SATHEE: Electrostatics Question 638

Electrostatics Question 638

Question: In the circuit shown in , $C_{1} = 6 μ F,$

$C_{2} = 3 μ F$ , and battery B = 20 V. The switch $S_{1}$ is first closed. It is then opened, and $S_{2}$ is closed. What is the final charge on $C_{2}$ ?

Options:

A) $120 μ C$

B) $80 μ C$

C) $40 μ C$

D) $20 μ C$

Show Answer

Answer:

Correct Answer: C

Solution:

[c] After closing $S_{1}$

charge on $C_{1}$ is $q = 6 \times 20 = 120 μ C$ . Now, $S_{1}$ is opened, on closing $S_{2}$ ,

charge q will be distributed between $C_{1}$ and $C_{2}$ according to their copacitances.

So charge on $C_{2}$ is $q_{2} = \frac{C_{2} q}{C_{1} + C_{2}} = \frac{3 \times 120}{3 + 6} = 40 μ C$

Electrostatics Question 638

Question: In the circuit shown in , $C_{1} = 6 μ F,$

Options:

Answer:

Solution:

Engineering Preparation

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NCERT Books & Solutions

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Sample Mock Test

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Electrostatics Question 638

Question: In the circuit shown in ,C1=6μF,

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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Question: In the circuit shown in , $C_{1} = 6 μ F,$