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Questions  >>  CBSE XII  >>  Physics  >>  Current Electricity
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Q)

Given the resistance of $1\Omega,2 \Omega$ and $3 \Omega$ how will be combine them to get an equivalent resistance of $\large\frac{11}{5}$$\Omega$

$\begin{array}{1 1} \text{$R_1$ and $R_2$ in series and the resultant in parallel with $R_3$} \\ \text{$R_1$ and $R_3$ in series and the resultant in parallel with $R_2$} \\ \text{$R_2$ and $R_3$ in parallel and the resultant series in with $R_1$} \\ \text{$R_1$ and $R_2$ in parallel and the resultant series with $R_3$} \end{array} $

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Solution :
Let $R_1= 1\Omega,R_2= 2 \Omega$ and $R_3 =3 \Omega$
If $R_2$ and $R_3$ are in parallel.
then $ \large\frac{1}{R_p} =\frac{1}{R_2} +\frac{1}{R_3}$
$\large\frac{1}{2}+\frac{1}{3} =\frac{3+2}{6}$
$\qquad= \large\frac{5}{6}$
$R_p =\large\frac{6}{5}$$\Omega$
$R_p$ and $ R_1$ are in series . hence
$R= R_p +R_1$
$\quad= \large\frac{6}{5} $$+1$
$\qquad= \large\frac{11}{5}$$\Omega$
Answer : $\text{$R_2$ and $R_3$ in parallel and the resultant series in with $R_1$} $
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