A volmeter coil has resistance `50.0 Omega` and a resistor of `1.15 k Omega` is connected in series. It can read potential differences upto 12 V. If this same coil is used to construct an ammeter which can measure currents up to 2.0 A, what should be the resistance of the shunt used?

Answer

To solve the problem, we need to determine the resistance of the shunt resistor that will allow the voltmeter coil to function as an ammeter capable of measuring currents up to 2.0 A. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the total resistance of the voltmeter circuit
The voltmeter coil has a resistance of \( R_v = 50 \, \Omega \) and is connected in series with a resistor \( R_s = 1.15 \, k\Omega = 1150 \, \Omega \).

The total resistance \( R_{total} \) of the voltmeter circuit is:
\[ R_{total} = R_v + R_s = 50 \, \Omega + 1150 \, \Omega = 1200 \, \Omega \]

### Step 2: Calculate the maximum current through the voltmeter
Using Ohm's law, we can find the maximum current \( I_{max} \) that can flow through the voltmeter when it reads the maximum voltage of \( 12 \, V \):
\[ V = I_{max} \cdot R_{total} \]
Rearranging gives:
\[ I_{max} = \frac{V}{R_{total}} = \frac{12 \, V}{1200 \, \Omega} = 0.01 \, A = \frac{1}{100} \, A \]

### Step 3: Set up the ammeter circuit
When converting the voltmeter to an ammeter, we will connect a shunt resistor \( R_{sh} \) in parallel with the voltmeter coil. The total current \( I \) that the ammeter can measure is given as \( 2.0 \, A \).

### Step 4: Calculate the current through the voltmeter coil
The current through the voltmeter coil \( I_v \) will be \( \frac{1}{100} \, A \) (as calculated in Step 2). Therefore, the current through the shunt resistor \( I_{sh} \) can be calculated as:
\[ I_{sh} = I - I_v = 2.0 \, A - \frac{1}{100} \, A = 2.0 \, A - 0.01 \, A = 1.99 \, A \]

### Step 5: Use the current division rule to find the shunt resistance
Using the current division rule, we can relate the currents through the voltmeter coil and the shunt resistor to their resistances:
\[ \frac{I_v}{I_{sh}} = \frac{R_{sh}}{R_v} \]
Substituting the known values:
\[ \frac{\frac{1}{100} \, A}{1.99 \, A} = \frac{R_{sh}}{50 \, \Omega} \]
Cross-multiplying gives:
\[ R_{sh} = 50 \, \Omega \cdot \frac{1}{100} \cdot \frac{1}{1.99} \]
Calculating \( R_{sh} \):
\[ R_{sh} = 50 \cdot \frac{1}{199} \approx 0.251 \, \Omega \]

### Step 6: Convert to milli-ohms
To express this in milli-ohms:
\[ R_{sh} \approx 251 \, m\Omega \]

### Final Answer
The resistance of the shunt used should be approximately \( 251 \, m\Omega \).

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