Physics Solution – Question 16

Solution to Question 16

Figure 2: The nodal configuration.

1. Nodal Analysis Principle

In a capacitor network, the potential of a junction (Node O) is determined by the conservation of charge on the isolated plates connected to that junction. The sum of charges on the plates meeting at O must equal the initial net charge $Q_{net}$ (usually zero if not specified).

$$ \sum Q_i = Q_{net} $$ $$ C_1(V_O – V_A) + C_2(V_O – V_B) + C_3(V_O – V_C) + C_4(V_O – 0) = Q_{net} $$

Note: The resistor $R$ in branch C does not affect steady-state potential as no DC current flows.

2. Analysis of Unknowns

Given $V_A = -10\text{V}$, $V_C = 10\text{V}$, $V_{ground} = 0$, and $C_1=C_2=C_3=C_4=5\mu\text{F}$. The equation becomes:

$$ 5(V_O + 10) + 5(V_O – V_B) + 5(V_O – 10) + 5V_O = Q_{net} $$ $$ 20V_O – 5V_B = Q_{net} $$ $$ 4V_O – V_B = \frac{Q_{net}}{5} $$

We have one equation with two unknowns: $V_O$ and $V_B$. Additionally, $Q_{net}$ is technically unknown (though often assumed zero). Because $V_B$ is unknown, we cannot determine $V_O$. This validates statement (b).

3. Effect of Specifying Charge on $C_2$

If the charge on capacitor $C_2$ ($Q_2$) is specified, we have an additional constraint: $$ |V_O – V_B| = \frac{Q_2}{C_2} $$ This relates $V_B$ directly to $V_O$. If we also assume the standard condition that the junction was initially neutral ($Q_{net}=0$), the system becomes solvable for both $V_O$ and $V_B$. This validates statement (d).

Correct Statements: (b) and (d)