# Electric Circuit Analysis/Simple Resistive Circuits

| Lesson Review: Lesson 1 The first lesson was about passive sign convention. The lesson introduced circuit components which will be encountered in electric circuit analysis. Active components Passive components Passive sign convention Guidelines for passive sign convention Lesson Preview This lesson is about simple resistive circuits. The student is expected to have understood the following at the end of the lesson: Voltage: (V or v - Volts)The electrical potential between two points in a circuit. Current: (I or i - Amperes)The amount of charge flowing through a part of a circuit. Power: (W - Watts)Simply P = IV. It is the current times the voltage. Source: A voltage or current source is the supplier for the circuit. Resistor: (R measured in Ω - Ohms)A circuit element that "constricts" current flow. | Lessons in Electric Circuit Analysis Lesson #1: Passive Sign Convention Lesson #2: Simple Resistive Circuits← You are here Lesson #3: Resistors in Series Lesson #4: Resistors in Parallel Quiz Test: Circuit Analysis Quiz 1 Lesson #5: Kirchhoff's Voltage Law Lesson #6: Kirchhoff's Current Law Lesson #7: Nodal Analysis Lesson #8: Mesh Analysis Quiz Test: Circuit Analysis Quiz 2 Home Laboratory: Circuit Analysis - Lab1 |
| Part 1 Voltage Source This is possibly the simplest circuit. The voltage source supplies a voltage to the circuit. When this voltage is applied over a resistor, R, there is a current. Equation 2.1 $V=I*R\!$ This equation explains the relation between all three elements in the circuit. In this case the voltage source has the same magnitude as the voltage drop across the resistor. We know that it is V. The resistor has a certain amount of Ohms depending on its rating. We now know R. With algebra I = V/R. So as long as you know two of the variables then you can find the third. Now comes the power part of the circuit analysis. Equation 2.2 $P=I*V\!$ Once Equation 2.1 is solved then this equation should follow quickly. The I and V are the same variables so insert them into the equation and solve for P (Watts). With these two equations, 1.1 and 1.2, and a little bit of algebra you get: Equation 2.3 $P=I^{2}*R\!$ Equation 2.4 $P={\frac {V^{2}}{R}}$ | Part 2 As an explanation, the power running through is the voltage times the current. This is instantaneous power rather than power used over time. Power has to be supplied and consumed. In a perfect world without heat-loss both are equal. The source supplies the required power that is consumed in this case by the resistor. Example 2.1 Figure 2.1: Voltage Source Given': $V_{s}~=~12V,~R~=~20\Omega$ Find: I, the current in Amps. The power produced by the source. The power consumed by the resistor. Solution: Using the equations: ${\begin{aligned}I&=V/R\\&=12V/20\Omega \\I&=.6A\\\end{aligned}}$ ${\begin{aligned}P&=I*V\\&=.6A*12V\\P&=7.2W\\\end{aligned}}$ Remember: The power supplied equals the power consumed. |
| Part 3 Current Source All that will happen here is that the givens will change. Rather than knowing what the voltage is across the resistor we now know what the current is flowing through the resistor. Don't forget in the description of resistors that a similar model in fluid physics is a smaller pipe that constricts the amount of flow. Well, current is flow of charge. With the fluid the side of the smaller section being supplied with fluid will have a greater pressure than the out flowing side. The difference between these is potential. In circuits this potential is known as voltage, but then again this is all review, right? So now we use equation 2.1 again. The current source gives us the current through the resistor. Given the resistor value it should be just a matter of multiplication. | Part 4 Example: 2.1 Figure 2.2: Current Source Given': $I_{s}~=~2A,~R~=~10\Omega$ Find: V, the voltage. The power produced by the source. The power consumed by the resistor. Solution: Using the equations: ${\begin{aligned}V&=I*R\\&=2A*10\Omega \\V&=20V\\\end{aligned}}$ ${\begin{aligned}P&=I*V\\&=2A*20V\\P&=40W\\\end{aligned}}$ Of course, power consumed equals power supplied in this perfect universe, which is only found in this course. |
| Part 5: More Examples Example 2.2 Given': $V_{s}~=~5V,~R~=~50\Omega$ Find: I, the current in Amps. The power produced by the source. The power consumed by the resistor. Solution: Using the equations: ${\begin{aligned}I&=V/R\\&=5V/50\Omega \\I&=.1A\\\end{aligned}}$ ${\begin{aligned}P&=I*V\\&=.1A*5V\\P&=.5W\\\end{aligned}}$ | Part 6 Example 2.3 Given': $I_{s}~=~.3A,~R~=~30\Omega$ Find: V, the voltage. The power produced by the source. The power consumed by the resistor. Solution: Using the equations: ${\begin{aligned}V&=I*R\\&=.3A*30\Omega \\V&=9V\\\end{aligned}}$ ${\begin{aligned}P&=I*V\\&=.3A*9V\\P&=2.7W\\\end{aligned}}$ |
| Part 7 Example 2.4 Given': $I_{s}~=~5A,~P_{consumed}~=~30W$ Find: V, the voltage. R, the resistance Solution: Using the equations: ${\begin{aligned}V&=P/I\\&=30W/5A\\V&=6V\\\end{aligned}}$ ${\begin{aligned}R&=V/I\\&=6V/5A\\R&=1.2\Omega \\\end{aligned}}$ | Part 8 Example 2.5 Given': $V_{s}~=~12V,~P~=~5W$ Find: I, the current in Amps. R, the resistance. Solution: Using the equations: ${\begin{aligned}I&=P/V\\&=5W/12V\\I&=0.4167A\\\end{aligned}}$ ${\begin{aligned}R&=V/I\\&=12V/0.4167A\\R&=28.798\Omega ,~rounded~to~29\Omega .\\\end{aligned}}$ |
| Part 9 Example 2.6 Given': $P=32W,~R=2\Omega$ Find: V, the voltage. I, the current. Solution: Using the equations: ${\begin{aligned}P&=I^{2}*R\\I&={\sqrt {P/R}}\\&={\sqrt {32/2}}\\I&=4A\\\end{aligned}}$ OR ${\begin{aligned}P&=V^{2}/R\\V&={\sqrt {P*R}}\\&={\sqrt {32*2}}\\V&=8V\\\end{aligned}}$ OR ${\begin{aligned}V&=I*R\\\end{aligned}}$ | Part 10: Exercise 2 If the given current is 300 Amps and the resistance is 2 Ohms, what is the Voltage across the resistor and how much power is being produced or consumed? An engineer measures the resistivity of a resistor before putting it into a simple circuit. It is 50 Ohms. After putting the resistor into place the engineer measures 2 Volts across the resistor. How many Amps are going through the resistor? A 60 Watt bulb is found to have 300 Ohms of resistance. What is the necessary voltage and current to have the bulb run optimally? A large voltage supply has 10,000V. The company wants to know how big a resistance can be put on the voltage supply along with how much power will be consumed. Is this a solvable problem? Practice drawing the elements of a simple resistive circuit. Draw a resistor and Ω 5 times. Draw 5 current sources and 5 voltage sources. Answers to Exercise 2 Completion list Once you finish your Exercises you can post your score here! To post your score just e-mail your course co-ordinator your name and score *Click Here. Ozzimotosan -- 100% Doldham -- 75% & Corrected … … |
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