1. Kirchhoff’s Laws Objective: to recall Kirchhoff’s laws and use them to solve problems. Starter: what is the current and voltage in these circuits? Mark the homework

2. Kirchhoff’s first law Kirchhoff’s first law states that: The sum of the currents leaving any junction is always equal to the sum of the currents that entered it. The German physicist Gustav Kirchhoff established two laws which help us to understand the function of electric circuits. This law is based upon the idea of the conservation of charge: no charge can be lost or made in a circuit. Thus the sum of the currents at a junction should be zero. I IN = I 1 + I 2 … + I n Σ I = 0 I IN I3I3 I2I2 I1I1

3. Using Kirchhoff’s first law

4. The energy in a circuit What happens to the energy supplied to a circuit? Batteries and power supplies supply electrical energy to a circuit. Devices within the circuit transduce this energy: bulbs produce heat and light, resistors produce heat. Energy cannot be created or destroyed. All of the energy provided by a power supply must be used by the circuit. Voltage is the energy transferred to the charge in a circuit. The battery’s voltage is shared between the components, which transduce this energy into different forms. What is the conservation of energy? How does the voltage of a battery relate to the voltage measured across the devices in a circuit?

5. Kirchhoff’s second law Kirchhoff’s second law is based upon the law of the conservation of energy. It states that: The total voltage across a circuit loop is equal to the sum of the voltage drops across the devices in that loop. Essentially, the energy you put into the circuit equals the energy you get out of each circuit loop. An equation can be produced for each loop in a circuit. For example: V IN = V 1 + V 2 V IN = I R 1 + I R 2 R1R1 I V IN V2V2 V1V1 R2R2

6. Simple uses of Kirchhoff’s second law

7. There are 3 loops in the circuit. Each has a voltage drop equal to the input voltage according to the 2nd law. Further uses for Kirchhoff’s law Use Kirchhoff’s laws to find the values for each current. 10 Ω 100 Ω 40 Ω I1I1 I2I2 I3I3 I E = 12 V Therefore: I = I 1 + I 2 + I 3 As I = I = 12 1010040 12 ++ I = 1.2 + 0.12 + 0.3 I = 1.6 A I 1 = 1.2 A I 2 = 0.1 A I 3 = 0.3 A The 1st law means that current entering each junction equals the current leaving. V R V3V3 V2V2 V1V1 Therefore: E = V 1 = V 2 = V 3

8. Current and drift velocity Current is a flow of charge. Electrical devices activate almost instantly once they are supplied with power, however the electrons actually move around a circuit quite slowly. Their velocity is called drift velocity. Current and drift velocity are linked by the following equation: I = nAve I = current (amps) n = charged particles per unit volume A = cross-sectional area (m 2 ) v = drift velocity (m/s) e = charge on an electron (1.6 x 10 -19 C)

9. Understanding I = nAve

10. Alternating current and direct current

11. RMS voltage The voltage of AC can be viewed using an oscilloscope. There are three common voltage measures, namely peak, peak-to-peak and RMS (root mean squared) voltage. RMS is a measure of the average magnitude of the voltage. V RMS = V PEAK √2 RMS voltage zero volts peak voltage peak-to-peak voltage

12. RMS current and RMS power To investigate voltage we use an oscilloscope connected across a resistor. As V  I, the equation for calculating RMS current is similar to the equation for RMS voltage: I RMS The equation for RMS power is a little different: P PEAK = I PEAK × V PEAK P RMS = I RMS × V RMS = P RMS = = I PEAK √2 × I PEAK √2 V PEAK √2 P PEAK 2

13. AC calculations

14. AC/DC summary