Question

Three resistances of $1\Omega ,2\Omega$ and $3\Omega$ are connected in series combination. Find out equivalent resistance of the combination. If this combination is connected by the battery of 12 V e.m.f. and negligible internal resistance then find out the voltage across each ends of resistance.

Answer 1

Connecting 1 $\Omega$, 2 $\Omega$ and 3 $\Omega$ in series
$R_{equi}=R_1+R_2+R_3=1+2+3=6\Omega$
Current flowing in circuit
$I=\frac{V}{R_{equi}}=\frac{12}{6}=2amp$
Potential on 1 $\Omega$ resistance = $V=IR=2\times 1=2V$
Potential on 2 $\Omega$ resistance $V=IR=2\times 2=4V$
Potential on 6 $\Omega$ resistance $\left(V\right)=IR=2\times 3=6V$