Kirchhoff’s Second Law

The sum of all the voltages, as you go around a circuit from some fixed point and return there from the opposite direction, and taking polarity into account, is always zero. Does this seem counter intuitive? Let’s think about it a little more carefully.

What Kirchhoff was expressing, when he wrote his second law, is the principle that voltage cannot appear out of nowhere, nor can it vanish. All the potential differences must ultimately cancel each other out in any closed dc circuit, no matter how complicated that circuit happens to be. This is Kirchhoff ’s Second Law.We can also call it Kirchhoff ’s Voltage Law or the principle of conservation of voltage. Remember the rule you’ve already learned about series dc circuits: The sum of the voltages across all the individual resistances adds up to the supply voltage. This statement is true as far as it goes, but it is an oversimplification, because it ignores polarity. The polarity of the potential difference across each resistance is opposite to the polarity of the potential difference across the battery. So when you add up the potential differences all the way around the circuit, taking polarity into account for every single component, you always get a net voltage of zero.

An example of Kirchhoff ’s Second Law is shown in below figure. The voltage of the battery, E, has polarity opposite to the sum of the potential differences across the resistors, E1 + E2 + E3 + E4. Therefore, E + E1 + E2 + E3 + E4 = 0.

Kirchhoff ’s Second Law. The sum of the voltages across the resistances is equal to, but has opposite polarity from, the supply voltage. Therefore, E + E1 + E2 + E3 + E4 = 0.

Problem 1

Refer to the diagram of above diagram. Suppose the four resistors have values of 50 Ω, 60 Ω, 70Ω, and 80 Ω, and that the current through each of them is 500 mA. What is the battery voltage, E?

Find the voltages E1, E2, E3, and E4 across each of the resistors. This can be done using Ohm’s Law. For E1, say with the 50-Ω resistor, calculate E1 = 0.500 × 50 = 25 V. In the same way, you can calculate E2 = 30 V, E3 = 35 V, and E4 = 40 V. The supply voltage is the sum E1 + E2 + E3 + E4 = 25 + 30 + 35 + 40 = 130 V. Kirchhoff ’s Second Law tells us that the polarities of the voltages across the resistors are in the opposite direction from that of the battery.

Problem 2

In the situation shown by above diagram, suppose the battery provides 20 V. Suppose the resistors labeled with voltages E1, E2, E3, and E4 have ohmic values in the ratio 1:2:3:4 respectively. What is the voltage E3?

This problem does not provide any information about current in the circuit, nor does it give you the exact resistances. But you don’t need to know these things to solve for E3. Regardless of what the actual ohmic values are, the ratio E1:E2:E3:E4 will be the same as long as the resistances are in the ratio 1:2:3:4. We can plug in any ohmic values we want for the values of the resistors, as long as they are in that ratio.

Let Rn be the resistance across which the voltage is En, where n can range from 1 to 4. Now that we have given the resistances specific names, suppose R1 = 1.0 Ω, R2 = 2.0 Ω, R3 = 3.0 Ω, and R4 = 4.0 Ω. These are in the proper ratio. The total resistance is R = R1 + R2 + R3 + R4 = 1.0 + 2.0 + 3.0 + 4.0 = 10 Ω. You can calculate the current as I = E/R = 20/10 = 2.0 A. Then the voltage E3, across the resistance R3, is given by Ohm’s Law as E3 = IR3 = 2.0 × 3.0 = 6.0 V.