The current in a simple series circuit is 5.0 A. When an additional resistance of `2.0 Omega` is inserted, the current drops to 4.0 A. The original resistance of the circuit in ohms was

Options

A. 1.25
B. 8
C. 10
D. 20

Answer

B

Explanation

Let the original resistance is `Romega`
V=l R
V =5 x R x=5 R …(i)
When `2Omega` resistance is inserted, then total resistance
= `(R + 2)Omega`
V = l'(R + 2) = 4 (R + 2) …(ii)
From Eqs.(i) and (ii), we get
5R = 4(R + 2)
`:." " R = 8Omega`