Physics Solution Q28

Solution to Question 28

Step 1: Model the Circuit and Ammeter

Let the external circuit consist of an EMF $E$ and a resistance $R$. The true current without any ammeter is: $$ I_0 = \frac{E}{R} $$ When an ammeter of resistance $R_A$ is inserted, the measured current is: $$ I_{meas} = \frac{E}{R + R_A} $$

Step 2: Relate Ammeter Resistance to Range

For a multi-range ammeter using a galvanometer with shunts, the full-scale voltage drop $V_{fs}$ across the ammeter terminals is constant regardless of the selected range. $$ V_{fs} = I_{range} \times R_{A,range} = \text{constant} $$ Therefore, the resistance of the ammeter is inversely proportional to its range: $$ R_A \propto \frac{1}{\text{Range}} $$

Let $R_{A1}$ be the resistance for the 1.0 mA range and $R_{A2}$ be the resistance for the 3.0 mA range. $$ \frac{R_{A1}}{R_{A2}} = \frac{\text{Range}_2}{\text{Range}_1} = \frac{3.0}{1.0} = 3 $$ So, $R_{A1} = 3R_{A2}$. Let $R_{A2} = x$, then $R_{A1} = 3x$.

Step 3: Formulate Equations for Measurements

Case 1 (1.0 mA range): The reading is $I_1 = 1.0 \text{ mA}$. $$ I_1 = \frac{E}{R + R_{A1}} \implies 1.0 = \frac{E}{R + 3x} \implies E = 1.0(R + 3x) \quad \dots(1) $$

Case 2 (3.0 mA range): The reading is $I_2 = 1.5 \text{ mA}$. $$ I_2 = \frac{E}{R + R_{A2}} \implies 1.5 = \frac{E}{R + x} \implies E = 1.5(R + x) \quad \dots(2) $$

Step 4: Solve for R in terms of x

Equating (1) and (2): $$ 1.0(R + 3x) = 1.5(R + x) $$ $$ R + 3x = 1.5R + 1.5x $$ $$ 1.5x = 0.5R $$ $$ R = 3x $$

Step 5: Calculate True Current

Substitute $R = 3x$ into equation (1) to find $E$: $$ E = 1.0(3x + 3x) = 6x $$ The true current $I_0$ is: $$ I_0 = \frac{E}{R} = \frac{6x}{3x} = 2.0 \text{ mA} $$

Final Answer

The true current $I_0$ in the circuit without the milli-ammeter is 2.0 mA.