lecture - 2 basic circuit laws

Lecture - 2

Basic circuit laws

Outline• Electrical resistance.

• Ohm’s law.

• Conductance.

• Power in resistor.

• Open circuit and short circuit.

• Branches, nodes, loops.

• Kirchhoff's current law

• Kirchhoff's voltage law.

Electrical Resistance• Resistance is the capacity of materials to impede the flow of

current or, more specifically, the flow of electric charge. • The circuit element used to model this behaviour is the

resistor.• Most materials exhibit measurable resistance to current. • The amount of resistance depends on the material.• Metals such as copper and aluminium have small values of

resistance, making them good choices for wiring used to conduct electric current.

• The resistance of the wire is so small compared to the resistance of other elements in the circuit that we can neglect the wiring resistance to simplify the diagram.

Ohm's Law• For purposes of circuit analysis, we must reference the current

in the resistor to the terminal voltage. We can do so in two ways: – in the direction of the voltage drop across the resistor,– in the direction of the voltage rise across the resistor.

Conductance• The reciprocal of the resistance is referred to as conductance,

is symbolized by the letter G, and is measured in Siemens (S).Thus:

• An 8 Ω resistor has a conductance value of 0.125 S.

SR

G ,1

Power in a resistor • Regardlessof voltage polarity and current direction, the power

at the terminals of a resistor is positive. Therefore, a resistor absorbs power from the circuit.

• Power equations:

Rvp

Rip2

2

GvpGip

2

2

vip

Example 1• In each circuit in the figure:a) Calculate the values of v and i.b) Determine the power dissipated in each resistor.

Example 1a)• The voltage va in (a) is a drop in the direction of the current in the resistor,

va= ( 1 ) ( 8 ) = 8 V.

• The current ib in the resistor with a conductance of 0.2 S in (b) is in the direction of the voltage drop across the resistor. ib = (50)(0.2) = 10 A.

• The voltage vc in (c) is a rise in the direction of the current in the resistor,vc = -(1)(20) = -20 V.

• The current id in the 25 Ω resistor in (d) in the direction of the voltage rise across the resistor,id =-50/25= - 2 A.

Example 1b) The power dissipated in each of the four resistors is:

Wp

Wp

Wp

Wp

S

10025

)50(

2020

)20(

500)2.0()50(

88)8(

2

25

2

20

22.0

2

8

Short Circuit as Zero Resistance

• An element or wire with R=0 is called a short circuit.

• Just drawn as a wire (line).

Open circuit

• An element or (wire) with is called open circuit.

• Just omitted.

R

Branches• Branch: a single two-terminal element in a circuit.

• Segments of wire are not counted as elements.

• Examples: voltage source, resistor, current source.

Nods• Node: the point of connection between two or more branches.

• May include a portion of the circuit (more than a single point).

• Essential Node: the point of connection between three or more branches.

Loops• Loop: any closed path in the circuit.

Overview of Kirchhoff’s Laws

• The foundation of circuit analysis is:– the defining equations for circuit elements (e.g. Ohm’s law).– Kirchhoff’s current law(KCL).– Kirchhoff’s voltage law(KVL).

• The defining equations tell us how the voltage and current within a circuit element are related.

• Kirchhoff’s laws tell us how the voltage and currents in different branches are related.

Kirchhoff’s Current Law (KCL)• KCL: The algebraic sum of all the currents at any node in a

circuit equals zero.• The sum of currents entering a node is equal to the sum of the

currents leaving a node.

Example 2• Sum the currents at each node in the circuit shown in the

figure. --------------------------------

In writing the equations, we use a positive sign for a current leaving a node. The four equations are

node a i1 + i4 - i2 – i5 = 0,node b i2 + i3 - i1 - ib - ia = 0,node c ib - i3 - i4 - ic = 0,node d i5 + ia + ic = 0.

Kirchhoff’s Voltage Law (KVL)• KVL: The algebraic sum of all the voltages around any closed

path in a circuit equals zero.

• 011 sc Vvvv

Example 3• Sum the voltages around each designated path in the circuit shown in

the figure.

--------------------------In writing the equations, we use a positive sign for a voltage drop. The four equations are

path a -v1 + v2 + v4 - vb - v3 = 0,

path b -va + v3 + v5 = 0,

path c vb - v4 - vc -v6 -v5 = 0,

path d -va - v1 + v2 -vc + v7 - vd = 0.

Example 4a) Use Kirchhoff's laws and Ohm's law to find io in the circuit shown in the

figure.b) Test the solution for io by verifying that the total power generated equals

the total power dissipated.

----------------------------a) We begin by redrawing the circuit and assigning an unknown current to

the 50Ω resistor and unknown voltages across the 10 Ω and 50 Ω resistors.

Example 4by applying KCL to either node b or c:

i1- io - 6 = 0

by applying KVL to the closed path cabc:-120 + 10io + 50i1 = 0

by solving the two equations:io = -3A and i1 = 3A

b) the power dissipated in the 50Ω and the 10Ω resistors are:p50Ω = (3)2 (50) = 450 Wp10Ω = (-3)2 (10) = 90 W

the power delivered to the 120V and the 6A sources are:p120V = -120 io = -120 (-3) = 360Wp6A = - v1 (6) = (-50 i1) (6) = (-150) (6) = -900 W

Example 5a) Use Kirchhoff's laws and Ohm's law to find the voltage vo in the figure.b) Show that your solution is consistent with the constraint that the total power

developed in the circuit equals the total power dissipated.

---------------------------------------Once io is known, we can compute vo and we need two equations for the two currents. Because there are two closed paths and both have voltage sources, we can apply Kirchhoff's voltage law to each to give the following equations:

10 = 6 is and 3 is = 2 io + 3 io

Solving for the current yields:is = 1.67 A and io = 1A

Example 5Applying Ohm's law to the 3 Ω resistor gives the desired voltage:

v0 = 3 io = 3 V

b) The power delivered to the independent voltage source is:p = (10)(-1.67) = -16.7 W.

and the power delivered to the dependent voltage source is:p= (3 is)(- io) = (5)(-1) = - 5 W.

The power delivered to the 6 Ω resistor is: p = (1.67)2(6) = 16.7 W

The power delivered to the 2 Ω resistor isp = (1)2(2) = 2 W

The power delivered to the 3 Ω resistor isp = (1)2(3) = 3 W

Summary• A resistor constrains its voltage and current to be

proportional to each other. The value of the proportional constant relating voltage and current in a resistor is called its resistance and is measured in ohms.

• Ohm's law establishes the proportionality of voltage and current in a resistor. Specifically,

v = iR• If the current flow in the resistor is in the direction of the

voltage drop across it, orv = -iR

• If the current flow in the resistor is in the direction of the voltage rise across it.

Summary• By combining the equation for power, p = vi, with Ohm's law,

we can determine the power absorbed by a resistor:

• Circuits are described by nodes and closed paths. • A node is a point where two or more circuit elements join.• When just two elements connect to form a node, they are said

to be in series. • A closed path is a loop traced through connecting elements,

starting and ending at the same node and encountering intermediate nodes only once each.

RVRip

22

Summary• The voltages and currents of interconnected circuit elements obey

Kirchhoff's laws:– Kirchhoff's current law states that the algebraic sum of all

the currents at any node in a circuit equals zero.– Kirchhoff's voltage law states that the algebraic sum of all

the voltages around any closed path in a circuit equals zero.

• A circuit is solved when the voltage across and the current in every element have been determined.

• By combining an understanding of independent and dependent sources, Ohm's law, and Kirchhoff's laws, we can solve many simple circuits.