SATHEE: Electrostatics Question 568

Electrostatics Question 568

Question: Two capacitors of 3pF and 6pF are connected in series and a potential difference of 5000 $V$ is applied across the combination. They are then disconnected and reconnected in parallel. The potential between the plates is [MP PMT 1992]

Options:

A) 2250 $V$

B) 2222 $V$

C) $2.25 \times 10^{6} V$

D) $1.1 \times 10^{6} V$

Show Answer

Answer:

Correct Answer: B

Solution:

$\frac{1}{C} = \frac{1}{3} + \frac{1}{6}$

therefore $C = 2 p F$

Total charge $= 2 \times 10^{- 12} \times 5000 = 10^{- 8} C$

The new potential when the capacitors are connected in parallel is

$V = \frac{2 \times 10^{- 8}}{(3 + 6) \times 10^{- 12}} = 2222 V$

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Electrostatics Question 568

Question: Two capacitors of 3pF and 6pF are connected in series and a potential difference of 5000 $V$ is applied across the combination. They are then disconnected and reconnected in parallel. The potential between the plates is [MP PMT 1992]

Options:

Answer:

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

Electrostatics Question 568

Question: Two capacitors of 3pF and 6pF are connected in series and a potential difference of 5000V is applied across the combination. They are then disconnected and reconnected in parallel. The potential between the plates is [MP PMT 1992]

Options:

Answer:

Solution:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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Question: Two capacitors of 3pF and 6pF are connected in series and a potential difference of 5000 $V$ is applied across the combination. They are then disconnected and reconnected in parallel. The potential between the plates is [MP PMT 1992]