Question

A galvanometer is used for making an ammeter of range 500 mA when a shunt of 2 \(\Omega\) is used. The same galvanometer is used for making a voltmeter of range 10V when a resistance of 470 \(\Omega\) is used in series. Then find the resistance of the galvanometer.

• A. 30 \(\Omega\)
• B. 50 \(\Omega\)
• C. 10 \(\Omega\)
• D. 100 \(\Omega\)

Hint:

For making an ammeter or voltmeter from a galvanometer, the resistance is adjusted by adding a shunt or a series resistance, respectively. Use the range formulas to calculate the required resistance for the galvanometer.

Answer 1

The Correct Option is B

For ammeter:
\[ \text{For ammeter: } \quad R_{\text{shunt}} = 2 \, \Omega \] The ammeter range is 500 mA, i.e., \(I_g = 500 \times 10^{-3} \, \text{A}\). Using the formula for the ammeter resistance: \[ I_g R_G = (500 \times 10^{-3} - I_g) \times 2 \] For voltmeter:
The same galvanometer is used for making a voltmeter with a 470 \(\Omega\) resistance in series. For the voltmeter, the total resistance is \(470 \, \Omega + R_G\). Using the formula for the voltage range: \[ I_g = \frac{470 + R_G}{R_G} = 10 \] Now, solving for \(R_G\) from the above equation: \[ 470 I_g + R_G I_g = 10 R_G \] Substitute \(I_g = \frac{9}{468}\): \[ 470 \times \frac{9}{468} + R_G \times \frac{9}{468} = 10 R_G \] \[ R_G = \frac{10 \times 468}{9} = 50 \, \Omega \] Thus, the resistance of the galvanometer \(R_G\) is 50 \(\Omega\). Final Answer: Option (B) 50 \(\Omega\).