Question

A galvanometer has a resistance of \( 50\,\Omega \) and gives a full-scale deflection for a current of \( 2\,\text{mA} \). To convert this galvanometer into an ammeter capable of measuring currents up to \( 2\,\text{A} \), what is the value of the shunt resistance that must be connected in parallel with it?

• A. \( 0.05\,\Omega \)
• B. \( 0.50\,\Omega \)
• C. \( 0.02\,\Omega \)
• D. \( 5.00\,\Omega \)

Hint:

To save time during competitive exams, whenever you see that \( I_g \) is less than \( 1% \) of the total target current \( I \), drop \( I_g \) from the denominator term entirely. This simplifies your calculation down to a basic division: \( S \approx \frac{I_g G}{I} \).

Answer 1

The Correct Option is A

Concept: A moving coil galvanometer is converted into an ammeter by connecting a very low resistance value, known as a shunt resistance (\( S \)), in parallel with its coil. This parallel setup creates an alternative low-resistance path, allowing the bulk of the main circuit current to bypass the delicate galvanometer coil. Since the galvanometer coil and the shunt resistor are connected in parallel, the electrical potential difference across both branches must be exactly equal: \[ V_g = V_s \quad \Rightarrow \quad I_g G = (I - I_g)S \] Rearranging this relationship allows us to solve directly for the unknown shunt resistance: \[ S = \frac{I_g G}{I - I_g} \] where \( G \) is the internal galvanometer coil resistance, \( I_g \) is the full-scale deflection current, and \( I \) is the maximum target current limit of the ammeter.

Step 1: Convert all given variables into standard SI units. Identify the given experimental values from the problem statement:
• Galvanometer resistance, \( G = 50\,\Omega \)
• Full-scale current, \( I_g = 2\,\text{mA} = 2 \times 10^{-3}\,\text{A} = 0.002\,\text{A} \)
• Maximum target ammeter current, \( I = 2\,\text{A} \)

Step 2: Substitute parameters into the parallel shunt equation. Plug the values into our isolated shunt formula: \[ S = \frac{0.002 \times 50}{2 - 0.002} \] Calculate the value in the numerator: \[ 0.002 \times 50 = 0.1\,\text{V} \] Calculate the value in the denominator: \[ 2 - 0.002 = 1.998\,\text{A} \] Substitute these numbers back into the fraction: \[ S = \frac{0.1}{1.998} \]

Step 3: Simplify the division to find the final resistance value. Because the full-scale current \( I_g \) is extremely small compared to the total current \( I \) (\( 0.002\,\text{A} \ll 2\,\text{A} \)), we can approximate the denominator difference as \( I - I_g \approx I \): \[ S \approx \frac{0.1}{2} = 0.05\,\Omega \] Performing the exact division confirms the approximation: \( \frac{0.1}{1.998} = 0.05005\,\Omega \), which rounds precisely to \( 0.05\,\Omega \).