EIT Review – Electric Circuits

Electric circuits are used in the generation, transmission and consumption of electric power and energy.

Electric circuits are used in the encoding, decoding, storage, retrieval, transmission and processing of information.

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Interpreting the circuit diagram

An electric circuit is an interconnection of electric circuit elements. (Circuit elements are also called devices or components.)

Each circuit element has at least two terminals, i.e. places where that element can be connected to other circuit elements. (Terminals are sometimes called leads.) The parts of the circuit where terminals are connected together are called nodes. (Nodes are also called vertices.)

This circuit is an interconnection of two terminal elements.

The shape of the element indicates it behavior.

Each element is characterized by a parameter, represented either as a value with units or as a variable.

Each element has two terminals that are connected to nodes of the network. The element is said to be incident to the nodes ate which its terminals are connected.

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Element Equations

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Current and Voltage

Two quantities are identified for each two terminal circuit element: the element current and the element voltage.

Current and voltage each have a direction as well as value. •

Changing the direction corresponds to multiplying the value by –1•

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Voltage and Current Waveforms

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Steady State and Transient Parts of a Complete Response

All of the inputs are constant voltages and currents.

The circuit is at steady state.

All of the voltages and currents are constant functions of time and can be represented using real numbers

DC Circuits:

All of the inputs are sinusoidal voltages and currents having the same frequency.

The circuit is at steady state.

All of the voltages and currents are sinusoidal functions of time at the input frequency and can be represented using complex numbers

AC Circuits:

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The Passive Convention

When the reference direction of a particular current points from the + toward the of the polarity of a particular voltage, that current and voltage are said to adhere to the passive convention.

ia and va adhere to the passive convention•

ib and vb adhere to the passive convention•

ia and vb adhere to the passive convention•

ib and va adhere to the passive convention•

The “element equations”, e.g. Ohm's law, use the passive convention:

va = R ia and vb = va = R ia

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The Passive Convention

Specification of power and energy use of the passive convention.

power uses the passive convention

supplied bysupplied to

absorbed byreceived bydelivered bydelivered to

NoYesYesYesNoYes

In cases a and b, the element receives 24 W. In cases c and d, the element supplies 24 W.

Power and Energy: andp vi w pdt

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Kirchhoff's Laws

KCL: The algebraic sum of the currents entering any node is zero.

1 2 3 4 4 5 4 5

1 2 3 5

0, 0 ,

0,...

i i i i i i i i

i i i i

KVL: The algebraic sum of the voltages in any loop are zero.

1 4 5 6 2 30, 0,...v v v v v v

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Informal Analysis of DC circuits

Problem Determine the voltage and current of each of the circuit elements in this circuit.

Solution We can label the circuit as follows:

Apply KCL at the top right node to get

4 40.25 0.75 0 1.0 Ai i

Next, apply KCL at the bottom right node to get

3 4 0.25 1.0 0.25 0.75 Ai i

Next, apply KVL to the loop consisting of the voltage source and the 60 W resistor to get

2 215 0 15 Vv v

Apply Ohm’s law to each of the resistors to get

2

2 3 3

150.25 A, 10 10 0.75 7.5 V,

60 60

vi v i

3 310 10 0.75 7.5 Vv i

and

4 420 20 1 20 Vv i

Next, apply KCL at the bottom left node to get

1 3 2 0.75 0.25 1.0 Ai i i

Next, apply KVL to the loop consisting of the 0.75 A current source and three resistors to get

6 4 3 2 20 ( 7.5) 15 12.5 Vv v v v

Finally, apply KVL to the loop consisting of the 0.25 A current source and the 20 W resistor to get

5 4 20 20 Vv v

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Informal Analysis of DC circuits

Problem Determine the voltage and current of each of the circuit elements in this circuit.

Solution We can label the circuit as follows:

Apply KCL at the top right node to get

4 40.25 0.75 0 1.0 Ai i

Next, apply KCL at the bottom right node to get

3 4 0.25 1.0 0.25 0.75 Ai i

Next, apply KVL to the loop consisting of the voltage source and the 60 W resistor to get

2 215 0 15 Vv v

Apply Ohm’s law to each of the resistors to get

2

2 3 3

150.25 A, 10 10 0.75 7.5 V,

60 60

vi v i

3 310 10 0.75 7.5 Vv i

and

4 420 20 1 20 Vv i

Next, apply KCL at the bottom left node to get

1 3 2 0.75 0.25 1.0 Ai i i

Next, apply KVL to the loop consisting of the 0.75 A current source and three resistors to get

6 4 3 2 20 ( 7.5) 15 12.5 Vv v v v

Finally, apply KVL to the loop consisting of the 0.25 A current source and the 20 W resistor to get

5 4 20 20 Vv v

http://people.clarkson.edu/~jsvoboda/Syllabi/ES250/dcCkts/More%20Ad%20Hoc%20Analysis.pdf

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Dependent Sources

Controlling and controlled elements•Gain•Units of the gain•Power•

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Node Voltages

Node (Voltage) Equations

Express everything in terms of the node voltages.1.

Apply KCL at all nodes except for the reference node.2.

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Example:

Emphasize and label the nodes:

3 24 Vv

Apply KCL at node 1 to get

1 1 22 0

8 25

v v v

We will simplify this equation by doing two things:

Multiplying each side by to eliminate fractions.1.Move the terms that don’t involve node voltages to the right side of the equation.

2.

The result is

1 233 8 400v v

Next, apply KCL at node 2 to get

2 2 1 224

9 14 25

v v v v

Simplify this equation:

1 2126 701 5400v v

Solving these simultaneous equations, perhaps using MATLAB, the node voltages are v1 = -10.7209 V and v2 = 5.7763 V.

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Mesh Currents

Mesh (Current) Equations

Express everything in terms of the mesh currents.1.Apply KVL to all meshes.2.

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Mesh (Current) Equations

Express everything in terms of the mesh currents.1.Apply KVL to all meshes.2.

Example:

Solution: Label the label the mesh currents. Then, label the element currents in terms of the mesh currents:

Notice that

3 2 Ai

1 3 1 2 125 9 8 0i i i i i

Apply KVL to mesh 1 to get

Substituting

3 2 Ai

and doing a little algebra gives

1 242 9 50i i

Next, apply KVL to mesh 2 to get

2 3 1 214 24 9 0i i i i

Substituting and doing a little algebra gives

1 29 23 24 14 2 4i i

Solving these simultaneous equations, perhaps using MATLAB, the mesh currents are i1 = 1.3401 A andi2 = 0.6983 A.

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Circuit Equivalence

Series and Parallel Resistors - Voltage and Current Division

Equivalence: Replacing some circuit elements by an equivalent element does not change the current or voltage of the remaining circuit elements.

Successful application of circuit equivalences involves recognizing opportunitiesto simplify the a circuit without changing the quantity of interest.

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Example

The input to this circuit is the voltage of the independent voltage source. The output is the voltage measured by the meter.

The output is proportional to the input. Determine the value of the constant of proportionality.

SolutionUsing equivalent resistance and voltage division in the right part of the circuit

a s s

20 20 1

20 20 20 3v v v

Using voltage division in the left part of the circuit

o a s s

12 3 110 10 2

12 8 5 3v v v v

So vo is proportional to vs and the constant of proportionality is 2 V/V.

Voltage Division

http://people.clarkson.edu/~jsvoboda/Syllabi/ES250/vDiv/VdivVCVSworksheet.pdf

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Source Transformations

Equivalence: Replacing some circuit elements by an equivalent element does not change the current or voltage of the remaining circuit elements.

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Superposition

The response of a linear circuit to several inputs working together is equal to the sum of the responses to each input working separately.

Circuit inputs and outputs

Inputs are usually the voltages of independent voltage sources and the currents of independent current sources.

Outputs can be any current or voltage.

The circuit designer designates the input and output of the circuit.

A circuit can have more than one input or output.

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Superposition

The response of a linear circuit to several inputs working together is equal to the sum of the responses to each input working separately.

Example

This circuit has one output, vo, and three inputs, v1, i2 and v3.

Express the output as a linear combination of the inputs.

Solution 1

Writing and solving node equations gives

1 3

o 285 5

v vv i

Solution 2

o 1 1

10 1

40 10 5v v v

o 2 2

40 108

40 10v i i

o 3 3

10 1

40 10 5v v v

http://people.clarkson.edu/~jsvoboda/Syllabi/ES250/Thev/SuperpositionHOsoln.pdf

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Thevenin Equivalent CircuitsSunday, January 23, 2011

11:42 AM

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Example

We want to find the Thevenin equivalent circuit for this circuit:

Solution

oc 2 Vv t

8

3R

sc

3 A

4i

Finally, here's the Thevenin equivalent circuit:

http://people.clarkson.edu/~jsvoboda/Syllabi/ES250/Thev/ThevVCVS_HOsoln.pdf

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5 Terminal Op AmpOp Amp IC

Ideal Op Amp

Op Amps

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Op Amp Circuits

Determine the values of the node voltages; v1, v2 and vo; of this circuit.

Solution

1 1

13 3

0.250 0.125 V 125 mV

20 10 20 10

v vv

1 2

2 13 30 125 mV

20 10 20 10

v vv v

2 2

23 3

o

o0 2 25 mV20 10 20 10

v v vv v

http://people.clarkson.edu/~jsvoboda/Syllabi/ES250/oaCkts/ioaHOsoln.pdf

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Capacitors

Capacitors act like open circuits in dc circuits.

The capacitor voltage is continuous unless the capacitor current is unbounded.

Inductors

Inductors act like short circuits in dc circuits.

The inductor current is continuous unless the inductor voltage is unbounded.

Circuits containing capacitors and inductors are represented by differential equations:

After the switch closes, this circuit is represented by the differential equation

2

2

10 1 20 for 0

3 10

d R di t i t i t t

R L dt CL CLdt

R= 10 W, L = 0.4 H and C = 0.25 mF1)The circuit is at steady state before the switch closes. 2)

When

The capacitor voltage, v(t), can be shown to be

2.516 16.525 cos 9.682 165.5 V for 0tv t e t t

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First Order Circuits

oc oc0 for 0atv t V v V e t

where

t

1a

R C

and the initial condition, v(0), is the capacitor voltage at time t = 0.

sc sc0 for 0ati t I i I e t

where

tRa

L

and the initial condition, i(0), is the inductor current at time t = 0.

http://people.clarkson.edu/~jsvoboda/Syllabi/ES250/foc/FOC_HOsoln.pdf

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First -Order Circuits

http://people.clarkson.edu/~jsvoboda/Syllabi/ES250/foc/FOCsteps.pdf

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First-Order Circuit Example

This diagram represents a circuit for t ≥ 0 . Given the initial condition i(0) = 2.4 A and the

inductance, L = 6 H, represent the inductor current , i(t), as a function of t for t 0.

SolutionSimplify the circuit using source transformations and equivalent resistance:

Now recognize that Isc = 0.8 A and Rt = 18 Ω. Then

t 18 13

6 s

Ra

L

Finally

3 30.8 2.4 0.8 0.8 1.6 for 0t ti t e e t

http://people.clarkson.edu/~jsvoboda/Syllabi/ES250/foc/FOC_HOsoln.pdf

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AC and DC Circuits

All of the inputs are constant voltages and currents.

The circuit is at steady state.

All of the voltages and currents are constant functions of time and can be represented using real numbers

DC Circuits:

All of the inputs are sinusoidal voltages and currents having the same frequency.

The circuit is at steady state.

All of the voltages and currents are sinusoidal functions of time at the input frequency and can be represented using complex numbers

AC Circuits:

Voltage and Current Waveforms

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Phasors

Impedances

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Time and Frequency Domains

Time Domain Frequency Domain

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DC Circuits (in the Time Domain)

1 2i i i

1

1

1 2

Rv v

R R

2

2

1 2

Rv v

R R

1 2v v v

2

1

1 2

Ri i

R R

1

2

1 2

Ri i

R R

AC Circuits (in the Frequency Domain)

1 2 I I I

1

1

1 2

ZV V

Z Z

2

2

1 2

ZV V

Z Z

1 2 V V V

2

1

1 2

ZI I

Z Z

1

2

2 2

ZI I

Z Z

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AC and DC Circuits:

c 2 2

c c2

c c

2 2

o c

KCL and Ohm's law: 4 4.5 4 4.5

KVL: 10 2.5 0

1 4 18 45

1.5 10 0 6.75

2.5 112.5 V

v i i

i v v

v v

i i

v v

c 2 2

c c2

c c

2 2

o c

KCL and Ohm's law: 4 4.5 0 4 18

KVL: 10 2.5 0

1 4 18 11.25 90

1.5 10 0 1.6875 0

2.5 28.125 90 V

j j j

j

j j

j

V I I

I V V

V V

I I

V V

DC Circuit:

AC Circuit:

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