Question

If any ammeter is shunted, then the total resistance of the circuit

• A. increases
• B. decreases
• C. remains same
• D. none of these

Hint:

Remember the rule for parallel resistors: adding a resistor in parallel to any part of a circuit always decreases the total equivalent resistance of that part. A shunt is simply a resistor added in parallel.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
An ammeter is a device used to measure current in a circuit and is always connected in series. A "shunted ammeter" refers to an ammeter (or more accurately, a galvanometer) that has a low-resistance resistor, called a shunt, connected in parallel with it. This is done to extend the range of the ammeter. The question asks what effect this has on the total resistance of the circuit.
Step 2: Key Formula or Approach:
When resistors are connected in parallel, the equivalent resistance (\(R_{eq}\)) is always less than the smallest individual resistance. The formula for two resistors in parallel is:
\[ R_{eq} = \frac{R_1 R_2}{R_1 + R_2} \] Here, \(R_1\) is the ammeter's internal resistance (\(R_A\)), and \(R_2\) is the shunt resistance (\(R_S\)). The effective resistance of the shunted ammeter is \(R'_{A} = \frac{R_A R_S}{R_A + R_S}\).
Step 3: Detailed Explanation:
1. An ammeter is placed in series in a circuit. Let the original total resistance of the circuit be \(R_{total} = R_{circuit} + R_A\), where \(R_A\) is the ammeter's resistance and \(R_{circuit}\) is the resistance of the rest of the circuit.
2. When the ammeter is shunted, a low-resistance shunt (\(R_S\)) is connected in parallel to the ammeter's internal resistance (\(R_A\)).
3. The new effective resistance of the measuring instrument, \(R'_{A}\), is the parallel combination of \(R_A\) and \(R_S\).
4. Since the equivalent resistance of a parallel combination is always smaller than the smallest of the individual resistances, and a shunt has very low resistance, we have \(R'_{A}<R_S\). And since \(R_S\) is chosen to be much smaller than \(R_A\), it is certain that \(R'_{A}<R_A\).
5. The new total resistance of the circuit becomes \(R'_{total} = R_{circuit} + R'_{A}\).
6. Since \(R'_{A}<R_A\), it follows that \(R'_{total}<R_{total}\).
Therefore, shunting an ammeter decreases its own effective resistance, which in turn decreases the total resistance of the circuit it is part of.
Step 4: Final Answer:
If an ammeter is shunted, the total resistance of the circuit decreases. Option (B) is correct.