Solution to Question 15
Circuit Analysis
Let the potential at the positive terminal of the battery (left side) be $V$ and the negative terminal (right side) be $0$. Let us denote the node between $R_1$ and $R_2$ as Node $D$ and the node between $R_2$ and $R_3$ as Node $C$.
Since the ammeters are ideal (zero resistance):
- The top ammeter connects the main source node (Potential $V$) directly to Node $C$. Therefore, the potential at Node $C$ is $V$.
- The bottom ammeter connects Node $D$ directly to the return node (Potential $0$). Therefore, the potential at Node $D$ is $0$.
This nodal analysis reveals that all three resistors are effectively connected in parallel across the potential difference $V$:
- Resistor $R_1$: Connected between source ($V$) and Node $D$ ($0$). Potential difference $= V$.
- Resistor $R_2$: Connected between Node $C$ ($V$) and Node $D$ ($0$). Potential difference $= V$.
- Resistor $R_3$: Connected between Node $C$ ($V$) and return ($0$). Potential difference $= V$.
Let $I_1, I_2,$ and $I_3$ be the currents flowing through $R_1, R_2,$ and $R_3$ respectively. Since they are in parallel across the same voltage:
$$ I_1 = \frac{V}{R_1}, \quad I_2 = \frac{V}{R_2}, \quad I_3 = \frac{V}{R_3} $$Ammeter Equations
Using Kirchhoff’s Current Law (KCL) at Nodes $C$ and $D$:
- Top Ammeter ($0.2\,\text{A}$): This ammeter supplies current to Node $C$. The current leaving Node $C$ flows into $R_2$ and $R_3$. $$ I_{A1} = I_2 + I_3 = 0.2\,\text{A} \quad \dots \text{(i)} $$
- Bottom Ammeter ($0.3\,\text{A}$): This ammeter collects current from Node $D$. The current entering Node $D$ comes from $R_1$ and $R_2$. $$ I_{A2} = I_1 + I_2 = 0.3\,\text{A} \quad \dots \text{(ii)} $$
The total current supplied by the battery, $I_{\text{total}}$, is the sum of currents through all parallel branches:
$$ I_{\text{total}} = I_1 + I_2 + I_3 $$Using the “Interchange” Condition
The problem states that interchanging the places of two resistances leaves the ammeter readings unaffected. This implies that the currents associated with the swapped resistors must satisfy a specific symmetry.
From equations (i) and (ii), we can express the total current as:
$$ I_{\text{total}} = (I_1 + I_2) + (I_2 + I_3) – I_2 = 0.3 + 0.2 – I_2 = 0.5 – I_2 $$We must analyze which pair of resistors can be swapped without changing the readings.
Case 1: Interchanging $R_1$ and $R_2$
If we swap $R_1$ and $R_2$, the new currents $I_1’$ and $I_2’$ simply swap values ($I_1′ = I_2$ and $I_2′ = I_1$). For the readings to remain unaffected:
- Reading 1 ($I_2′ + I_3$) must equal original Reading 1 ($I_2 + I_3$). This requires $I_1 + I_3 = I_2 + I_3 \implies I_1 = I_2$.
- Reading 2 ($I_1′ + I_2’$) becomes $I_2 + I_1$, which is automatically equal to the original $I_1 + I_2$.
Thus, this swap is valid if $I_1 = I_2$.
Substituting into equation (ii):
$$ 2I_1 = 0.3 \implies I_1 = 0.15\,\text{A} \implies I_2 = 0.15\,\text{A} $$
Substituting $I_2$ into equation (i):
$$ 0.15 + I_3 = 0.2 \implies I_3 = 0.05\,\text{A} $$
Total Battery Current:
$$ I_{\text{total}} = 0.15 + 0.15 + 0.05 = \mathbf{0.35\,\text{A}} $$
Case 2: Interchanging $R_2$ and $R_3$
If we swap $R_2$ and $R_3$, we require $I_2 = I_3$.
- Reading 1 ($I_2 + I_3$) becomes $I_3 + I_2$ (automatically satisfied).
- Reading 2 ($I_1 + I_2$) becomes $I_1 + I_3$. For this to equal $0.3$, we need $I_3 = I_2$.
Thus, this swap is valid if $I_2 = I_3$.
Substituting into equation (i):
$$ 2I_2 = 0.2 \implies I_2 = 0.10\,\text{A} \implies I_3 = 0.10\,\text{A} $$
Substituting $I_2$ into equation (ii):
$$ I_1 + 0.10 = 0.3 \implies I_1 = 0.20\,\text{A} $$
Total Battery Current:
$$ I_{\text{total}} = 0.20 + 0.10 + 0.10 = \mathbf{0.40\,\text{A}} $$
Case 3: Interchanging $R_1$ and $R_3$
This would require $I_1 = I_3$.
From original equations: $I_1 = 0.3 – I_2$ and $I_3 = 0.2 – I_2$.
If $I_1 = I_3$, then $0.3 – I_2 = 0.2 – I_2$, which implies $0.3 = 0.2$. This is impossible.
Answer
Depending on which pair of resistances allows the interchange (the condition specified in the problem), there are two possible values for the supply current:
- If $R_1$ and $R_2$ are the swappable pair: 0.35 A
- If $R_2$ and $R_3$ are the swappable pair: 0.4 A