Question:
A galvanometer of resistance $27\ \Omega$ is converted into an ammeter of range $(0 - 10\text{ mA})$ using a resistance of $3\ \Omega$. The galvanometer will show full scale deflection for a current of about
A galvanometer of resistance $27\ \Omega$ is converted into an ammeter of range $(0 - 10\text{ mA})$ using a resistance of $3\ \Omega$. The galvanometer will show full scale deflection for a current of about
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Ammeter conversion always reduces galvanometer current using a shunt resistor.
Updated On: Jul 2, 2026
- $10\text{ mA}$
- $100\text{ mA}$
- $1\text{ mA}$
- $3\text{ mA}$
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The Correct Option is C
Solution and Explanation
Concept:
For conversion of galvanometer into ammeter, a shunt resistance is connected in parallel.
Step 1: Given data \[ G = 27\ \Omega,\quad S = 3\ \Omega,\quad I = 10\ \text{mA} = 0.01\ \text{A} \]
Step 2: Current division principle At full scale: \[ I = I_g + I_s \] Using parallel relation: \[ I_g G = I_s S \Rightarrow I_s = \frac{I_g G}{S} \]
Step 3: Total current \[ I = I_g + \frac{I_g G}{S} \] \[ I = I_g\left(1 + \frac{G}{S}\right) \]
Step 4: Substitute values \[ I_g = \frac{0.01}{1 + \frac{27}{3}} = \frac{0.01}{1 + 9} \] \[ I_g = \frac{0.01}{10} = 0.001\ \text{A} \] \[ I_g = 1\ \text{mA} \] Final Answer: \[ \boxed{(C)} \]
Step 1: Given data \[ G = 27\ \Omega,\quad S = 3\ \Omega,\quad I = 10\ \text{mA} = 0.01\ \text{A} \]
Step 2: Current division principle At full scale: \[ I = I_g + I_s \] Using parallel relation: \[ I_g G = I_s S \Rightarrow I_s = \frac{I_g G}{S} \]
Step 3: Total current \[ I = I_g + \frac{I_g G}{S} \] \[ I = I_g\left(1 + \frac{G}{S}\right) \]
Step 4: Substitute values \[ I_g = \frac{0.01}{1 + \frac{27}{3}} = \frac{0.01}{1 + 9} \] \[ I_g = \frac{0.01}{10} = 0.001\ \text{A} \] \[ I_g = 1\ \text{mA} \] Final Answer: \[ \boxed{(C)} \]
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