Question:medium

A galvanometer of resistance $50\text{ }\Omega$ has $30$ divisions and a current sensitivity of $1\text{ mA/div}$. What should be the shunt resistance so that it can be converted into an ammeter of range $10\text{ A}$?

Show Hint

Since $I_g$ ($0.03\text{ A}$) is extremely small compared to the target range $I$ ($10\text{ A}$), you can approximate the denominator $(I - I_g) \approx I$. This simplifies the calculation to $S \approx \frac{I_g \cdot G}{I} = \frac{1.5}{10} = 0.15\text{ }\Omega$, giving the answer instantly.
Updated On: May 20, 2026
  • $3.15\text{ }\Omega$
  • $1.55\text{ }\Omega$
  • $0.15\text{ }\Omega$
  • $2.50\text{ }\Omega$
Show Solution

The Correct Option is C

Solution and Explanation

Understanding the Concept: To convert a delicate galvanometer of coil resistance $G$ into an ammeter capable of measuring large currents up to range $I$, a small shunt resistor ($S$) must be connected in parallel across it. The value of $S$ is calculated using the current division relationship: \[ I_g \cdot G = (I - I_g) \cdot S \implies S = \frac{I_g \cdot G}{I - I_g} \] where $I_g$ represents the full-scale deflection current capacity of the galvanometer.
Step 1: Calculate full-scale deflection current limit ($I_g$).
We are given:
Total scale divisions $= 30\text{ div}$
Deflection factor $= 1\text{ mA/div} = 1 \times 10^{-3}\text{ A/div}$
Coil internal resistance, $G = 50\text{ }\Omega$
Evaluating total safe full-scale operational current boundary lines: \[ I_g = 30\text{ div} \times 1\text{ mA/div} = 30\text{ mA} = 0.03\text{ A} \]
Step 2: Substitute parameters into the parallel shunt expression.
Target extended line current limit, $I = 10\text{ A}$: \[ S = \frac{0.03 \times 50}{10 - 0.03} = \frac{1.5}{9.97} \approx 0.15045\text{ }\Omega \approx 0.15\text{ }\Omega \]
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