Electric Circuits I Circuit Elements

Dr. Firas Obeidat

1 Independent Voltage Sources

An independent voltage source is characterized by a terminal voltage which is completely independent of the current through it or other circuit elements.

 The independent voltage source is an ideal source and does not represent exactly any real physical device, because the ideal source could theoretically deliver an infinite amount of energy from its terminals.  Physical sources such as batteries and generators may be regarded as approximations to ideal voltage sources.

2 Dr. Firas Obeidat – Philadelphia University Independent Current Sources

Independent current source is characterized by a current which is completely independent of the voltage across it or other circuit elements.

 The independent current source is at best a reasonable approximation for a physical element. In theory it can deliver infinite power from its terminals because it produces the same finite current for any voltage across it, no matter how large that voltage may be.  It is, however, a good approximation for many practical sources, particularly in electronic circuits.

3 Dr. Firas Obeidat – Philadelphia University Dependent Voltage & Current Sources Dependent, or controlled, source, in which the source quantity is determined by a voltage or current existing at some other location in the system being analyzed. (a) current-controlled current source;

(b) voltage-controlled current source;

(c) voltage-controlled voltage source;

(d) current-controlled voltage source.  Sources such as these appear in the equivalent electrical models for many electronic devices, such as transistors, operational amplifiers, and integrated circuits.  Dependent and independent voltage and current sources are active elements; they are capable of delivering power to some external device. 4 Dr. Firas Obeidat – Philadelphia University Examples

Calculate the power supplied or absorbed by each element in Figures.

1. p1= 20×5=100W (Supplied power). 1. p1= 45W (supplied power). 2. P2= 5×12=60W (absorbed power). 2. P2= 18W (absorbed power). 3. P3=6×8=48W (absorbed power). 3. P3=12W (absorbed power). 4. P4= 8×(0.2×5)= 8W (supplied power). 4. P4= 15W (absorbed power).

5 Dr. Firas Obeidat – Philadelphia University Resistance

Materials in general have a characteristic behavior of resisting the flow of electric charge. This physical property, or ability to resist current, is known as resistance (R) measured in ohms (Ω).

Resistor is a passive element which is capable only of receiving power. 6 Dr. Firas Obeidat – Philadelphia University Resistance

R is the resistance (Ω). = ρ is the resistivity (Ω.m). is length of the conductor (m). A=πr2 A is cross sectional area (m2). G is the conductance (Siemens (S)).

=

7 Dr. Firas Obeidat – Philadelphia University Resistance

Example: certain cable is made from Example: What is the resistance of a 50m certain material which has length of copper wire with a radius of 0. ρ=0.01Ω.m, l=xm, A=ym2, and 5cm at 20°C? R=10Ω. another cable is made from 2 2 -5 2 A=πr = π×(0.005) =7.85×10 m the same material, If (l) is increased .by 20٪, and (A) is deceased by 50٪ = =1.72×10-8× =0.01Ωv Calculate R for the new cable? 7.85×10−5 . = ⇒ = Example: material (a) has ρa=0.005 Ω.m, = . 2 la=x m, and Aa=k m . material (b) has lb=2x 2 m, and Ab=0.5k m . Find ρb so that Ra=Rb? =+0.2=1.2

=-0.5=0. Ra=Rb ⇒ = .×. = = .× × . = . .×. = =24Ω .×. = × . = . . 8 Dr. Firas Obeidat – Philadelphia University Ohm’s Law Ohm’s law states that the voltage (v) across a resistor is directly proportional to the current i flowing through the resistor.

∝ then =  The value of R can range from zero to infinity.  An element with R=0 is called a short circuit, For short circuit V=iR=0 A short circuit is a circuit element with resistance approaching zero.  An element with R=∞ is called a open circuit, For open circuit = lim = → A open circuit is a circuit element with resistance approaching infinity.

9 Dr. Firas Obeidat – Philadelphia University Ohm’s Law From equation of power and the above equation, we can find that = = = = = = Example 1: An electric iron draws 2 A at 120 V. Find its resistance. Solution: From Ohm’s law, R=v/i=120/2=60 Ω

Example 2: The essential component of a toaster is an electrical element (a resistor) that converts electrical energy to heat energy. How much current is drawn by a toaster with resistance 15 Ω at 110 V? Solution: i=v/R=110/15=7.33 A Example 3: for the circuit shown, calculate the current i, the conductance G, and the power p. Solution: = = = × = = . = = × × =

10 Dr. Firas Obeidat – Philadelphia University Nodes, Paths, Loops, and Branches

 Node: A point at which two or more elements have a common connection. Or, it is the point of connection between two or more branches.

 Branch: represents a single element such as a voltage source or a resistor.

 Loop: is any closed path in a circuit.

 Path: is a set of nodes and elements that we have passed through.

11 Dr. Firas Obeidat – Philadelphia University Kirchhoff’s Laws  Kirchhoff’s current law (KCL) states that the algebraic sum of currents entering a node is zero.

= + + + ⋯ + () =

+ + (−) + (−) =

 Kirchhoff’s voltage law (KVL) states that the algebraic sum of all voltages around a closed path (or loop) is zero.

=

+ + + ⋯ + () =

(−) + + − = 12 Dr. Firas Obeidat – Philadelphia University Kirchhoff’s Laws - Examples

Ex: In the circuit there are eight

circuit elements. Find vR2 (the voltage across R2) and the voltage labeled vx.

4− + = =32V

− + 12 + 14 + vx = 0

− + 12 + 14 + vx = 0

vx =6V

Ex: Determine and vo i in the circuit shown in Figure. applying KVL around the loop − + + − + = 0

From the figure vo = -6i − + − = 0 = -8A

vo = -6i=48V 13 Dr. Firas Obeidat – Philadelphia University Kirchhoff’s Laws - Examples

Ex: Find current and voltage o in the circuit shown in Figure.

applying KCL at node a.

5 + − = =6A

o=4i=24V

Ex: Find current and voltage o in the circuit shown in Figure.

Answer: 12 V, 6 A.

14 Dr. Firas Obeidat – Philadelphia University Kirchhoff’s Laws - Examples

Ex: which of these elements (A,B, and C) is not resistor?

-2×5-vC-3×6=0

vC=-28V C is supply element. Because of that, we have to change the polarity of C. 4×7-28+vB=0

vB=0

vB could be short circuit.

-vA-0+2×5=0

vA=10V Current enter from positive side of element (A). (A) is a resistance. 15 Dr. Firas Obeidat – Philadelphia University Kirchhoff’s Laws - Examples

Ex: Find currents and voltages in the circuit shown in the figure. At node a, KCL gives i1-i2-i3=0 eq (1) Applying KVL to loop 1 then 3i =6i ⇒ i = eq(5) -30+v1+v2=0 eq(2) 2 3 3 By Ohm’s law Substituting eq(3)&eq(5) in eq(1) gives v1=8i1, v2=3i2, v3=6i3

then () ( ) -i2- =0 -30+8i +3i =0 ⇒i = eq(3) 1 2 1 i2=2A Applying KVL to loop 2 i =3 A, i =1 A, -v +v =0 ⇒v =v eq(4) 1 3 2 3 2 3 v1=24 V, v2=6 V, v3=6 V 16 Dr. Firas Obeidat – Philadelphia University Series and Parallel Connected Sources

 some of the equations for series and parallel circuits can be avoided by combining sources without changing all the current, voltage, and power relationships in the remainder of the circuit.  sources in series may be replaced by an equivalent voltage source having a voltage equal to the algebraic sum of the individual sources.  Parallel current sources may also be combined by algebraically adding the individual currents.

17 Dr. Firas Obeidat – Philadelphia University 18