Question

What is the circuit (amps) for a series circuit containing a 12 V potential and three resistors R1= 3 ohms,

R2= 2 ohms and R3= 1 ohm.

Answer 1

Gives

R₁=3 $\Omega$

R₂=2 $\Omega$

R₃=1$\Omega$

E=12 V

Now

$R_{net}=R_{1}+R_{2}+R_{3}$ $R_{net}=(3+2+1) \Omega$

$R_{net}=6\Omega$

$I_{net}=\frac{V}{R_{net}}\rightarrow(1)$

$I_{net}=\frac{12}{6} =2 A$

$I_{net}=2 A$