IPhO 1971 · Theory problem 3

Cube-shaped circuit of batteries, capacitors and resistors

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Four batteries of EMF \(E_1 = 4\ \mathrm{V}\), \(E_2 = 8\ \mathrm{V}\), \(E_3 = 12\ \mathrm{V}\), and \(E_4 = 16\ \mathrm{V}\), four capacitors with the same capacitance \(C_1 = C_2 = C_3 = C_4 = 1\ \mu\mathrm{F}\), and four equivalent resistors are connected in the circuit shown in Fig. 3. The internal resistance of the batteries is negligible.

  • Calculate the total energy \(W\) accumulated on the capacitors when a steady state of the system is established.
  • The points H and B are short connected. Find the charge on the capacitor \(C_2\) in the new steady state.
Figure 1

Fig. 3

Official English text of the problem.

Answers

The two unnumbered bullets are designated (1) and (2) below.

  1. Total energy stored: \(\boxed{W=26\ \mu\mathrm{J}}\).
  2. Charge on capacitor \(C_2\) after shorting H and B: \(\boxed{Q_2=0\ \mathrm{C}}\).

Solution

Let each resistor have resistance \(R\). The resistors occupy edges \(AB\), \(BF\), \(DH\), and \(HG\); crossings in the projection do not create additional junctions.

The long battery plates indicate the positive terminals, so \[ V_D-V_A=E_1,\qquad V_C-V_B=E_2, \] \[ V_H-V_E=E_3,\qquad V_G-V_F=E_4. \] In a steady state, capacitor voltages are constant and no current flows through the capacitors.

(1) Total energy stored

With the capacitor branches open, the only conducting loop is \[ F\longrightarrow G\longrightarrow H\longrightarrow D \longrightarrow A\longrightarrow B\longrightarrow F. \] The branches containing \(E_2\) and \(E_3\) carry no steady current.

Kirchhoff’s voltage law around the conducting loop gives \[ E_4-E_1-4IR=0. \] Thus the voltage drop across each resistor is \[ IR=\frac{E_4-E_1}{4} =\frac{16-4}{4}\ \mathrm{V} =3\ \mathrm{V}. \]

Taking \(V_F=0\), the battery rise and successive voltage drops give \[ V_G=16\ \mathrm{V},\quad V_H=13\ \mathrm{V},\quad V_D=10\ \mathrm{V}, \] \[ V_A=6\ \mathrm{V},\qquad V_B=3\ \mathrm{V}. \] The remaining batteries fix \[ V_C=V_B+E_2=11\ \mathrm{V},\qquad V_E=V_H-E_3=1\ \mathrm{V}. \]

Consequently, the capacitor voltages are:

Capacitor Voltage magnitude Positive plate
\(C_1\) \(\lvert V_D-V_C\rvert=1\ \mathrm{V}\) C
\(C_2\) \(\lvert V_C-V_G\rvert=5\ \mathrm{V}\) G
\(C_3\) \(\lvert V_A-V_E\rvert=5\ \mathrm{V}\) A
\(C_4\) \(\lvert V_E-V_F\rvert=1\ \mathrm{V}\) E

Using the electrostatic energy \(W_i=\tfrac12 C_iU_i^2\), \[ W=\frac12(1\times10^{-6}\ \mathrm{F}) \left(1^2+5^2+5^2+1^2\right)\mathrm{V}^2. \] Therefore, \[ \boxed{W=2.6\times10^{-5}\ \mathrm{J}=26\ \mu\mathrm{J}}. \]

(2) Charge on \(C_2\) after shorting H and B

The short circuit makes \(V_H=V_B\). Choose their common potential as zero: \[ V_H=V_B=0. \] Battery \(E_2\) then fixes \[ V_C=E_2=8\ \mathrm{V}. \]

In the new steady state, the capacitors again carry no current. Battery \(E_4\) drives a loop through the two equal resistors \(GH\) and \(BF\), so \[ I_4=\frac{E_4}{2R}. \] The potential of G relative to H is therefore \[ V_G-V_H=I_4R=\frac{E_4}{2}=8\ \mathrm{V}. \] Hence \(V_G=V_C\), and the voltage across \(C_2\) vanishes: \[ U_2=V_C-V_G=E_2-\frac{E_4}{2}=0. \] Its final charge is consequently \[ \boxed{Q_2=C_2\lvert U_2\rvert=0\ \mathrm{C}}. \]

The solution was checked against the official solution.