© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Understand the Circuit
The circuit consists of a battery of 30 V, a resistor of 40.8 Ω, and an ammeter. The ammeter itself has a coil of resistance 480 Ω and a shunt of 20 Ω connected in parallel internally. All components (the 40.8 Ω resistor and the ammeter) are connected in series with the battery.
Step 2: Calculate the Ammeter’s Effective Resistance
Since the coil and the shunt are connected in parallel inside the ammeter, their combined (effective) resistance is given by the parallel combination formula:
$R_\text{ammeter} = \frac{R_\text{coil} \times R_\text{shunt}}{R_\text{coil} + R_\text{shunt}}$
Substituting $R_\text{coil} = 480\,\Omega$ and $R_\text{shunt} = 20\,\Omega$:
$R_\text{ammeter} = \frac{480 \times 20}{480 + 20} = \frac{9600}{500} = 19.2\,\Omega$
Step 3: Determine the Total Circuit Resistance
The total resistance in the circuit is the sum of the external resistor and the effective resistance of the ammeter:
$R_\text{total} = 40.8\,\Omega + 19.2\,\Omega = 60\,\Omega$
Step 4: Apply Ohm’s Law to Find the Current
Ohm’s law states $I = \frac{V}{R}$. Therefore, the current in the circuit is:
$I = \frac{30\,\text{V}}{60\,\Omega} = 0.5\,\text{A}$
Step 5: Final Answer
Hence, the reading on the ammeter is 0.5 A.
Reference Diagram:
