In the circuit shown, the internal resistance of the cell is negligible. The steady state current in the 2`Omega` resistor is

Options

A. 0.6A
B. 1.2A
C. 0.9A
D. 1.5A

Answer

C

Explanation

No current will flow through branch CD.
Therefore, from figure (ii), the resultant resistance is `R=(2xx3)/(2+3)+2.8Omega=1.2+2.8=4Omega`.
`therefore` The current in circuit is
=`i=(6V)/(4Omega)=1.5A`.
Thus, from figure (iii),
`2xxi_(1)=(i-i_(1))3`
`rArr`   `2i_(1)+3i_(1)=3i`
`rArr`   `i_(1)=(3xx1.5)/(5)=0.9A`