ELECTRICITY & MAGNETISM LECTURE # 7 BY MOEEN GHIYAS

ELECTRICITY & MAGNETISM

LECTURE # 7

BY

MOEEN GHIYAS

TODAY’S LESSON

(Series – Parallel Networks – Chapter 7)

Introductory Circuit Analysis by Boylested (10th Edition)

Today’s Lesson Contents

• Introduction – Series-Parallel Networks

• General Approach

• Reduce and Return Approach

• Ladder Networks

• Voltage Divider Supply (Loaded and Unloaded)

• Potentiometer Loading

Introduction – Series-Parallel Networks

• Series-parallel networks are networks that contain both

series and parallel circuit configurations.

• A firm understanding of the basic principles is sufficient

to begin an investigation of any single-source dc (or

multi-sources connected only in simple series or

parallel) network having a combination of series and

parallel elements or branches.

General Approach

• Take a moment to study the problem “in total” and make

a brief mental sketch of the overall approach.

• Next examine each region of the network independently

before tying them together in series-parallel

combinations

• Redraw the network as often as possible with the

reduced branches towards source keeping unknown

quantities undisturbed or have provision for the trip back

to unknown quantities from the source.

General Approach

• Example – For the network of fig, determine the voltages V1

and V2 and the current I.

• Solution:

• Redraw the circuit

• By observation

• . By KVL in right loop

• . or

• . or

General Approach

• Apply KCL at node a

Reduce and Return Approach

• Used with single-source

(or multi-sources

connected only in simple

series or parallel) series-

parallel networks.

• In this analytical approach

we first reduce network

towards the source.


• Reduce network to single

element (RT) towards the

source to determine the

source current (IS).

• Followed by expanding the

circuit backwards.


• Then find the desired

unknowns by expanding

the circuit back to original

network.

Ladder Network

• Ladder network appears in fig. The reason for the

terminology is quite obvious for the repetitive structure

• Applying reduce and return approach (starting farthest

from source)

Ladder Network

Ladder Network

By Current Divider law

By Ohm’s Law

Voltage Divider Supply (Loaded & Unloaded)

• Through a voltage divider network such as

the one in fig, a number of terminal voltages

can be made available from a single supply.

• The voltage levels shown (with respect to

ground) are determined by a direct

application of the voltage divider rule.

• Figure reflects a no load situation due to the

absence of any current-drawing elements

connected between terminals a, b, or c and

ground.


• The application of a load can affect the

terminal voltage of the supply.

1k 1k ΩΩ

1k 1k ΩΩ

1k 1k ΩΩ


• The larger the resistance level of the applied loads

compared to the resistance level of the voltage divider

network, the lower the current demand from a supply,

closer the terminal characteristics are to the no-load

levels.

1k 1k ΩΩ

1k 1k ΩΩ

1k 1k ΩΩ


• Let us consider the network of fig with resistive loads

that are the average value of the resistive elements of

the voltage divider network.


• The voltage Va is unaffected by the load RL1 since the

load is in parallel with the supply voltage E.

• Thus Va = 120 V, same as the no-load level.


• Now remaining load situation create a series-parallel effect

• R′3 = R3 || RL3 = 30 Ω || 20 Ω =12 Ω .

• R′2 = (R2 + R′3) || RL2 = 32Ω || 20Ω = 12.31Ω .


• Applying voltage divider law

versus 100 V under no-load

conditions


• Applying voltage divider law

versus 60 V under no-load

conditions


• If the load resistors are changed to the 1kΩ level, the terminal

voltages will all be relatively close to the no-load values

1k 1k ΩΩ

1k 1k ΩΩ

1k 1k ΩΩ


• Comparing current levels

• With 20Ω load

• With 1kΩ

1k 1k ΩΩ

1k 1k ΩΩ

1k 1k ΩΩ


• Example – Determine R1, R2, and R3 for the voltage divider

supply of fig. Can 2W resistors be used in the design?

• Solution: For R3:

Yes! 2W resistor Yes! 2W resistor

can be usedcan be used


• For R1: Apply KCL at node a,

• Note: Va ≠ VR1

• But VR1 = Vab

Yes! 2W resistor can be usedYes! 2W resistor can be used


• For R2: Apply KCL at node b,

Yes! 2W resistor Yes! 2W resistor

can be usedcan be used

Potentiometer Loading

• For the unloaded potentiometer of fig, the output

voltage is determined by the voltage divider rule, with

RT in the figure representing the total resistance of the

potentiometer.


• When a load is applied as shown in fig (right), the output

voltage VL is now a function of the magnitude of the load

applied since R1 is not as shown in fig (left) but is instead the

parallel combination of R1 and RL.


• If it is desired to have good control of output voltage VL

through the controlling dial or knob (Design Parameter),

it is advisable to choose a load or potentiometer that

satisfies the following relationship:


• For example, if we disregard eq. and choose

a 1MΩ potentiometer with a 100Ω load and set the wiper

arm to 1/10 of total resistance, as shown, then

which is extremely small compared which is extremely small compared

to the expected level of 1 V.to the expected level of 1 V.


• Using the reverse situation of RT = 100Ω and RL = 1 MΩ and

the wiper arm at the 1/10 position, as in fig, we find

which is the desired voltage i.e. which is the desired voltage i.e.

1/10 of source voltage E = 10V1/10 of source voltage E = 10V

Summary / Conclusion

• Introduction – Series-Parallel Networks

• General Approach

• Reduce and Return Approach

• Ladder Networks

• Voltage Divider Supply (Loaded and Unloaded)

• Potentiometer Loading

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ELECTRICITY & MAGNETISM LECTURE # 7 BY MOEEN GHIYAS