In the circuit shown, ratio of capacitors is C 1 C 2 = 3 4 and terminal voltage of the battery connected at the left end is V 0 . Given that the potentials of nodes 1 , 2 , 3 , 4 , 5 and 6 a

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In the circuit shown, ratio of capacitors is $\frac{C _{2}}{C _{1}} = \frac{4}{3}$ and terminal voltage of the battery connected at the left end is $V_{0}$ . Given that the potentials of nodes $1, 2, 3, 4, 5$ and 6 are in geometric progression, when the terminal voltage of the battery connected at the right most end is V. (in steady state condition) Then :-

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Apply nodal at P: ​ ⇒ C 1 ​ ( v 0 ​ r − v 0 ​ r 2 ) + C 1 ​ ( v 0 ​ r − v 0 ​ ) + C 2 ​ v 0 ​ r = 0 ⇒ 3 r 2 − 10 r + 3 = 0 ⇒ r = 3 1 ​ or 3 ⇒ v ′ = 3 5 v 0 ​ ​ or v 0 ​ ⋅ 3 5 ​

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