Solution to Question 39
Step 1: Steady State Logic
In the steady state, fully charged capacitors act as open circuits (infinite resistance). No current flows through the branches containing capacitors ($P-T$, $Q-S$, $R-U$). The current must find a continuous resistive path from the source node $A$ (12V) to the sink node $B$ (0V).
Based on the diagram and connections provided:
- Node $A$ connects to $P, Q, R$ via $3\Omega$ resistors.
- Node $B$ connects to $S, T, U$ via $1\Omega$ resistors.
- The middle layer has connections between these groups. Specifically, the resistive connections (paths that can actually carry current) are:
- $P \to S$ ($8\Omega$)
- $Q \to U$ ($8\Omega$)
- $R \to T$ ($8\Omega$)
Step 2: Analysis of Resistive Paths
Current splits into three independent paths because the cross-links (the capacitors) are open.
Path 1: $A \to P \to S \to B$
Total Resistance $R_1 = 3\Omega + 8\Omega + 1\Omega = 12\Omega$.
Path 2: $A \to Q \to U \to B$
Total Resistance $R_2 = 3\Omega + 8\Omega + 1\Omega = 12\Omega$.
Path 3: $A \to R \to T \to B$
Total Resistance $R_3 = 3\Omega + 8\Omega + 1\Omega = 12\Omega$.
Since all paths have equal resistance, the total current divides equally. The voltage drop across the entire circuit is 12V.
Current in each path: $i = \frac{12\,\text{V}}{12\,\Omega} = 1\,\text{A}$.
Step 3: Calculating Potentials
We calculate the potential at each intermediate node to find the potential difference across the capacitors.
Potentials at First Layer Nodes (P, Q, R):
These nodes are after the $3\Omega$ resistor from A (12V).
$$V_P = V_Q = V_R = 12\,\text{V} – (1\,\text{A} \times 3\,\Omega) = 9\,\text{V}$$
Potentials at Second Layer Nodes (S, U, T):
These nodes are after the $8\Omega$ resistor from P, Q, or R.
$$V_S = V_P – (1\,\text{A} \times 8\,\Omega) = 9\,\text{V} – 8\,\text{V} = 1\,\text{V}$$
$$V_U = V_Q – (1\,\text{A} \times 8\,\Omega) = 9\,\text{V} – 8\,\text{V} = 1\,\text{V}$$
$$V_T = V_R – (1\,\text{A} \times 8\,\Omega) = 9\,\text{V} – 8\,\text{V} = 1\,\text{V}$$
(Check: From these nodes to B, drop is $1\,\text{A} \times 1\,\Omega = 1\,\text{V}$, so $V_B = 0\,\text{V}$. Correct.)
Step 4: Capacitor Charges
Now we find the potential difference across each capacitor.
Capacitor 1 (Between P and T):
$$\Delta V_{PT} = V_P – V_T = 9\,\text{V} – 1\,\text{V} = 8\,\text{V}$$
Capacitor 2 (Between Q and S):
$$\Delta V_{QS} = V_Q – V_S = 9\,\text{V} – 1\,\text{V} = 8\,\text{V}$$
Capacitor 3 (Between R and U):
$$\Delta V_{RU} = V_R – V_U = 9\,\text{V} – 1\,\text{V} = 8\,\text{V}$$
Since all capacitors have the same capacitance $C = 5\,\mu\text{F}$ and the same potential difference $\Delta V = 8\,\text{V}$:
$$q = C \Delta V = 5\,\mu\text{F} \times 8\,\text{V} = 40\,\mu\text{C}$$