## In This post you will get answer of Q.

A current of 0.25 A flows in the main circuit. Now, the resistance R is disconnected and then connected across the $4,Omega$ resistance. The current in the circuit now is

Solution:

## Here 0.25 =$frac{2}{left(frac{2R}{2 + R} right) +4 +3}$ i.e. R = 2 $Omega$

In the second case, I =$frac{2}{2+ left(frac{2 times 4}{4 +2} right) + 3} = frac{6}{19} A$