Kirchhoff's 2nd Law (The Loop or Mesh Law)

Kirchoff's 2nd Law - The sum of the potential differences around any closed circuit or network is zero. - This is the case because the sum of the ohms each resistor has is added together and then the amount of each resistor is then a fraction of that total. The fraction of this total is then multiplied by the voltage amount and then that is the amount the resistor reduces the voltage. - In this case resistor 1 (R1) is 10ohms, resistor 2 (R2) is 30ohms and the DC voltage is 6V. R1 + R2 = Total Resistance 10ohms + 30ohms = 40ohms R1 is a quarter of the total resistance and R2 is three quarters of the total resistance. Therefore the resistance amount just past the first resistor is a quarter of the starting voltage 6V. 1/4 of 6V = 1.5V Then just past the first resistor is 4.5V because the resistor resisted 1.5V. Using this knowledge we can figure out the resistance of R2. 3/4 of 6V = 4.5V That means that past both resistors the voltage is cut back to 0V, just like the Kirchhoff's Law says.