Q = C \times \frac{E R_{2}}{(R_{2} + r)}$

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Electrostatics Question 285

Question: A network of four capacitors of capacity equal to C1=C,C2=2C,C3=3C and C4=4C are conducted in a battery as shown in the . The ratio of the charges on C2 and C4 is [CBSE PMT 2005]

Options:

Answer:

Solution:

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Question: A network of four capacitors of capacity equal to $C_{1} = C, C_{2} = 2 C, C_{3} = 3 C$ and $C_{4} = 4 C$ are conducted in a battery as shown in the . The ratio of the charges on $C_{2}$ and $C_{4}$ is [CBSE PMT 2005]