CURRENT BYU 40

We are given an infinite ladder network (or sufficiently long finite section) of capacitors. The potentials of the nodes $1, 2, 3, \dots, 6$ are in a geometric progression. Let the potential at the $n$-th node be $V_n$. Since they form a geometric progression, we can write:

$$V_{n+1} = k V_n$$

where $k$ is the common ratio.

Consider a general node $n$ in the ladder. It is connected to node $n-1$ via $C_1$, node $n+1$ via $C_1$, and to the ground via $C_2$. Applying Kirchhoff’s Law (conservation of charge) at node $n$ for an isolated section:

$$ q_{\text{left}} + q_{\text{right}} + q_{\text{down}} = 0 $$ $$ C_1(V_n – V_{n-1}) + C_1(V_n – V_{n+1}) + C_2(V_n – 0) = 0 $$

Rearranging the terms:

$$ C_1(2V_n – V_{n-1} – V_{n+1}) + C_2 V_n = 0 $$

Dividing by $C_1$ and substituting the given ratio $\frac{C_2}{C_1} = \frac{4}{3}$:

$$ 2V_n – V_{n-1} – V_{n+1} + \frac{4}{3}V_n = 0 $$ $$ V_{n+1} – \left(2 + \frac{4}{3}\right)V_n + V_{n-1} = 0 $$ $$ 3V_{n+1} – 10V_n + 3V_{n-1} = 0 $$

Since the potentials are in geometric progression, we substitute $V_n = V_0 k^n$. This gives the characteristic equation:

$$ 3k^{n+1} – 10k^n + 3k^{n-1} = 0 $$

Dividing by $k^{n-1}$ (assuming $k \neq 0$):

$$ 3k^2 – 10k + 3 = 0 $$

Solving the quadratic equation:

$$ (3k – 1)(k – 3) = 0 $$

Thus, the possible values for the ratio $k$ are $k = 3$ or $k = \frac{1}{3}$.

The problem asks for the terminal voltage of the battery at the rightmost end (connected to node 6) given the battery at the left end (connected to node 1) is $V_0$.

Let the potential at node 1 be $V_1 = V_0$. The potential at node 6 is $V_6$.

$$ V_6 = V_1 \cdot k^{6-1} = V_0 k^5 $$

Substituting the two possible values for $k$:

1. If $k = 3$, $V_6 = V_0 \cdot 3^5$.
2. If $k = \frac{1}{3}$, $V_6 = V_0 \left(\frac{1}{3}\right)^5 = \frac{V_0}{3^5}$.