Question

In the network shown, find the potential difference between $A$ and $B$.
$(R=r_1=r_2=r_3=1\Omega ,E_1=3V,E_2=2V,E_3=1V)$.

Answer 1

Given,

$R=r_1=r_2=r_3=1\Omega$

$E_1=3V,E_2=2V$ and $E_3=1V$

We know,

$V=I\cdot R$

$\therefore I=\frac{V}{R}=\frac{E}{R}$

So,

$I_1=\frac{E_1}{r_1}=\frac{3V}{1\Omega }=3Amp$

$I_2=\frac{E_2}{r_2}=\frac{2V}{1\Omega }=2Amp$

$I_3=\frac{E_3}{r_3}=\frac{1V}{1\Omega }=1Amp$

So, $I=I_1+I_2+I_3=3+2+1=6Amp$

Since $r_1,r_2$ and $r_3$ are in parallel

$\therefore \frac{1}{R^{\prime }}=\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}=\frac{1}{1}+\frac{1}{1}+\frac{1}{1}=\frac{1+1+1}{1}=\frac{3}{1}\Omega$

$\Rightarrow R^{\prime }=\frac{1}{3}\Omega$

and $R^{\prime }$ and $R$ are in series

$\therefore R_{eq}=R^{\prime }+R=\frac{1}{3}+1=\frac{1+3}{3}=\frac{4}{3}\Omega$

So, potential difference, $V=\frac{I}{R}=\frac{I}{R_{eq}}=\frac{6}{\frac{4}{3}}=\frac{18}{4}=4.5V$

The potential difference between A and B is $4.5V$