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

Understanding the Problem

We are given a galvanometer with a resistance of 54 Ω and a full-scale deflection current of 1 mA. We need to find the resistance R to convert it into a voltmeter reading 50 V, and the shunt resistance r to convert it into an ammeter reading 10 A.

Solution

For Voltmeter:

1. Current Through Voltmeter:

The current flowing through the voltmeter is given by \( I = \frac{V}{R + G} \), where \( G \) is the galvanometer resistance.

2. Substitute Values:

We are given \( V = 50 \, \text{V} \), \( G = 54 \, \Omega \), and \( I = 1 \, \text{mA} = 0.001 \, \text{A} \).

\( 0.001 = \frac{50}{R + 54} \)

3. Solve for R:

\( R + 54 = \frac{50}{0.001} = 50000 \)

\( R = 50000 - 54 = 49946 \, \Omega \)

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

For Ammeter:

1. Shunt Current:

The total current is \( I = 10 \, \text{A} \). The galvanometer current is \( I_g = 1 \, \text{mA} = 0.001 \, \text{A} \). The shunt current is \( I_s = I - I_g \).

\( I_s = 10 - 0.001 \approx 10 \, \text{A} \)

2. Voltage Across Galvanometer and Shunt:

The voltage across the galvanometer and the shunt resistor is the same.

\( I_g G = I_s r \)

3. Substitute Values and Solve for r:

\( (0.001)(54) = (10) r \)

\( r = \frac{0.001 \times 54}{10} = 0.0054 \, \Omega \)

\( r = 5.4 \, \text{m}\Omega \)

Corrected Answers:

The resistance R is approximately \( 50 \, \text{k}\Omega \).

The shunt resistance r is approximately \( 5.4 \, \text{m}\Omega \).