We can change that formula around to find out the current of the entire circuit, aswell as current through single resistors
$$
\begin{flalign}
I &= I*R_{t} = \frac{V_{S}}{R_{t}} = \frac{V_{S}}{R_{1}+R_{2}}\\
\\
VR_{1} &= V_{S}(\frac{R1}{R1+R2})\\
VR_{2} &= V_{S}(\frac{R2}{R1+R2})\\
\\
VR_{1} &= 5(\frac{500}{500+1500})\\
VR_{1} &= 1.25v&\\
\\
VR_{2} &= 5(\frac{1500}{500+1500})\\
VR_{2} &= 3.75v
\end{flalign}
$$
### Example 2
```circuitjs
$ 1 0.000005 10.20027730826997 50 5 43 5e-11
v 192 288 192 112 0 0 40 10 0 0 1
x 125 208 167 211 4 24 10V
r 192 112 336 112 0 10
r 336 112 480 112 0 20
w 480 112 480 288 0
r 336 112 336 288 0 50
w 480 288 336 288 0
w 336 288 192 288 0
x 304 204 319 207 4 12 R3
x 256 140 271 143 4 12 R1
x 400 138 415 141 4 12 R2
x 332 104 340 107 4 12 A
x 332 304 340 307 4 12 B
x 205 102 215 105 4 12 I1
x 467 105 477 108 4 12 I2
x 346 276 356 279 4 12 I3
```
In this circuit we have three major loops we can apply [[formulas#Conservation of Energy Second Law|Kirchhoffs Second Law]] to, the one on the left, the one on the right and the most outer one. We can also use [[formulas#Conservation of Charge First Law | Kirchhoffs First Law]] for the node title **A**.
$$
\begin{flalign}
&\textit{Node A:} \\\
& I_{1} = I_{2}+I_{3} \\
\\
&\textit{Left Loop:} \circlearrowright &\\
& 10_{v} - I_{1}*R_{1} - I_{3}*R_{3} = 0& \\
\\
&\textit{Right Loop} \circlearrowleft &\\
&I_{2}*R_{2} - I_{3}*R_{3} = 0; &\\
\\
&\textit{Outer Loop:} \circlearrowright &\\
& 10_{v} - I_{1}*R_{1} - I_{2}*R_{2} = 0& \\
\\
\end{flalign}
$$
Now if we would like to find out I2 for example we can use the Right Loop Formula to do so:
