Capacitors in Parallel and in Series — Question 7

Question 7

A balanced capacitive bridge carries no charge on its cross branch

Between A at V and B at zero, the upper path is C from A to M then 2C from M to B; the lower path is 2C from A to N then 4C from N to B. A capacitor Cx joins M and N. Both internal nodes are initially neutral.

1. Find VM and VN and show that Cx does not affect the result.

2. Find the equivalent capacitance seen between A and B.

Question 7 - Solution

A balanced capacitive bridge carries no charge on its cross branch

(a) Method and calculation

$$V_M=\frac{C}{C+2C}V=V/3,\quad V_N=\frac{2C}{2C+4C}V=V/3.$$$$Q_x=C_x(V_M-V_N)=0.$$

(b) Interpretation and further calculation

$$C_{\mathrm{upper}}=\frac{C(2C)}{3C}=2C/3,\quad C_{\mathrm{lower}}=\frac{(2C)(4C)}{6C}=4C/3.$$$$C_{\mathrm{eq}}=2C.$$

Physical check

The zero cross-branch voltage follows equal divider ratios, not a need for equal individual capacitances.