Question

Three resistors of resistances $2\Omega ,3\Omega$ and $4\Omega$ are combined in series. What is the total resistance of the combination?

Answer 1

$V=iR$

In series ,current remains smc

$v\alpha R$

$v=v_1+v_2+v_3$

$i{R}_{tot}=i{R}_1+i{R}_2+i{R}_3$

$i{R}_{tot}=1+2+3=6\Omega$

$v=v_1+v_2+v_3$

also $v=i{R}_{tot}$

$=i\left(6\right)$

$i=2$

$\therefore v_1=i{R}_1=2\left(1\right)=2v$

$v_2=i{R}_2=2\left(2\right)=4v$

$v_3=i{R}_3=2\left(3\right)=6v$