## Working principle

According to Kirchoff’s Voltage Law (KVL), the sum of all voltages around a loop is equal to zero. When going around the loop, intuitively, you can treat the voltage source as a positive value, and the resistors as negative, voltage-consuming, values. In this simulation, the input voltage equals the sum of the voltage drops across R_{1} and R_{2}: V_{in} - V_{R1} - V_{R2} = 0. In other words, V_{in} = V_{R1} + V_{R2}.

You can find the voltage across R_{2} by using the voltage divider rule. First, use the equation for determining R_{eq }for two unequal resistors from the resistor network model (this also applied for equal valued resistors, though those can be solved without this equation):

Next, use the voltage divider equation to find V_{R2:}

Additionally, the voltage across R_{2 }and R_{3} is equal because these resistors are connected in parallel: V_{R2} = V_{R3}.

According to Kirchoff’s Current Law (KCL), the sum of all currents entering a node equals to the sum of all currents leaving it. The current I_{R1} in this simulation divides into two - I_{R2} and I_{R3} – and is, thus, equal to their sum: I_{R1} - I_{R2} - I_{R3} = 0. In other words, I_{R1} = I_{R2} + I_{R3}.

By Ohm's Law, current through each resistor will be equal to the voltage across the resistor divided by its resistance. This simulation shows that current flows through the path of least resistance (there is more current flowing through R_{2} than R_{3}): V = IR_{1} = I_{2}R_{2} = I_{3}R_{3}.

This model also lists the amount of power dissipated by each resistor. You can verify that the power dissipated equals to the current running through a resistor times the voltage across it.

### Experiments

- Equate R
_{2}and R_{3}values. What is the current through these resistors in relation to the current through R_{1 }now? - Change the R
_{2 }or R_{3 }value to zero ohms. What is the current through the remaining two non-zero resistors now?