A galvanometer having full scale deflection current of 10 mA and resistance `8 Omega` is joined in circuit as shown. When the terminals A and B are used and C remains open, the range of the ammeter is I. When the terminals A and C are used and B remains open, then the range of the ammeter is

Options

A. `I`
B. `(I)/(2)`
C. `(I)/(3)`
D. `(2I)/(3)`

Answer

To solve the problem step by step, we will analyze the circuit configurations and apply the principles of current division and resistance in series and parallel.

### Step 1: Understand the Circuit Configuration
When terminals A and B are used with terminal C open, the galvanometer is connected in such a way that it measures the current flowing through the circuit. The galvanometer has a full-scale deflection current of 10 mA and a resistance of 8 ohms.

### Step 2: Calculate the Total Resistance in the Circuit (AB Configuration)
In the configuration with terminals A and B, we have:
- The galvanometer resistance (R_g) = 8 ohms
- Two resistors of 2 ohms each in parallel with the galvanometer.

The total resistance (R_total) when using A and B can be calculated as follows:
1. The two 2-ohm resistors in parallel give a combined resistance (R_parallel):
\[ R_{\text{parallel}} = \frac{R_1 \cdot R_2}{R_1 + R_2} = \frac{2 \cdot 2}{2 + 2} = 1 \text{ ohm} \]
2. The total resistance in the circuit when A and B are used:
\[ R_{\text{total}} = R_g + R_{\text{parallel}} = 8 + 1 = 9 \text{ ohms} \]

### Step 3: Determine the Current through the Galvanometer (AB Configuration)
Using Ohm's law, the current flowing through the galvanometer (I_g) when the total current (I) is flowing through the circuit can be expressed as:
\[ I_g = \frac{I \cdot R_{\text{parallel}}}{R_g + R_{\text{parallel}}} \]
Substituting the values:
\[ I_g = \frac{I \cdot 1}{9} = \frac{I}{9} \]

### Step 4: Analyze the Second Configuration (AC Configuration)
When terminals A and C are used and terminal B is open, the configuration changes:
- The galvanometer is still 8 ohms.
- The two 2-ohm resistors are now in series, giving a total resistance of 4 ohms.

### Step 5: Calculate the Total Resistance in the Circuit (AC Configuration)
The total resistance (R_total) when using A and C:
\[ R_{\text{total}} = R_g + R_{\text{series}} = 8 + 4 = 12 \text{ ohms} \]

### Step 6: Determine the Current through the Galvanometer (AC Configuration)
Using Ohm's law again, the current through the galvanometer (I_g) when the total current (I') is flowing through the circuit can be expressed as:
\[ I_g = \frac{I' \cdot R_{\text{series}}}{R_g + R_{\text{series}}} \]
Substituting the values:
\[ I_g = \frac{I' \cdot 4}{12} = \frac{I'}{3} \]

### Step 7: Equate the Galvanometer Currents from Both Configurations
Since the galvanometer measures the same current (I_g) in both configurations, we can set the two equations equal:
\[ \frac{I}{9} = \frac{I'}{3} \]

### Step 8: Solve for I'
Cross-multiplying gives:
\[ 3I = 9I' \implies I' = \frac{I}{3} \]

### Conclusion
The range of the ammeter when terminals A and C are used is:
\[ \text{Range} = I' = \frac{I}{3} \]