1. 11.6 Kirchhoff’s Laws In 1845, German physicist Gustov Kirchhoff developed two important laws when investigating circuits. These laws describe the behavior of electric potential difference (voltage) and current in an electric circuit. When engineers design circuits, they use Kirchhoff’s laws to understand what will happen to voltage and current as the circuit operates.

2. 11.6 Kirchhoff’s Voltage Law Kirchhoff’s Voltage Law (KVL): In any complete loop in an electric circuit, the total electric potential increase (voltage gain) at the source is equal to the total electric potential decreases (voltage drops) throughout the rest of the circuit loop. A series circuit has only one complete path, so the loads must share the voltage: A parallel circuit has more than one path, so voltage drops at each load must be the same as the voltage gain at the source:

3. 11.6 Kirchhoff’s Current Law Kirchhoff’s Current Law (KCL): In any closed circuit, the amount of current entering a junction is equal to the amount of current exiting the junction. In electric circuits, junctions are points where the current can split to follow more than one path. A series circuit has only one path, so only one possible current: A parallel circuit has more than one path, so the current can split, depending on the number of paths:

4. 11.6 Applying Kirchhoff’s Laws Series Circuit: Voltages add up & currents remain constant. If three identical light bulbs are connected in series, the voltage is divided equally among the three light bulbs. So for a 6.0 V battery, each light bulb would have a voltage of 2.0 V. There is no junction in a series, thus there is only one complete path with which all of the current must flow through. If the battery has a current of 0.30 A, all of the light bulbs will have a current of 0.30 A.

5. 11.6 Applying Kirchhoff’s Laws Parallel Circuit: Voltages remain constant & currents add up. If three identical light bulbs are connected in parallel, then the voltage must be the same for all three bulbs. For a 6.0 V battery, each light bulb would have a voltage of 6.0 V. There are three independent paths for the parallel circuit, thus the current is divided equally along each path. If the battery has a current of 0.30 A, all of the light bulbs will have a current of 0.10 A. SP #1,2 p.521

6. 11.6 Homework Practice # 1, 2 p.522 Questions # 1, 2 p.522