Condition for balanced wheatstone-bridge,
$\frac{P}{Q}=\frac{R}{S} $
$\Rightarrow \frac{5}{1}=\frac{50}{10}$
$\Rightarrow \frac{5}{1}=\frac{5}{1}$
So, it is a balanced wheatstone-bridge and hence no current will flow through the arm $G$. The, circuit can be redrawn as given below,
As, we know that in a series branch, same current flows.
Hence, from the KVL loop rule,
$V_{P Q}=I_{1}(5+1)=6l_1$...(i)
and $V_{R S}=(50+10)\left(1.1-l_{1}\right)$(ii)
Since, for the parallel branches, the voltage drop across them remains same, So, $V_{P Q}=V_{R S}$
$\Rightarrow 6l_1 =60\left(1.1-l_{1}\right)$ (using Eqs. (i) and(ii))
$\Rightarrow l_{1}(60+6) =66$
$\Rightarrow l_{1}=1 A$
Hence, current of 1 A passes through $1 \Omega$ resistance.
