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Q.
The equivalent resistance between $A$ and $B$ for the circuit shown in figure is
Current Electricity
Solution:
For equivalent resistance between $A$ and $B$. $5\,\Omega$ and $8\,\Omega$ resistances are connected in series. $R$', their equivalent resistance is parallel to $6\,\Omega$
$\therefore R'=5+8=13\,\Omega$ and $\frac{1}{R''}=\frac{1}{13}+\frac{1}{6}$
$=\frac{6+13}{78}=\frac{19}{78}$
$R''=\frac{78}{19}$
Now $4\,\Omega$, $R$'' and $5\,\Omega$ resistances are connected in series equivalent resistance between $A$ and $B$.
$\therefore R_{eq}=4+\frac{78}{19}+5$
$=\frac{76+78+95}{19}=13.1\,\Omega$
