A galvanometer of resistance 100 Ω when connected in series with 400 Ω measures a voltage of upto 10 V. The value of resistance required to convert the galvanometer into ammeter to read upto 10 A is x × 10⁻² Ω. The value of x is :
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Question

A galvanometer of resistance 100 Ω when connected in series with 400 Ω measures a voltage of upto 10 V. The value of resistance required to convert the galvanometer into ammeter to read upto 10 A is x × 10⁻² Ω. The value of x is :

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G1 Physics · Accepted Answer

Galvanometer current for full-scale deflection: I(g) = 10/(400 + 100) = 0.02 A = 20 × 10⁻³ A Let shunt resistance be S. For conversion into an ammeter: I(g)G = [I - I(g)]S Since I = 10 A and I(g) is very small compared with 10 A: 20 × 10⁻³ × 100 ≈ 10S S ≈ 20 × 10⁻² Ω Therefore: x = 20