Irodov 3.115: Two Dividers Stacked in a Row – Amit & Deepali Sinha Foundation

Problem Irodov 3.115

A source of emf $\mathcal{E} = 110$ V feeds a two-link capacitor ladder. Its positive terminal is joined to the top wire, which runs through a capacitor $C_1$ to a first junction, through a second capacitor $C_1$ to a second junction, and on to the open terminal $A$ . Its negative terminal is joined to the bottom wire, which runs on to the open terminal $B$ . From each of the two junctions a capacitor $C_2$ goes down to the bottom wire. The capacitance ratio is $C_2/C_1 = \eta = 2.0$ . Find the potential difference between $A$ and $B$ .

The two-link ladder: the source's long (+) plate faces the top wire; series capacitors C_1 above, shunts C_2 down to the bottom wire, and A, B left open. — The two-link ladder: the source’s long ( $+$ ) plate faces the top wire; series capacitors $C_1$ above, shunts $C_2$ down to the bottom wire, and $A$ , $B$ left open.

Given

$\mathcal{E} = 110$ V
$C_2/C_1 = \eta = 2.0$
$A$ , $B$ open
capacitors uncharged before connection (assumed)

Find

\varphi_A - \varphi_B

Solution

Step 1The last link is a two-capacitor divider

Call $U_1$ the voltage across the first shunt $C_2$ , from the first junction down to the bottom wire. Nothing is connected across $A$ and $B$ , so the second $C_1$ and the second $C_2$ form a series pair across that same voltage $U_1$ ; the plates between them and the second junction make an isolated, neutral conductor, so both carry one charge $q_2$ :

q_2 = \frac{C_1C_2}{C_1 + C_2}\,U_1

The terminals $A$ and $B$ are the plates of the second $C_2$ , so the voltage asked for is the voltage across it:

U = \frac{q_2}{C_2} = \frac{C_1}{C_1 + C_2}\,U_1 = \frac{U_1}{1 + \eta} \tag{1}

Step 2The capacitance hanging on the first junction

Seen from the first junction and the bottom wire, the last link has the series capacitance $C_1C_2/(C_1 + C_2)$ , and it stands in parallel with the first shunt $C_2$ . Writing $C_2 = \eta C_1$ :

\begin{aligned} C' &= C_2 + \frac{C_1C_2}{C_1 + C_2} = \eta C_1 + \frac{\eta C_1}{1 + \eta}\\[2pt] &= \eta C_1\,\frac{(1 + \eta) + 1}{1 + \eta} \end{aligned}

C' = \frac{\eta(2 + \eta)}{1 + \eta}\,C_1

(2)

Step 3The first

C_1

divides the emf with

C'

The first series $C_1$ and the block $C'$ again form a series pair, now across the source: one charge on both, so the emf splits inversely to the capacitances, and $C'$ receives

U_1 = \frac{C_1}{C_1 + C'}\,\mathcal{E}

Substituting (2) and clearing the fraction by $(1 + \eta)$ :

\begin{aligned} U_1 &= \frac{\mathcal{E}}{1 + \dfrac{\eta(2 + \eta)}{1 + \eta}} = \frac{(1 + \eta)\,\mathcal{E}}{1 + \eta + 2\eta + \eta^2} \end{aligned}

U_1 = \frac{(1 + \eta)\,\mathcal{E}}{1 + 3\eta + \eta^2}

(3)

Step 4The voltage at the terminals

Putting (3) into (1), the factor $1 + \eta$ cancels:

U = \frac{1}{1 + \eta}\cdot\frac{(1 + \eta)\,\mathcal{E}}{1 + 3\eta + \eta^2}

Numerically, with $\eta = 2.0$ the denominator is $1 + 6.0 + 4.0 = 11$ :

U = \frac{110~\text{V}}{11}

\varphi_A - \varphi_B = \frac{\mathcal{E}}{1 + 3\eta + \eta^2} = 10~\text{V}

(4)

The source’s positive terminal is on the top wire, so $A$ is the higher of the two points.

Check

Walk the ladder with $C_1 = 1$ and $C_2 = 2$ in any unit. The last $C_2$ holds $10$ V and so the charge $20$ ; the second $C_1$ carries the same $20$ and holds $20$ V, so $U_1 = 30$ V, in line with (3): $3 \times 110/11 = 30$ . The first $C_2$ then holds $2 \times 30 = 60$ , and the first junction is neutral only if the first $C_1$ carries $60 + 20 = 80$ , that is $80$ V. Around the loop, $80 + 30 = 110$ V, the emf. Limits: as $\eta \to 0$ the shunts take no charge, nothing drops across the $C_1$ , and $U \to \mathcal{E}$ ; as $\eta \to \infty$ the shunts short the signal and $U \to 0$ .

Remark

Each link is a divider that passes on the fraction $1/(1 + \eta)$ of what it receives only when nothing loads it; the second link loads the first, which is where the extra $\eta$ in $1 + 3\eta + \eta^2$ , against the unloaded $(1 + \eta)^2 = 1 + 2\eta + \eta^2$ , comes from. With more links the denominators keep growing in the same chained way, which is why a long ladder of this kind attenuates a voltage roughly geometrically along its length.