For the mixed connection shown in the figure,derive the equation for the equivalent resistance.

(N/A) As shown in the figure,$R_{2}$ and $R_{3}$ are connected in parallel between points $B$ and $C$,and $R_{1}$ is connected in series with this parallel combination between points $A$ and $B$.
Let $R^{\prime}$ be the equivalent resistance of $R_{2}$ and $R_{3}$ connected in parallel.
Therefore,$\frac{1}{R^{\prime}} = \frac{1}{R_{2}} + \frac{1}{R_{3}}$.
$R^{\prime} = \frac{R_{2} R_{3}}{R_{2} + R_{3}}$ $.....(1)$
The equivalent resistance of the entire circuit is the sum of $R_{1}$ and $R^{\prime}$ as they are in series:
$R_{eq} = R_{1} + R^{\prime}$
$R_{eq} = R_{1} + \frac{R_{2} R_{3}}{R_{2} + R_{3}}$
$R_{eq} = \frac{R_{1} R_{2} + R_{1} R_{3} + R_{2} R_{3}}{R_{2} + R_{3}}$
If the voltage between $A$ and $C$ is $V$,the total current $I$ flowing through the circuit is:
$I = \frac{V}{R_{eq}} = \frac{V(R_{2} + R_{3})}{R_{1} R_{2} + R_{1} R_{3} + R_{2} R_{3}}$