If the capacitance ratio C 1 C 2 = η , then the potential difference V between points M and N in the given figure is

Loop ABEF c q 1 + ηc q 1 − q 2 = E .....(1) Loop ABCDEF c q 1 + c q 2 + ηc q 2 = E ....(2) Bu (1) and (2) we get V MN = ηc q 2 = η 2 + 3 η + 1 E

Physics - Capacitor Question with Solution | TestHub

PhysicsCapacitorCircuit AnalysisMedium2 min

PhysicsMediummultiple choice

If the capacitance ratio $\frac{C _{2}}{C _{1}} = η$ , then the potential difference V between points M and N in the given figure is

Answer:

A, B, C

Solution:

Loop ABEF

$\frac{q _{1}}{c} + \frac{q _{1} - q _{2}}{ηc} = E$ .....(1)

Loop ABCDEF

$\frac{q _{1}}{c} + \frac{q _{2}}{c} + \frac{q _{2}}{ηc} = E$ ....(2)

Bu (1) and (2) we get

$V_{MN} = \frac{q _{2}}{ηc} = \frac{E}{η ^{2} + 3 η + 1}$

Stream:JEESubject:PhysicsTopic:CapacitorSubtopic:Circuit Analysis

⏱ 2m

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