Physics Solution – Question 27

Solution to Question 27

1. Circuit Diagram & Analysis

First, let’s visualize the circuit and identify the key nodes. The circuit consists of two main parallel branches connected across a battery of voltage $V$. Let us define the potential at the positive terminal (left rail) as $V$ and the negative terminal (right rail) as $0$.

Figure 1: Simplified schematic showing potentials nodes X, Y, Z, and P.

2. Calculating Potentials of Connection Points

We treat the voltmeters as ideal. However, since the problem implies a unique potential distribution for the node $Z$ where the three voltmeters meet, we consider them to be “identical” with a very high resistance $R_V$ ($R_V \to \infty$). Because the resistance is infinite, they do not draw significant current from the main circuit branches.

We can therefore calculate the potentials at nodes $X$, $Y$, and $P$ using simple voltage divider rules, unaffected by the voltmeters.

Top Branch (Nodes X and Y)

The top branch contains three identical resistors $R$ in series connected across $V$. The total resistance is $3R$. The potential drops equally across each resistor.

Potential at the Left Rail = $V$
Potential at node $X$ (after one $R$): $$V_X = V – \frac{V}{3} = \frac{2V}{3}$$
Potential at node $Y$ (after two $R$s): $$V_Y = V – \frac{2V}{3} = \frac{V}{3}$$

Bottom Branch (Node P)

The bottom branch contains a resistor $R$ and a resistor $2R$ in series. The total resistance is $R + 2R = 3R$.

Node $P$ is located between the resistor $R$ (connected to $V$) and the resistor $2R$ (connected to $0$).
Using the voltage divider rule: $$V_P = V \times \frac{2R}{R + 2R} = V \times \frac{2}{3} = \frac{2V}{3}$$

3. Analyzing the Voltmeter Network (Finding $V_Z$)

The three identical voltmeters meet at a central node $Z$. Since they are identical, we can model them as three equal resistors $R_V$ connecting node $Z$ to nodes $X$, $Y$, and $P$.

Applying Kirchhoff’s Current Law (KCL) at the floating node $Z$ (sum of currents leaving is zero):

$$ \frac{V_Z – V_X}{R_V} + \frac{V_Z – V_Y}{R_V} + \frac{V_Z – V_P}{R_V} = 0 $$

Since $R_V$ is common (and non-zero), we can cancel it out:

$$ (V_Z – V_X) + (V_Z – V_Y) + (V_Z – V_P) = 0 $$ $$ 3V_Z = V_X + V_Y + V_P $$ $$ V_Z = \frac{V_X + V_Y + V_P}{3} $$

Substitute the values we found earlier:

$$ V_Z = \frac{\frac{2V}{3} + \frac{V}{3} + \frac{2V}{3}}{3} = \frac{\frac{5V}{3}}{3} = \frac{5V}{9} $$

4. Final Calculations of Readings

The reading of an ideal voltmeter is the magnitude of the potential difference across it.

Voltmeter A

Connected between $X$ and $Z$:

$$ \text{Reading A} = |V_X – V_Z| = \left| \frac{2V}{3} – \frac{5V}{9} \right| = \left| \frac{6V}{9} – \frac{5V}{9} \right| = \frac{V}{9} $$

Voltmeter B

Connected between $Z$ and $P$:

$$ \text{Reading B} = |V_Z – V_P| = \left| \frac{5V}{9} – \frac{2V}{3} \right| = \left| \frac{5V}{9} – \frac{6V}{9} \right| = \left| -\frac{V}{9} \right| = \frac{V}{9} $$

Voltmeter C

Connected between $Z$ and $Y$:

$$ \text{Reading C} = |V_Z – V_Y| = \left| \frac{5V}{9} – \frac{V}{3} \right| = \left| \frac{5V}{9} – \frac{3V}{9} \right| = \frac{2V}{9} $$

Answer

The readings of the voltmeters are:

A: $\frac{V}{9}$
B: $\frac{V}{9}$
C: $\frac{2V}{9}$