Question

When a resistance of 100 Ω is connected in series with a galvanometer of resistance R, its range is y. To double its range a resistance of 1000 Ω is connected in series. Find R.

• A. 700 Ω
• B. 800 Ω
• C. 900 Ω
• D. 100 Ω

Answer 1

Answer: (B)

800 Ω

Answer 2

Let the galvanometer resistance be $R$ and the full-scale current be $I_g$.

1. With $100\,\Omega$ series resistance:

   $$y = I_g (R + 100).$$

2. With $1000\,\Omega$ series resistance, to double the range:

   $$2y = I_g (R + 1000).$$

Dividing the second equation by the first: $\frac{2y}{y} = \frac{R + 1000}{R + 100} \quad \Longrightarrow \quad 2 = \frac{R + 1000}{R + 100}.$

Cross-multiplying: $2(R + 100) = R + 1000 \quad \Longrightarrow \quad 2R + 200 = R + 1000.$

Solving for $R$: $2R - R = 1000 - 200 \quad \Longrightarrow \quad R = 800\,\Omega.$

The galvanometer resistance R is 800 Ω.