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

Options

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

Answer

To solve the problem step by step, we will follow these calculations:

### Step 1: Identify the components of the circuit
- The circuit consists of:
- A battery with a voltage of \( V = 30 \, \text{V} \)
- A resistor with a resistance of \( R = 40.8 \, \Omega \)
- An ammeter with a coil resistance of \( R_a = 480 \, \Omega \) and a shunt resistance of \( R_s = 20 \, \Omega \)

### Step 2: Calculate the equivalent resistance of the ammeter
The ammeter has a coil and a shunt connected in parallel. The equivalent resistance \( R_{a,eq} \) of the ammeter can be calculated using the formula for resistors in parallel:

\[ \frac{1}{R_{a,eq}} = \frac{1}{R_a} + \frac{1}{R_s} \]

Substituting the values:

\[ \frac{1}{R_{a,eq}} = \frac{1}{480} + \frac{1}{20} \]

Calculating the right side:

\[ \frac{1}{R_{a,eq}} = \frac{1}{480} + \frac{24}{480} = \frac{25}{480} \]

Now, taking the reciprocal to find \( R_{a,eq} \):

\[ R_{a,eq} = \frac{480}{25} = 19.2 \, \Omega \]

### Step 3: Calculate the total equivalent resistance in the circuit
The total resistance \( R_{total} \) in the circuit is the sum of the resistance of the ammeter (equivalent resistance) and the external resistor:

\[ R_{total} = R_{a,eq} + R = 19.2 + 40.8 = 60 \, \Omega \]

### Step 4: Calculate the current in the circuit
Using Ohm's law, the current \( I \) in the circuit can be calculated as:

\[ I = \frac{V}{R_{total}} = \frac{30 \, \text{V}}{60 \, \Omega} = 0.5 \, \text{A} \]

### Conclusion
The reading in the ammeter will be \( 0.5 \, \text{A} \).

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