Question

When a galvanometer is shunted with a 4 Ω resistance, the deflection is reduced to one-fifth. If the galvanometer is further shunted with a 2 Ω wire. The further reduction (find the ratio of decrease in current to the previous current) in the deflection will be (the main current remains the same)

Options

• `(8/13)` of the deflection when shunted with 4 Ω only

• `(5/13)` of the deflection when shunted with 4 Ω only

• `(3/4)` of the deflection when shunted with 4 Ω only

• `(3/13)` of the deflection when shunted with 4 Ω only

Answer 1

`(8/13)` of the deflection when shunted with 4 Ω only.

Explanation:

Case I:

`"Rg"xx1/5 = ("I"-"I"/5)xx4`

⇒ R_g = 16 Ω

Case II:

16 I₁ = `(4xx2)/6`(I - I₁) ⇒ I₁ = I/13

So decrease in current to previous current

= `("I"//5-"I"//13)/("I"//5) = 8/13`