Question

For designing a voltmeter of range 50 V and an ammeter of range 10 mA using a galvanometer which has a coil of resistance 54Ω showing a full scale deflection for 1 mA as in figure.

(A) for voltmeter R ≈ 50 kΩ  
(B) for ammeter r ≈ 0.2Ω  
(C) for ammeter r ≈ 6Ω  
(D) for voltmeter R ≈ 5 kΩ (E) for voltmeter R ≈ 500Ω 
Choose the correct answer from the options given below :
 

• (A) and (C)
• (A) and (B)
• (C) and (D)
• (C) and (E)

Answer 1

The Correct Option is A

To solve this problem, we need to find the appropriate resistance values for converting a galvanometer into a voltmeter of range 50 V and an ammeter of range 10 mA.

Step 1: Calculating Resistance for Voltmeter

To convert a galvanometer into a voltmeter, a high resistance \( R \) needs to be connected in series with the galvanometer.

The formula for the series resistance \( R \) required for increasing the range to \( V \) volts is given by:

R = \frac{V}{I_g} - G

Where:

• V = 50 \, \text{V} (desired voltage range)
• I_g = 1 \, \text{mA} = 0.001 \, \text{A} (full-scale deflection current of galvanometer)
• G = 54 \, \Omega (resistance of the galvanometer)

Plugging in the values, we get:

R = \frac{50}{0.001} - 54 = 50000 - 54 \approx 49946 \, \Omega

This is approximately equal to 50 \, \text{k}\Omega, so the correct option for the voltmeter is:

(A) for voltmeter R \approx 50 \, \text{k}\Omega

Step 2: Calculating Resistance for Ammeter

To convert a galvanometer into an ammeter, a shunt resistance \( r \) needs to be connected in parallel with the galvanometer.

The formula for the shunt resistance \( r \) required is given by:

r = \frac{I_g \times G}{I - I_g}

Where:

• I = 10 \, \text{mA} = 0.01 \, \text{A} (desired current range)

Plugging in the values, we get:

r = \frac{0.001 \times 54}{0.01 - 0.001} = \frac{0.054}{0.009} = 6 \, \Omega

Thus, the correct option for the ammeter is:

(C) for ammeter r \approx 6 \, \Omega

Conclusion

The correct answer is:

(A) and (C)