DC Circuits Notes PDF

Module 3: Electrical Fundaments Topic 3.6: DC Circuits  INTRODUCTION On completion of this topic you should be able to: 3.6.1 Describe the following laws: Ohms Law Kirchoff’s voltage and current Laws...

Module 3: Electrical Fundaments Topic 3.6: DC Circuits  INTRODUCTION On completion of this topic you should be able to: 3.6.1 Describe the following laws: Ohms Law Kirchoff’s voltage and current Laws 3.6.2 Using Ohms Law and Kirchoff’s Laws, perform calculations to find resistance voltage and current in DC circuits. 3.6.3 Describe the significance of the internal resistance of a DC power supply. 30-04-2024 Slide No. 2  OHM’S LAW Dr Georg Simon Ohm ( 1787 - 1854 ) German physicist of 19th century Ohm discovered that: Voltage, Current, and Resistance are related to each other in a definite way. He developed a law that governs the operation of all DC circuits: “The amount of current flowing in a circuit made up of pure resistances is directly proportional to the potential difference or voltage impressed on the circuit and inversely proportional to the total resistance of the circuit.” Where V = Voltage, I = Current and R = Resistance. 30-04-2024 Slide No. 3  OHM'S LAW Equivalent expressions of Ohm's law are: V = IR or R = V/I. An easy way to remember the formulas is by using this diagram. Horizontal line in middle means to divide 2 remaining values. "X" in bottom section of circle means to multiply I and R values: If calculating voltage, cover V and you have I X R remaining (V= I x R) If calculating current, cover I, and you have V divided by R (I=V/R) If calculating resistance, cover R, and you have V divide by I left (R=V/I) Note: The letter E is sometimes used instead of V for voltage – E is for EMF. 30-04-2024 Slide No. 4  OHM'S LAW Problem: A nine volt battery supplies power to an iron with a resistance of 18 ohms. How much current is flowing through the curling iron? V = 9 Volt R = 18 ohm battery Curling iron Solution: 0.5 Amp 30-04-2024 Slide No. 5  OHM'S LAW Problem: A 110 volt wall outlet supplies power to a light with a resistance of 2200 ohms. How much current is flowing through the light? V = 110 Volts R = 2200 ohms Solution: 0.05 Amp A component has a resistance of 40 ohms and a current flow of 0.1 amps. How many volts are supplied to the component? Solution: 4 Volts 30-04-2024 Slide No. 6  KIRCHOFF’S CURRENT LAW Gustav Robert Kirchhoff (1824-1887) German physicist of 19th century Kirchoff's Current Law The sum of currents flowing into a junction equals the sum of currents flowing away from the junction I1 = I2 + I3 I1 is flowing into the junction whereas I2 and I3 are flowing out If I1 was 20 Amps and I3 was 5 Amps, what would I2 equal? 15 Amps 30-04-2024 Slide No. 7  KIRCHOFF’S CURRENT LAW The sum of currents flowing into a junction equals the sum of currents flowing away from the junction 8 Amps enter junction and 3 and 5 Amps leave junction. Makes a total of 8 Amps entering and 8 Amps leaving. 8 Amps and 1 Amp enter junction and 9 Amps leave. Makes a total of 9 Amps entering and 9 Amps leaving. 8 Amps and 1 Amp enter junction and 7 Amps and 2 Amps leave. Makes a total of 9 Amps entering and 9 Amps leaving. The sum of currents flowing into a junction equals the sum of currents flowing away from the junction. 30-04-2024 Slide No. 8  KIRCHOFF’S VOLTAGE LAW 10V Kirchoff's Voltage Law The algebraic sum of the voltage drops in any closed path in a circuit and the electromotive forces in that path is equal to zero. Also stated as the sum of voltage drops around a closed circuit is equal to the sum of voltage sources. Input Voltage's) = sum of voltage drops. In above circuit – Voltage drops across R1, R2 and R3 must equal 10 V 10 V = V1 + V2 + V3 30-04-2024 Slide No. 9  KIRCHOFF’S VOLTAGE LAW Voltage drop across 3 resistors equals the supply voltage: E1 + E2 + E3 = EA 30-04-2024 Slide No. 10  SERIES AIDING AND OPPOSING SOURCES Circuit may contain more than one source of emf. Emf sources in same direction – series aiding. Emf sources in opposite direction – series opposing. Effective source voltage is algebraic sum. When 2 opposing sources are inserted into circuit current flow is in direction determined by larger source. 30-04-2024 Slide No. 11  SERIES AIDING AND OPPOSING SOURCES Using Kirchhoff's voltage equation, what is the total current is flowing in the circuit? 1.5 Amps 30-04-2024 Slide No. 12  KIRCHOFF’S LAW Current flow at a junction will divide into 2 parts. Current through the respective branch can be worked out as shown. 30-04-2024 Slide No. 13  KIRCHOFF’S LAW 30-04-2024 Slide No. 14  KIRCHOFF’S AND OHM’S LAWS I2, I3 and unknown resistance RX can all be calculated, using basic DC theory. Voltage drop on R1 20 ohm resistor is (I1 x R1 ) or 8 volts. By Kirchoff’s voltage law, p.d. across R2 is thus (20 – 8 or) 12 volts. Using Ohms law, current through R2 10 ohm resistor (V ÷ R) is 1.2 Amps (I3). 30-04-2024 Slide No. 15  KIRCHOFF’S AND OHM’S LAWS Using Kirchoffs current law and now knowing I1 and I3, I2 is found. I3 = I1 + I2 therefore I2 = 800 mA. Using Kirchoffs voltage law, p.d. across RX can be calculated. 20 = (I2 x RX) + 12 or same voltage over parallel legs (R1//RX). Voltage across RX (I2 x RX) is then 8 volts. Value of RX is (V ÷ I) or 8 ÷ 0.8 or 10 ohms. 30-04-2024 Slide No. 16  DC POWER SUPPLY INTERNAL RESISTANCE All power sources have an internal resistance (usually small). As more current is drawn – voltage drop across internal resistance increases. This decreases output voltage from supply. So when more current is drawn from power source – voltage provided drops. Higher the internal resistance – the greater the connected load will affect output. 30-04-2024 Slide No. 17  DC POWER SUPPLY INTERNAL RESISTANCE 0.2 Ω 12 Volts 1 kΩ If a 1 kΩ load is connected to power supply, what is the voltage supplied to it? Solution = 11.99 Volts (circuit current draw of 0.01199 amps) 30-04-2024 Slide No. 18  DC POWER SUPPLY INTERNAL RESISTANCE 0.2 Ω 12 Volts 1.8 Ω If a 1.8 Ω load is connected to power supply, what is the voltage supplied to it? Solution = 10.8 Volts (circuit current draw of 6 amps) So when more current is drawn from power source – voltage provided drops. Higher the internal resistance – the greater the connected load will affect output. 30-04-2024 Slide No. 19  CONCLUSION Now that you have completed this topic, you should be able to: 3.6.1 Describe the following laws: Ohms Law Kirchoff’s voltage and current Laws 3.6.2 Using Ohms Law and Kirchoff’s Laws, perform calculations to find resistance voltage and current in DC circuits. 3.6.3 Describe the significance of the internal resistance of a DC power supply. 30-04-2024 Slide No. 20  This concludes: Module 3: Electrical Fundaments Topic 3.6: DC Circuits