What is the current in `10 Omega` resistance in the following network ?

Options

A. 1A
B. 1.2A
C. 0.8A
D. 2A

Answer

A

Explanation

This is a balanced wheatstone's network. Current in the branch ABC is `I_(1)` and current in the branch ADC is `I_(2)`.
`(I_(1))/(I_(2))=(30)/(12)=(5)/(2) therefore I_(1)=2.5 I_(2) and I_(1)+I_(2)=1.4`
`therefore I_(2)=0.4 and I_(1)=1 A`