Question

plwase i want the answer please

Answer 1

let
$V_{1} = 10V$
$V_{2} = 2V$
$R_{1} = 10 \ohm$
$R_{2} = 5 \ohm$

The circuit has to be simplified such that there will be placed only single cell of emf and single internal resistance.

We have,
From Effective Resistance Formula,
$\frac{1}{R_{eff}}= \frac{1}{R_{1}}+\frac{1}{R_{2}}$

$\frac{1}{R_{eff}} = \frac{1}{10}+ \frac{1}{5}$

$\frac{1}{R_{eff}} = \frac{1+2}{10}$

$\frac{1}{R_{eff}} = \frac{3}{10}$

$R_{eff} = \frac{10}{3}$

We have,
From Effective emf formula,

$I_{eff} = I_{1} + I_{2}$

We have,
From Ohms Law,
$V = IR$
$\implies I= \frac{V}{R}$

$\frac{V_{eff}}{R_{eff}} = \frac{V_{1}}{R_{1}}+\frac{V_{2}}{R_{2}}$

$\frac{V_{eff}}{R_{eff}} = \frac{10}{10} + \frac{2}{5}$

$\frac{V_{eff}}{\frac{10}{3}} = 1+\frac{2}{5}$

$\frac{V_{eff}}{\frac{10}{3}} = \frac{7}{5}$

$V_{eff} = \frac{10}{3} × \frac{7}{5}$

$V_{eff} = \frac{70}{15}$

$V_{eff} = 4.67V$