For a given circuit- charge on capacitor C 1 and C 2 in steady state will be equal toA- C 1 V A - V C - C 2 V C - V B respectivelyB- C 1 V A - V B - C 2 V A - V B respectivelyC- C 1- C 2 V A - V B on each capacitorD- C 1 C 2- C 1- C 2 V A - V B on each capacitor

No current through the resistance. nbsp;Potential across nbsp;capacitor C_1 is (V_A V_C) Hence charge on capacitor C_1 is Q_1=C_1(V_A V_C) Potential across ...

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Byju's Answer

Standard XII

Physics

Kirchhoff's Junction Law

For a given c...

Question

For a given circuit, charge on capacitor $C_{1}$ and $C_{2}$ in steady state will be equal to

C_{1} (V_{A} - V_{C}), C_{2} (V_{C} - V_{B})

respectively

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C_{1} (V_{A} - V_{B}), C_{2} (V_{A} - V_{B})

respectively

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(C_{1} + C_{2}) (V_{A} - V_{B})

on each capacitor

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(\frac{C_{1} C_{2}}{C_{1} + C_{2}}) (V_{A} - V_{B})

on each capacitor

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Solution

The correct option is A $C_{1} (V_{A} - V_{C}), C_{2} (V_{C} - V_{B})$ respectively
No current through the resistance.
Potential across capacitor $C_{1}$ is $(V_{A} - V_{C})$
Hence charge on capacitor $C_{1}$ is $Q_{1} = C_{1} (V_{A} - V_{C})$
Potential across capacitor $C_{2}$ is $(V_{C} - V_{B})$
Hence charge on capacitor $C_{2}$ is $Q_{2} = C_{2} (V_{C} - V_{B})$

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For a given circuit, charge on capacitor C1 and C2 in steady state will be equal to

The correct option is A C1(VA−VC),C2(VC−VB) respectivelyNo current through the resistance. Potential across capacitor C1 is (VA−VC) Hence charge on capacitor C1 is Q1=C1(VA−VC) Potential across capacitor C2 is (VC−VB) Hence charge on capacitor C2 isQ2=C2(VC−VB)

For a given circuit, charge on capacitor $C_{1}$ and $C_{2}$ in steady state will be equal to