Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsMiscellaneousEasy2 minPYQ_2020

PhysicsEasysingle choice

A series combination of $N_{1}$ is charged to a potential difference $3 V$ . Another parallel combination of $N_{2}$ (each of capacity $C_{2}$ is charged to a potential difference $V$ . The total energy stored in both the combination is the same. The value of $C_{1}$ in terms of $C_{2}$ is

Options:

Answer:

Solution:

$C_{e q} = \frac{C_{1}}{N_{1}}$
$E = \frac{1}{2} C V^{2}$

$= \frac{1}{2} \frac{C_{1}}{N_{1}} 9 V^{2}$

$= \frac{9}{2} \frac{C_{1}}{N_{1}} V^{2}$

$C_{e q} = N_{2} C_{2}$

$E = \frac{1}{2} C V^{2} = \frac{1}{2} C_{2} N_{2} V^{2}$

$\frac{9}{2} \frac{C_{1}}{N_{1}} V^{2} = \frac{C_{2} N_{2} V^{2}}{2}$

$C_{1} = C_{2} \frac{N_{2} N_{1}}{9}$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:ElectrostaticsSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

Doubts & Discussion

Loading discussions...