A galvanometer of internal resistance 50 and full scale deflection of 20 mA is converted into a voltmeter by…

Analyze the question: A galvanometer is turned into a voltmeter by adding a large resistor in series (a multiplier). We need the maximum voltage the voltmeter can read, which is when the galvanometer carries its full-scale current.

Concept: In a series circuit, the same current flows through both resistors. The total resistance is the sum of the two, and V = I × R_total.

Solution:

• Galvanometer resistance: G = 50 Ω
• Series resistor: R = 1950 Ω
• Total resistance: R_total = 50 + 1950 = 2000 Ω
• Full-scale current: I = 20 mA = 20 × 10⁻³ A = 0.02 A

V = I × R_total V = 0.02 × 2000 V = 40 V

Check: The voltage across the galvanometer alone is 0.02 × 50 = 1 V, and across the resistor it is 0.02 × 1950 = 39 V. Together that gives 1 + 39 = 40 V. ✓

Tip: Do not forget to include the galvanometer's own resistance. Option C (38 V) is a trap for students who forget a step or miscalculate. Always use the total resistance (G + R) and convert mA to A before multiplying.