Question

A circuit contains an ammeter, a battery of $30\, V$ and a resistance $40.8\, ohm$ all connected in series. If the ammeter has a coil of resistance $480\, ohm$ and a shunt of $20\, ohm$, the reading in the ammeter will be

• A. 2 A
• B. 1 A
• C. 0.5 A
• D. 0.25 A

Answer 1

The Correct Option is C

To solve this problem, we need to determine the reading on the ammeter in the given circuit configuration. Here, the circuit consists of a battery, a resistance, and an ammeter in series. Let's analyze the situation step by step.

1. The total resistance of the ammeter is given by its coil resistance and the shunt resistance connected in parallel. The formula for parallel resistance \( R_{\text{parallel}} \) is: \[ R_{\text{parallel}} = \frac{R_1 \times R_2}{R_1 + R_2} \] where \( R_1 = 480 \, \text{ohm} \) (coil resistance) and \( R_2 = 20 \, \text{ohm} \) (shunt resistance).
2. Substituting the given values: \[ R_{\text{parallel}} = \frac{480 \times 20}{480 + 20} = \frac{9600}{500} = 19.2 \, \text{ohm} \]
3. Now, consider the total resistance in the circuit which includes the resistance of 40.8 ohms and the combined resistance of the ammeter: \[ R_{\text{total}} = 40.8 + 19.2 = 60 \, \text{ohm} \]
4. Using Ohm’s Law, the current \( I \) flowing through the circuit is given by: \[ I = \frac{V}{R_{\text{total}}} \] \] where \( V = 30 \, \text{V} \) is the voltage of the battery.
5. Substituting the values: \[ I = \frac{30}{60} = 0.5 \, \text{A} \]

Therefore, the reading on the ammeter will be 0.5 A. This matches the given correct answer.