In the adjoining circuit, the current I drawn from the 5 volt source will be

Options

A. 0.67 A
B. 0.17A
C. 0.33A
D. 0.5A

Answer

D

Explanation

The given circuit is redrawn as follows.

This is a balanced Wheatstone's network. No current flows between B and D. Hence the resistance of `10 Omega` is not effective. Thus two resistance of `5+10=15 Omega and 10+20=30 Omega` are in parallel.
`therefore` Their effective resistance `R_(Ef)=(15xx30)/(15+30)=10 Omega`
`therefore` Current drawn from the source
`I=(V)/(R )=(5)/(10)=0.5 A