Solution 3 – Experimental Analysis

Problem 3: Determining Circuit Parameters

To determine the internal circuit, we analyze the Voltage ($V$) vs Current ($I$) graphs for three configurations. We look for the “simplest possible circuit,” which typically implies branches connecting a common node.

1. Analyze Experiment II

Setup: Input at Terminal B. Terminals A and C are Grounded (0 V).

Graph Analysis: The V-I graph is a straight line passing through the origin. This indicates a purely resistive path.

Calculation: Looking at the graph (middle), when $V = 5\text{ V}$, the current $I$ is approximately $0.83\text{ A}$ (5/6). The slope gives the equivalent resistance between node B and the ground.

$$ R_{eq(B)} = \frac{\Delta V}{\Delta I} \approx \frac{5}{0.83} = 6.0\ \Omega $$

Interpretation: Since A and C are both grounded, this $6.0\ \Omega$ is the resistance between B and the common ground point (A).

2. Analyze Experiment III

Setup: Input at Terminal C. Terminals A and B are Grounded (0 V).

Graph Analysis: The V-I graph is linear but starts with a non-zero Y-intercept.

Intercept (EMF): At $I=0$, voltage $V = 3.0\text{ V}$. This intercept represents an internal battery opposing the input (positive terminal at C).
$\Rightarrow E = 3.0\text{ V}$.
Slope (Resistance): The graph rises from $3\text{ V}$ to $5\text{ V}$. The corresponding change in current is $\Delta I = 1.0\text{ A}$.

$$ R_{eq(C)} = \frac{\Delta V}{\Delta I} = \frac{5 – 3}{1 – 0} = \frac{2}{1} = 2.0\ \Omega $$

Interpretation: The path from C to ground contains a $3.0\text{ V}$ battery and a $2.0\ \Omega$ resistor in series.

3. Analyze Experiment I

Setup: Input at Terminal A. Terminals B and C are Grounded.

Graph Analysis: The graph shows a straight line with a low slope.

Calculation: Using the graph data points:

$$ \text{For } \Delta V = 3\text{ V}, \Delta I = 2\text{ A} $$ $$ R_{eq(A)} = \frac{\Delta V}{\Delta I} = \frac{3\text{ V}}{2\text{ A}} = 1.5\ \Omega $$

Consistency Check: Let’s verify if this matches our findings from Experiments II and III. If we assume the components found above connect to a common node A:

Branch A-B: $6\ \Omega$ (from Exp II)
Branch A-C: $2\ \Omega$ (from Exp III – considering differential resistance/slope)

When measuring resistance from A with B and C grounded, these two branches appear in parallel.

$$ R_{parallel} = \frac{R_{AB} \times R_{AC}}{R_{AB} + R_{AC}} = \frac{6 \times 2}{6 + 2} = \frac{12}{8} = 1.5\ \Omega $$

Conclusion: The calculated slope of $1.5\ \Omega$ perfectly matches the theoretical parallel resistance. This confirms the internal components are correctly identified.

4. Final Circuit Synthesis

Combining the results, we construct the simplest circuit:

Terminal B: Connected to Node A via a $6.0\ \Omega$ resistor.
Terminal C: Connected to Node A via a series combination of a $2.0\ \Omega$ resistor and a $3.0\text{ V}$ Battery.
Node A: Acts as the common connection point.