The circuit shown in the figure continues to infinity. The potential difference between points 1 and 2 is V2, that between points 3 and 4 is V4and so on; ie., the potential difference becomes 1/2 after every step of the ladder. Find the ratio C1/ C2

pd across C1 is V2 so charge on C1⇒q0=C1V2But total charge supplied by battery is Q0 = Ceq V. Hence C1 = 2CeqCeq=Ceq+C2C1Ceq+C2+C1 ⇒C1C2=2

The circuit shown in the figure continues to infinity. The potential difference between points 1 and 2 is V2, that between points 3 and 4 is V4and so on; ie., the potential difference becomes 1/2 after every step of the ladder. Find the ratio C1/ C2

pd across C1 is V2 so charge on C1⇒q0=C1V2But total charge supplied by battery is Q0 = Ceq V. Hence C1 = 2CeqCeq=Ceq+C2C1Ceq+C2+C1 ⇒C1C2=2

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The circuit shown in the figure continues to infinity. The potential difference between points 1 and 2 is $\frac{V}{2}$ , that between points 3 and 4 is $\frac{V}{4}$ and so on; ie., the potential difference becomes 1/2 after every step of the ladder. Find the ratio C₁/ C₂

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pd across C₁ is $\frac{V}{2}$ so charge on C₁

$\Rightarrow q_{0} = \frac{C_{1} V}{2}$

But total charge supplied by battery is Q₀ = C_eq V.

Hence C₁ = 2C_eq

$C_{eq} = \frac{(C_{eq} + C_{2}) C_{1}}{C_{eq} + C_{2} + C_{1}} \Rightarrow \frac{C_{1}}{C_{2}} = 2$

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The circuit shown in the figure continues to infinity. The potential difference between points 1 and 2 is V2, that between points 3 and 4 is V4and so on; ie., the potential difference becomes 1/2 after every step of the ladder. Find the ratio C1/ C2

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The circuit shown in the figure continues to infinity. The potential difference between points 1 and 2 is $\frac{V}{2}$ , that between points 3 and 4 is $\frac{V}{4}$ and so on; ie., the potential difference becomes 1/2 after every step of the ladder. Find the ratio C₁/ C₂