Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsMiscellaneousEasy2 minPYQ_2020

PhysicsEasysingle choice

Initially $K_{1}$ is closed, now if $K_{2}$ is also closed, find the heat dissipated in the resistances of connecting wires

Options:

Answer:

Solution:

When $K_{1}$ is closed upper circuit is connected to source

$Q = C V$

Energy $U_{1} = \frac{1}{2} C V^{2}$

When $K_{2}$ is also closed complete circuit is connected to source

Therefore the Equivalent circuit

$C_{e q u i} = C + \frac{2 C}{3} = \frac{5 C}{3}$

$U_{f} = \frac{1}{2} \frac{5 C}{3} V^{2} = \frac{5 C}{6} V^{2}$

Charge supplied by battery after closing $K_{2}$ is

$Q = \frac{5}{3} C V - C V = \frac{2}{3} C V$

Energy supplied by battery $= U_{f} - U_{i} + H$

$\frac{2}{} C V^{2} = \frac{5}{6} C V^{2} - \frac{1}{2} C V^{2} + H$

$3$

$H = \frac{1}{3} C V^{2}$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:ElectrostaticsSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

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