Basic Laws of Electric Cicuits Lesson 2

Independent Voltage and Current Sources

Dependent Voltage Sources Dependent voltage-controlled voltage source Dependent current-controlled voltage source

Dependent Current Sources Dependent voltage-controlled current source Dependent current-controlled current source

Series and Parallel Connections In a series connection, current remains the same I I I I In a parallel connection, voltage remains the same + V - + V - + V -

Resistance Resistance is denoted by “R” Its unit is “Ohm ()” In a circuit, it is shown as: Resistance is the capacity of materials to obstruct the flow of current

Property of Resistance: Resistivities of some basic materials Material Resistivity (ohm meters) Common Use silver 1.6x10-8 conductor copper 1.7x10-8 conductor aluminum 2.8x10-8 conductor gold 2.5x10-8 conductor carbon 4.1x10-5 semiconductor germanium 47x10-2 semiconductor silicon 6.4x102 semiconductor paper 1x1010 insulator mica 5x1011 insulator glass 1x1012 insulator teflon 3x1012 insulator

Basic Laws of Circuits Ohm’s Law Ohm’s Law: The voltage across a resistor is directly proportional to the current moving through the resistor.   Ohm’s Law

Basic Laws of Circuits Ohm’s Law: Directly proportional means a straight line relationship. v(t) R v(t) = Ri(t) i(t)

Basic Laws of Circuits Ohm’s Law: About Resistors: The unit of resistance is ohms( ). So, the current equals to the voltage divided by the resistance The reciprocal of the resistance is referred to as conductance S (Siemens)  

Power calculations  

Example 2.3

Assessment problem 2.3

Useful Circuit Symbols (a) Short circuit. (b) Open circuit. Resistance and current in (a) and (b)? What would happen if you connect the two holes of an electrical socket with a wire? Why an electrical appliance starts working when plugged into a socket? (c) Switch.

Flashlight + –

Circuit Model for a Flashlight + –

Basic Laws of Circuits Kirchhoff’s Current Law The algebraic sum of all the currents at any node in a circuit equals zero.

Example 2.6

Basic Laws of Circuits Kirchhoff’s Voltage Law The algebraic sum of all the voltages around any closed path in a circuit equals zero  

Example 2.7 Write KVL equation for all the loops shown in the figure

Kirchhoff’s Current Law Ohm’s Law:   The algebraic sum of all the currents at any node in a circuit equals zero. Kirchhoff’s Current Law The algebraic sum of all the voltages around any closed path in a circuit equals zero Kirchhoff’s Voltage Law

Assessment Problem 2.5

Assessment Problem 2.6 Use Ohm’s law and Kirchhoff’s laws to find the value of R in the circuit shown

Problem 2.18    

H.W. # 2 2.19 and 2.20 2A 0.5A 40V 25W, 80W, 20W 125W 8mA 10mA -160mW

Analysis of circuit containing dependent sources   (1) (2) Solving 1 & 2

Analysis of circuit containing dependent sources Example 2.10 Use Kirchhoff’s laws and Ohm’s law to find the voltage vo as shown in the following circuit. Show that the total power developed in the circuit equals the total power dissipated.

Solution to Example 2.10(a) To find vo, we need io so that we could apply Ohm’s law (vo = 3io) Apply KVL in the right loop 2io + 3io – 3is = 0 => io = 3/5is Since 10V voltage source and the 6  resistor are in parallel, the voltage across the 6  resistor is also 10 V. Therefore, using Ohm’s law, is = 10/6 = 5/3 A io = 3/5is = 3/5 × 5/3 = 1 A Using Ohm’s law, vo = 3io= 3 V

Solution to Example 2.10(b) Power from 10 V independent source = p = – v . is = – (10)(5/3)= –16.7 W Power from the dependent source = p = –v .io = – (3×5/3)(1) = –5 W Power dissipated in the three resistors: p6 = (is)2(6) = 16.7 W p2 = (io)2(2) = 2 W p3 = (io)2(3) = 3 W Power delivered = 16.7 + 5 = 21.7 W Power dissipated = 16.7 + 2 + 3 = 21.7 W

H.W#3: Assessment Problem 2.9 Find: Current i1 in A Voltage v in volts Total power generated Total power absorbed