1. Basic Laws of Electric Cicuits
Lesson 2

2. Independent Voltage and Current Sources

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

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

5. 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 -

6. 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

7. Property of Resistance:
Resistivities of some basic materials
Material Resistivity (ohm meters) Common Use
silver x conductor
copper x conductor
aluminum x conductor
gold x conductor
carbon 4.1x semiconductor
germanium 47x semiconductor
silicon 6.4x semiconductor
paper 1x insulator
mica 5x insulator
glass 1x insulator
teflon 3x insulator

8. 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

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

10. 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)

11. Power calculations

12. Example 2.3

13. Assessment problem 2.3

15. 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.

16. Flashlight + –

17. Circuit Model for a Flashlight+ –

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

19. Example 2.6

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

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

22. 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

24. Assessment Problem 2.5

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

26. Problem 2.18

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

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

29. 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.

30. 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

31. 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 = = 21.7 W
Power dissipated = = 21.7 W

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