Question

plz anyone help in this numerical.....

Answer 1

Section ABOC is a wheatstone bridge. Therefore,
$\frac{2}{3}=\frac{X}{6}\\ \Rightarrow X=4\Omega$
Now, calculate net resistance, ignoring resistor AO as no current is flowing through it.
$R_{BAC}=2+4=6\Omega\\ R_{BOC}=3+6=9\Omega\\ R_{BC}=\frac{9\times6}{9+6}\\=\frac{54}{15}=3.6\Omega\\ R=2.4+3.6=6\Omega\\ I=\frac{V}{R}=\frac{6}{6}=1A$

Answer 2

Section ABOC is a wheatstone bridge. Therefore,
\begin{lgathered}\frac{2}{3}=\frac{X}{6}\\ \Rightarrow X=4\Omega\\\end{lgathered}32​=6X​⇒X=4Ω​ 
Now, calculate net resistance, ignoring resistor AO as no current is flowing through it.
\begin{lgathered}R_{BAC}=2+4=6\Omega\\ R_{BOC}=3+6=9\Omega\\ R_{BC}=\frac{9\times6}{9+6}\\=\frac{54}{15}=3.6\Omega\\ R=2.4+3.6=6\Omega\\ I=\frac{V}{R}=\frac{6}{6}=1A\\\end{lgathered}RBAC​=2+4=6ΩRBOC​=3+6=9ΩRBC​=9+69×6​=1554​=3.6ΩR=2.4+3.6=6ΩI=RV​=66​=1A​