Two identical galvanometers are converted into an ammeter and a milliammeter. If the shunt, which has more resistance, the current passing through the coil will be

Options

A. less
B. equal
C. more
D. zero

Answer

To solve the problem, we need to understand how a galvanometer is converted into an ammeter and a milliammeter using shunt resistors. Here’s a step-by-step solution:

### Step 1: Understand the setup
Two identical galvanometers are used, one converted into an ammeter (which measures larger currents) and the other into a milliammeter (which measures smaller currents). A shunt resistor is used in parallel with the galvanometer to allow most of the current to bypass the galvanometer.

### Step 2: Define the variables
Let:
- \( I \) = total current entering the circuit
- \( I_g \) = current through the galvanometer
- \( R_g \) = resistance of the galvanometer
- \( R_s \) = resistance of the shunt resistor

### Step 3: Apply Kirchhoff’s Law
According to Kirchhoff's law, the potential difference across the galvanometer and the shunt resistor must be the same when they are connected in parallel. Therefore, we can write:

\[ I_g \cdot R_g = (I - I_g) \cdot R_s \]

### Step 4: Analyze the effect of shunt resistance
If the shunt resistor \( R_s \) has a very large resistance compared to the galvanometer's resistance \( R_g \), then:

\[ R_s \gg R_g \]

This implies that most of the current \( I \) will pass through the galvanometer. As \( R_s \) increases, the term \( (I - I_g) \) becomes smaller, which means that \( I_g \) must increase to keep the equation balanced.

### Step 5: Conclude the relationship
Thus, if the shunt resistance \( R_s \) is very large, the current passing through the galvanometer \( I_g \) will be larger. Therefore, the current passing through the coil (the galvanometer) will increase as the shunt resistance increases.

### Final Answer
The current passing through the coil will be more if the shunt resistance is more.

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