Possibility 1: Delta ($\Delta$) Configuration

Assume three resistors $R_{AB}, R_{BC}, R_{AC}$ form a triangle.

Analyze Experiment 1

Input: $20\text{ V}$ across A-B. Output: $8\text{ V}$ measured across B-C.

The voltmeter is across $R_{BC}$. The total resistance across the source is $(R_{AC} + R_{BC})$ in parallel with $R_{AB}$. Since the source is applied directly to A-B, the branch A-C-B acts as a potential divider.

$$ V_{BC} = V_{in} \times \frac{R_{BC}}{R_{AC} + R_{BC}} $$ $$ 8 = 20 \times \frac{R_{BC}}{R_{AC} + R_{BC}} \implies \frac{2}{5} = \frac{R_{BC}}{R_{AC} + R_{BC}} $$ $$ 2(R_{AC} + R_{BC}) = 5 R_{BC} \implies 2 R_{AC} = 3 R_{BC} \implies R_{AC} = 1.5 R_{BC} \quad \dots(1) $$

Analyze Experiment 2

Input: $20\text{ V}$ across B-C. Output: $15\text{ V}$ measured across A-C.

The source is across B-C. The branch B-A-C acts as a potential divider. We measure across $R_{AC}$.

$$ V_{AC} = V_{in} \times \frac{R_{AC}}{R_{AB} + R_{AC}} $$ $$ 15 = 20 \times \frac{R_{AC}}{R_{AB} + R_{AC}} \implies \frac{3}{4} = \frac{R_{AC}}{R_{AB} + R_{AC}} $$ $$ 3(R_{AB} + R_{AC}) = 4 R_{AC} \implies 3 R_{AB} = R_{AC} \implies R_{AC} = 3 R_{AB} \quad \dots(2) $$

Resulting Ratios

Let $R_{AB} = R$.

• From (2): $R_{AC} = 3R$.
• From (1): $3R = 1.5 R_{BC} \implies R_{BC} = 2R$.

Result: $R_{AB}=R, R_{BC}=2R, R_{AC}=3R$.

Possibility 2: Star (Y) Configuration

Assume three resistors $R_A, R_B, R_C$ meet at a central node $N$. Terminals A, B, C connect to the free ends of these resistors.

Analyze Experiment 1

Input: Source connected to A and B. Output: Measured across B and C.

• Current flows through $R_A$ and $R_B$.
• Terminal C is open (voltmeter has infinite resistance), so no current flows in $R_C$.
• Therefore, Potential at C = Potential at N ($V_C = V_N$).
• The measured voltage $V_{BC}$ is the potential difference between B and C. Since $V_C = V_N$, we are measuring $V_{BN}$ (the voltage drop across $R_B$).

$$ V_{drop(R_B)} = V_{in} \times \frac{R_B}{R_A + R_B} $$ $$ 8 = 20 \times \frac{R_B}{R_A + R_B} \implies \frac{2}{5} = \frac{R_B}{R_A + R_B} $$ $$ 2 R_A + 2 R_B = 5 R_B \implies 2 R_A = 3 R_B \implies R_A = 1.5 R_B \quad \dots(3) $$

Analyze Experiment 2

Input: Source connected to B and C. Output: Measured across A and C.

• Current flows through $R_B$ and $R_C$.
• Terminal A is open, so no current in $R_A$. Thus $V_A = V_N$.
• Measured voltage $V_{AC}$ is $|V_A – V_C| = |V_N – V_C|$. This is the voltage drop across $R_C$.

$$ V_{drop(R_C)} = V_{in} \times \frac{R_C}{R_B + R_C} $$ $$ 15 = 20 \times \frac{R_C}{R_B + R_C} \implies \frac{3}{4} = \frac{R_C}{R_B + R_C} $$ $$ 3 R_B + 3 R_C = 4 R_C \implies 3 R_B = R_C \quad \dots(4) $$

Resulting Ratios

We express all resistances in terms of $R_A$. From (3), let $R_A = R$.

• Then $R = 1.5 R_B \implies R_B = \frac{R}{1.5} = \frac{2}{3}R$.
• From (4): $R_C = 3 R_B = 3 \times \frac{2}{3}R = 2R$.

Result: Resistor at A = $R$, Resistor at B = $2R/3$, Resistor at C = $2R$.