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.

EIT_1_29 Page 1 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.

EIT_1_29 Page 2 Element Equations

EIT_1_29 Page 3 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

EIT_1_29 Page 4 Voltage and Current Waveforms

EIT_1_29 Page 5 Steady State and Transient Parts of a Complete Response

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

AC 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

EIT_1_29 Page 6 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

EIT_1_29 Page 7 The Passive Convention

Specification of power and energy use of the passive convention.

Power and Energy: p vi and w pdt

power uses the passive convention supplied by No supplied to Yes absorbed by Yes received by Yes delivered by No delivered to Yes

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

EIT_1_29 Page 8 Kirchhoff's Laws

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

i1 i 2  i 3  i 4 0, i 4  i 5  0  i 4  i 5 ,

i1 i 2  i 3  i 5  0,...

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

v1 v 4  v 5  v 6 0, v 2  v 3  0,...

EIT_1_29 Page 10 Informal Analysis of DC circuits

Problem Determine the voltage and current of each of the circuit Next, apply KCL at the bottom right node to get elements in this circuit. ii34 0.25  1.0  0.25  0.75 A Next, apply KVL to the loop consisting of the voltage source and the 60 W resistor to get

vv2215  0   15 V Apply Ohm’s law to each of the resistors to get v 15 i2  0.25 A, v  10 i  10  0.75  7.5 V, 260 60 3 3

Solution We can label the circuit as follows: vi3310  10  0.75  7.5 V and

vi4420  20  1  20 V

Next, apply KCL at the bottom left node to get

i1 i 3  i 2 0.75  0.25  1.0 A Next, apply KVL to the loop consisting of the 0.75 A current source and three resistors to get

v6 v 4  v 3  v 2 20  (  7.5)  15  12.5 V

Apply KCL at the top right node to get Finally, apply KVL to the loop consisting of the 0.25 A current source and the 20 W resistor to get ii440.25  0.75  0   1.0 A

vv54    20  20 V

EIT_1_29 Page 11 Informal Analysis of DC circuits

Problem Determine the voltage and current of each of the circuit Next, apply KCL at the bottom right node to get elements in this circuit. ii34 0.25  1.0  0.25  0.75 A Next, apply KVL to the loop consisting of the voltage source and the 60 W resistor to get

vv2215  0   15 V Apply Ohm’s law to each of the resistors to get v 15 i2  0.25 A, v  10 i  10  0.75  7.5 V, 260 60 3 3

Solution We can label the circuit as follows: vi3310  10  0.75  7.5 V and

vi4420  20  1  20 V

Next, apply KCL at the bottom left node to get

i1 i 3  i 2 0.75  0.25  1.0 A Next, apply KVL to the loop consisting of the 0.75 A current source and three resistors to get

v6 v 4  v 3  v 2 20  (  7.5)  15  12.5 V

Apply KCL at the top right node to get Finally, apply KVL to the loop consisting of the 0.25 A current source and the 20 W resistor to get ii440.25  0.75  0   1.0 A

vv54    20  20 V http://people.clarkson.edu/~jsvoboda/Syllabi/ES250/dcCkts/More%20Ad%20Hoc%20Analysis.pdf

EIT_1_29 Page 12 Dependent Sources

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

EIT_1_29 Page 13 Node Voltages

Node (Voltage) Equations

1. Express everything in terms of the node voltages. 2. Apply KCL at all nodes except for the reference node.

EIT_1_29 Page 14 Example: Apply KCL at node 1 to get v v v 1 1 2 20  8 25 We will simplify this equation by doing two things:

1. Multiplying each side by to eliminate fractions. 2. Move the terms that don’t involve node voltages to the right side of the equation.

The result is Emphasize and label the nodes: 33vv12 8   400

Next, apply KCL at node 2 to get v v24 v v 2 2 1 2 9 14 25 Simplify this equation:

126vv12  701  5400 Solving these simultaneous equations, perhaps using MATLAB, the node voltages are v1 = -10.7209 V and v2 = 5.7763 V.

v 3  24 V

EIT_1_29 Page 15 Mesh Currents

Mesh (Current) Equations

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

EIT_1_29 Page 16 Mesh (Current) Equations

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

Example:

Notice that

i3  2 A Apply KVL to mesh 1 to get

25i1 i 3  9 i 1  i 2  8 i 1  0 Substituting

i3  2 A Solution: Label the label the mesh currents. Then, label the element currents in terms of and doing a little algebra gives the mesh currents: 42ii12 9 50 Next, apply KVL to mesh 2 to get

14i2 i 3  24  9 i 1  i 2   0

Substituting and doing a little algebra gives

9ii12  23   24  14 2  4 Solving these simultaneous equations, perhaps using MATLAB, the mesh currents are i1 = 1.3401 A andi2 = 0.6983 A.

EIT_1_29 Page 17 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 opportunities to simplify the a circuit without changing the quantity of interest.

EIT_1_29 Page 18 Voltage Division

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.

Solution Using equivalent resistanceand voltage division in the right part of the circuit

20 20 1 v v v a20 20 20 s 3 s

Using voltage division in the left part of the circuit

12 3 1 vo10 v a   10  v s  2 v s 12 8 5 3

So vo is proportional to vs and the constant of proportionality is 2 V/V. http://people.clarkson.edu/~jsvoboda/Syllabi/ES250/vDiv/VdivVCVSworksheet.pdf

EIT_1_29 Page 19 Source Transformations

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

EIT_1_29 Page 20 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.

EIT_1_29 Page 21 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

vv vi13 8  o255

Solution 2

10 1 vo v 1 v 1 10 1 40 10 5 40 10 vo i 28 i 2 vo  v 3   v 3 40 10 40 10 5

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

EIT_1_29 Page 22 Sunday, January 23, 2011 11:42 AM Thevenin Equivalent Circuits

EIT_1_29 Page 23 Example

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

Solution

v 2 V 3 8 oc i  A Rt    sc 4 3

Finally, here's the Thevenin equivalent circuit:

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

EIT_1_29 Page 24 Op Amps

Ideal Op Amp

Op Amp IC 5 Terminal Op Amp

EIT_1_29 Page 25 Op Amp Circuits

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

Solution

vo  v22 v 0.25vv11 33 0 vvo  22  25 mV 33 0 v1  0.125 V  125 mV 20 10 20 10 20 10 20 10

vv 12 0 vv   125 mV 20 1033 20 10 21

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

EIT_1_29 Page 26 Capacitors Inductors

Capacitors act like open circuits in dc circuits. Inductors act like short circuits in dc circuits.

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

d2 10 R d  1 20 i t i t  i t  for t  0 2       dt 3R 10 L dt CL CL When 1) R = 10 W, L = 0.4 H and C = 0.25 mF 2) The circuit is at steady state before the switch closes. The capacitor voltage, v(t), can be shown to be v t16  16.525 e2.5t cos 9.682 t  165.5  V for t  0

EIT_1_29 Page 27 First Order Circuits

at i t  I  i0  I eat for t  0 v t Voc  v0  V oc e for t  0 sc sc  where where 1 a  R t RCt a  L and the initial condition, v(0), is the capacitor and the initial condition, i(0), is the inductor voltage at time t = 0. current at time t = 0.

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

EIT_1_29 Page 28 First -Order Circuits

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

EIT_1_29 Page 29 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.

Solution Simplify the circuit using source transformations and equivalent resistance:

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

R 18 1 at  3 L 6s Finally i t0.8  2.4  0.8 e33tt  0.8  1.6 e for t  0

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

EIT_1_29 Page 30 AC and DC Circuits

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

AC 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

Voltage and Current Waveforms

EIT_1_29 Page 31 Phasors

Impedances

EIT_1_29 Page 32 Time and Frequency Domains

Time Domain Frequency Domain

EIT_1_29 Page 33 DC Circuits (in the Time Domain) AC Circuits (in the Frequency Domain)

i12 i i III12

R Z1 1 VV vv1  1 ZZ12 RR12

Z 2 R2 VV vv 2 2 ZZ12 RR12

VVV v12 v v 12

R 2 Z2 ii1  II1  RR12 ZZ12

R1 Z1 ii2  II2  RR12 ZZ22

EIT_1_29 Page 35 AC and DC Circuits:

DC Circuit:

KCL and Ohm's law: vc  4 i22  4.5  4 i  4.5

KVL:  10i2  2.5 vcc  v  0

1 4 vvcc    18      45             1.5 10 ii22   0     6.75 

vvoc2.5   112.5 V

AC Circuit:

KCL and Ohm's law: VIIc  j 4 22  4.5  0    j 4  j 18

KVL: j 10IVV2  2.5cc   0

1jj 4 VVcc   18     11.25 90             1.5j 10 II22   0     1.6875  0  

VVoc2.5  28.125  90  V

EIT_1_29 Page 36