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A circuit contains an ammeter, a battery of 30 V30\text{ V} and a resistance 40.8Ω40.8\,\Omega all connected in series. If the ammeter has a coil of resistance 480Ω480\,\Omega and a shunt of 20Ω20\,\Omega then reading in the ammeter will be :

A

0.5A

B

0.25A

C

2A

D

1A

Step-by-Step Solution

To find the ammeter reading, we first determine the total resistance of the circuit. According to the sources, an ammeter consists of a galvanometer coil and a shunt resistance connected in parallel .

  1. Calculate the ammeter resistance (RaR_a): The coil (480Ω480\,\Omega) and shunt (20Ω20\,\Omega) are in parallel. Their effective resistance is: Ra=Rcoil×rsRcoil+rs=480×20480+20=9600500=19.2ΩR_a = \frac{R_{\text{coil}} \times r_s}{R_{\text{coil}} + r_s} = \frac{480 \times 20}{480 + 20} = \frac{9600}{500} = 19.2\,\Omega .

  2. Calculate the total circuit resistance (RtotalR_{\text{total}}): The ammeter and the external resistor (40.8Ω40.8\,\Omega) are connected in series . Rtotal=Rext+Ra=40.8+19.2=60.0ΩR_{\text{total}} = R_{\text{ext}} + R_a = 40.8 + 19.2 = 60.0\,\Omega.

  3. Apply Ohm's Law to find the current (II): The ammeter reading represents the total current flowing through the series circuit . I=VRtotal=30 V60Ω=0.5 AI = \frac{V}{R_{\text{total}}} = \frac{30\text{ V}}{60\,\Omega} = 0.5\text{ A}.

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