A galvanometer of full-scale deflection 10 mA has a resistance of 5$\Omega$. Calculate the shunt needed to convert it to measure currents of 2 A.

Analyze the question

A galvanometer only handles a small current (10 mA). To measure 2 A, we connect a low-resistance shunt in parallel with it so that the extra current bypasses the coil.

Concept

In parallel, both branches have the same voltage:

I_g × R_g = I_s × R_s

where I_s = I − I_g

Solution

• Full-scale current: I_g = 10 mA = 0.010 A
• Galvanometer resistance: R_g = 5 Ω
• Total current wanted: I = 2 A
• Shunt current: I_s = 2 − 0.010 = 1.99 A

R_s = (I_g × R_g) / I_s = (0.010 × 5) / 1.99 = 0.05 / 1.99 ≈ 0.0251 Ω

So R_s ≈ 0.025 Ω.

Answer

The shunt needed is about 0.025 Ω, which is option A.

Tip

Quick approximation: since 2 A is much larger than 10 mA, nearly all the current goes through the shunt, so R_s ≈ (0.01 × 5)/2 = 0.025 Ω. Remember the shunt is always parallel and very small, while a multiplier for a voltmeter is series and large.