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Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsMiscellaneousEasy2 minPYQ_2020

PhysicsEasysingle choice

A condenser of capacitance $10 μF$ has been charged to $100 V$ . It is now connected to another uncharged condenser in parallel. The common potential becomes $40 V$ . The capacitance of another condenser is

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Answer:

Solution:

$15 μF$

$C_{1} = 10 μF C_{2} = ?$

$V_{1} = 100 V V_{2} = 0$

$V_{c o m m o n} = 40 V$

$V_{c o m m o n} = \frac{C_{1} V_{1} + C_{2} V_{2}}{C_{1} + C_{2}}$

$40 = \frac{10 (100) + 0}{10 + C_{2}}$

$C_{2} = 15 μF$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:ElectrostaticsSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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