ELECTRICITY & MAGNETISM (Fall 2011) LECTURE # 13 BY MOEEN GHIYAS

ELECTRICITY & MAGNETISM (Fall 2011)

LECTURE # 13

BY

MOEEN GHIYAS

TODAY’S LESSON

(Parallel Circuits – Chapter 6)

Introductory Circuit Analysis by Boylested (10th Edition)

Today’s Lesson Contents

• Parallel Elements

• Total Conductance and Total Resistance

• Parallel Circuits

• Kirchhoff’s Current Law (KCL)

• Solution to Problems

Parallel Elements

• Two elements, branches, or networks are in

parallel if they have two points in common.

Parallel Elements

• Different ways in which three parallel elements may

appear.

Parallel Elements

• In fig,

• Elements 1 and 2 are in parallel because they have

terminals a and b in common. The parallel

combination of 1 and 2 is then in series with element

3 due to the common terminal point b.

Parallel Elements

• In fig, elements 1 and 2 are in series because they

have only terminal b in common. The series

combination of 1 and 2 is then in parallel with element

3 due to the common terminals point b and c.

Total Conductance and Total Resistance

• For parallel elements,

the total conductance

is the sum of the

individual

conductances

• Note that the equation

is for 1 divided by the

total resistance rather

than total resistance.

Total Conductance and Total Resistance

• Example – Determine the total conductance and

resistance for the parallel network of Fig

• Solution:

Total Conductance and Total Resistance

• Example – Determine the effect on total conductance

and resistance of the network of fig if another resistor of

10Ω were added in parallel with the other elements

• Solution:

Note that adding additional terms Note that adding additional terms

increases the conductance level increases the conductance level

and decreases the resistance level.and decreases the resistance level.

Total Conductance and Total Resistance

• Recall for series circuits that the total resistance will

always increase as additional elements are added in

series.

• For parallel resistors, the total resistance will always

decrease as additional elements are added in parallel.

Total Conductance and Total Resistance

• Example – Determine the total resistance for the

network of Fig

• Solution:

Total Conductance and Total Resistance

• The total resistance of parallel resistors is always less

than the value of the smallest resistor.

• The wider the spread in numerical value between two

parallel resistors, the closer the total resistance will be

to the smaller resistor.

Total Conductance and Total Resistance

• For equal resistors in parallel, the equation becomes,

• For same conductance levels, we have

Total Conductance and Total Resistance

• For two parallel resistors,

• For three parallel resistors,

Total Conductance and Total Resistance

• Example – Find the total resistance of the network of

Fig

• Solution:

Total Conductance and Total Resistance

• Example – Calculate the total resistance for the network

of Fig

• Solution:

Total Conductance and Total Resistance

• Parallel elements can be interchanged without changing

the total resistance or input current.

Total Conductance and Total Resistance

• Example – Determine the values of R1, R2, and R3 in fig

if R2 = 2R1 and R3 = 2R2 and total resistance is 16 kΩ.

• Solution:

• . =

• Since

Parallel Circuits

• The voltage across parallel elements is the same.

• or

• But and

• Take the equation for the total resistance and multiply

both sides by the applied voltage, For single-source For single-source

parallel networks, the parallel networks, the

source current (Isource current (Iss) is ) is

equal to the sum of the equal to the sum of the

individual branch individual branch

currents.currents.

Parallel Circuits

• The power dissipated by the resistors and delivered

by the source can be determined from

Parallel Circuits

• Example – Given the information provided in fig:

a) Determine R3.

b) Calculate E.

c) Find Is.

d) Find I2.

e) Determine P2.

Parallel Circuits

a) Determine R3.

Solution:

Parallel Circuits

b) Calculate E.

c) Find Is.

Solution:

Parallel Circuits

d) Find I2.

e) Determine P2.

Solution:

Kirchhoff’s Current Law (KCL)

• Kirchhoff’s current law (KCL) states that

the algebraic sum of the currents

entering and leaving an area, system, or

junction is zero.

• In other words, the sum of the currents

entering an area, system, or junction

must equal the sum of the currents

leaving the area, system, or junction.

Kirchhoff’s Current Law (KCL)

• In technology the term node is commonly used to

refer to a junction of two or more branches. Therefore,

this term will be used frequently in future.

Kirchhoff’s Current Law (KCL)

• At node a:

• At node b:

• At node c:

• At node d:

Kirchhoff’s Current Law (KCL)

• Example – Determine unknown current I1.

• Solution:

• I1 is 5mA and leaving system.

Kirchhoff’s Current Law (KCL)

• Example – Determine the currents I3 and I5 of fig using

Kirchhoff’s current law (KCL).

• Solution:

• . At node a: I1 + I2 = I3

• . At node b: I3 = I4 + I5

Summary / Conclusion

• Parallel Elements

• Total Conductance and Total Resistance

• Parallel Circuits

• Kirchhoff’s Current Law (KCL)

• Solution to Problems