Capacitors in Parallel and in Series — Question 8

Question 8

An unbalanced bridge is solved by two floating-node equations

A is at V and B at zero. Capacitors A-M, M-B, A-N and M-N each equal C. The N-B capacitor equals 2C. Internal nodes M and N have zero net charge.

1. Find VM and VN.

2. Find the equivalent capacitance and the signed charge on the M plate of the bridge capacitor.

Question 8 - Solution

An unbalanced bridge is solved by two floating-node equations

(a) Method and calculation

$$C(V_M-V)+CV_M+C(V_M-V_N)=0\Rightarrow3V_M-V_N=V.$$$$C(V_N-V)+2CV_N+C(V_N-V_M)=0\Rightarrow4V_N-V_M=V.$$

(b) Interpretation and further calculation

$$V_M=5V/11,\quad V_N=4V/11,\quad Q_{M\to N}=CV/11.$$$$Q_A=C(V-V_M)+C(V-V_N)=13CV/11\Rightarrow C_{\mathrm{eq}}=13C/11.$$

Physical check

An internal node with three incident capacitors prevents a direct series reduction; charge conservation remains valid.