plz anyone help in this numerical.....
Attachments:
Answers
Answered by
1
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
\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
sona561:
thanks
please mark as brainliest
let the BRAINLIEST option to come ....
i will :-)
Similar questions