JEE Main · 2024 · 05 Apr 2024 · Shift 2 · Current Electricity

A galvanometer of resistance

100Ω

when connected in series with

400Ω

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 \times 1 0^{- 2} Ω

. The value of x is :

Answer: (C)

Solution

Full-scale current through the galvanometer, when in series with

400 Ω

i_{g} = \frac{10}{100 + 400} = 0.02 A

To convert to an ammeter reading up to

10

A, a shunt

S

is placed in parallel so that

i_{g} R_{g} = (i - i_{g}) S ⟹ 0.02 \times 100 = (10 - 0.02) S ⟹ S \approx \frac{2}{9.98} \approx 0.2 Ω = 20 \times 1 0^{- 2} Ω

x = 20

JEE Main · 2024 · 05 Apr 2024 · Shift 2 · Single correct option