engg 1015 tutorial

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ENGG 1015 Tutorial

Circuit Analysis 5 Nov Learning Objectives

Analysis circuits through circuit laws (Ohm’s Law, KCL and KVL)

News HW1 deadline (5 Nov 23:55)

Ack.: HKU ELEC1008 and MIT OCW 6.01

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Quick CheckingNOT always true

Always True

If , then

If , then

32

4 5

RRR R

6 0i

2 3 4 5i i i i

2 6 3i i i

4

1 02 4

Re VR R

6 0i

32

2 4 3 5

RRR R R R

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What is a Circuit?

Circuits are connects of components Through which currents flow Across which voltages develop

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Rules Governing Flow and Voltages Rule 1: Currents flow in loops

The same amount of current flows into the bulb (top path) and out of the bulb (bottom path)

Rule 2: Like the flow of water, the flow of electrical current (charged particles) is incompressible Kirchoff’s Current Law (KCL): the sum of the currents into a

node is zero Rule 3: Voltages accumulate in loops

Kirchoff’s Voltage Law (KVL): the sum of the voltages around a closed loop is zero

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Rules Governing Components Each component is represented by a

relationship between the voltage (V) across the component to the current (I) through the component

Ohm’s Law (V = IR) R: Resistance

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Question 1: Current and Voltage

If R = 0 ohm, I1 =

If R = 1 ohm, V1 =

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Solution 1

If R = 0 ohm, I1 = 6V/3 ohm = 2A If R = 1 ohm, 1 1 1

1

6 50 3

3 1 1

V V VV V

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Parallel/Series Combinations

To simplify the circuit for analysis

1 2

1 2

s

s

v R i R i

v R i

R R R

1 2

1 21 2

1 2

11 1

//

p

R RR

R RR R

R R

Series

Parallel

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Voltage/Current Divider

1 2

11 1

1 2

22 2

1 2

VI

R R

RV R I V

R R

RV R I V

R R

1 2

1 2 21

1 1 1 2

12

1 2

//

//

V R R I

R R RVI I I

R R R R

RI I

R R

Voltage Divider

Current Divider

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Question 2a: Voltage Calculation Find V2 using single loop analysis

Without simplifying the circuit Simplifying the circuit

1 2 3 1 2 32 , 2 , 2 , 1 , 2 , 4s s sV v V v V v R R R

R1

Vs1

Vs3

Vs2

R3

-R2

+

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Solution 2a

Choose loop current

Apply KVL Replace V2 by R2I

Find V2

R1

Vs1

Vs3

Vs2

R3

-R2

+

2 1 2 3 3 1 0

2

7

s s sV R I R I R I V V

I A

2 2

4

7V R I v

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Solution 2b

Simplify the circuit with one voltage source and one resistor

Req. = R1 + R2 + R3 = 7 ohm

Veq. = Vs1 + Vs2 + Vs3 = -2 + 2 + 2 = 2 V

I = Veq. / Req. = 2/7 A

V2 = 4/7 v Veq.

Req.

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Question 3: Potential Difference Assume all resistors have the same resistance,

R. Determine the voltage vAB.

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Solution 3

Determine VAB

We assign VG=0

2

1 2

4

3 4

5 2.5

3 1.5

A

B

RV V

R R

RV V

R R

2.5 1.5 4AB A BV V V V

For the circuit in the figure, determine i1 to i5.

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Question 4: Current Calculation using Parallel/Series Combinations

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Solution 4

21 // 2

3

2 44 //

3 7

4 253 //

7 7

40V

3Ω

4Ω 1Ω 2Ω

40V

3Ω

4Ω 2/3Ω

40V

3Ω

4/7Ω

40V 25/7Ω

(i)

(iii)

(ii)

(iv)

We apply: V = IR Series / Parallel Combinations Current Divider

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Solution 4

1 1

2540 11.2

7

V IR

i i A

1 2 3

2 1

3

2 13 11.2 1.62 74 3

11.2 1.6 9.6

i i i

i i A

i A

3 4 5

4

5

2 9.6 6.43

1 9.6 3.23

i i i

i A

i A

40V 25/7Ω4Ω 2Ω

i2

i1

i3

4Ω 2Ω

i4

i3

i5(v) (vi) (vii)

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Question 5: Resistance Calculation using Parallel/Series CombinationsFind Req and io in the circuit of the figure.

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Solution 5

12 // 6 4 20 // 80 16

4 16 20

40V

5Ω

15Ω6Ω

12Ω

60Ω

20Ω 80Ω

i0

Req

40V

5Ω

15Ω

4Ω

60Ω

16Ω

i0

Req

(i)

(ii)

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Solution 5

0 0

15 // 20 // 60 7.5

40 7.5 5 3.2

eqR

V IR i i A

40V

5Ω

15Ω 60Ω20Ω

i0

Req

(iii)

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Analyzing Circuits

Assign node voltage variables to every node except ground (whose voltage is arbitrarily taken as zero)

Assign component current variables to every component in the circuit

Write one constructive relation for each component in terms of the component current variable and the component voltage

Express KCL at each node except ground in terms of the component currents

Solve the resulting equations

Power = IV = I2R = V2/R

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R1 = 80Ω, R2 = 10Ω, R3 = 20Ω,R4 = 90Ω, R5 = 100Ω

Battery: V1 = 12V, V2 = 24V, V3 = 36V

Resistor: I1, I2, …, I5 = ? P1, P2, …, P5 = ?

Question 6: Circuit Analysis (I)

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Solution 6a

VN = 0

I1: M R5 V1 R1 B

I2: M V3 R3 R2 B

I4: M V2 R4 B

Step 1, Step 2

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Solution 6a

VM – VB = R5I1 + V1 + R1I1

I1 = (VM – VB – V1)/(R5 + R1) = (24 – VB)/180Step 3

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Solution 6a

VN – VB = R2I2 + R3I2

I2 = (VN – VB)/(R2 + R3) = – VB/30

Step 3

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Solution 6a

VM – VB = V2 + R4I4

I4 = (VM – VB – V2)/R4 = (12 – VB)/90

We get three relationships now (I1, I2, I4)Step 3

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Solution 6a

KCL of Node B: I1 + I4 + I2 = 0

(24 – VB)/180 + (12 – VB)/90 – VB/30 = 0

VB = 16/3 V Step 4, Step 5

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Solution 6a

I1 = (24 – VB)/180 = 14/135 A = 0.104A

I4 = (12 – VB)/90 = 2/27 A = 0.074A

I2 = – VB/30 = – 8/45 A = – 0.178AStep 5

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Solution 6a

P = I2R = P1 = (0.104)2 80 = 0.86528W

P4 = (0.074)2 90 = 0.49284W = VR42 / R

(6.66V, 90Ω)

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Solution 6b

VM = 0

I1: B R1 V1 R5 M

I2: B R2 R3 V3 M

I4: B R4 V2 M

Let’s try another reference ground

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Quick Checking

I1: B R1 V1 R5 M

I2: B R2 R3 V3 M

I4: B R4 V2 M Different direction, different result?

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Solution 6b

KCL of Node B: I1 + I2 + I4 = 0

VB – VM = R1I1 – V1 + R5I1

I1 = (VB – VM + V1)/(R1 + R5) = (VB + 12)/180

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Solution 6b

VB – VM = R2I2 + R3I2 – V3

I2 = (VB – VM + V3)/(R2 + R3) = (VB + 36)/30

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Solution 6b

VB – VM = R4I4 – V2

I4 = (VB – VM + V2)/R4 = (VB + 24)/90

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Solution 6b

KCL of Node B: I1 + I2 + I4 = 0

(VB + 12)/180 + (VB + 36)/30 + (VB + 24)/90 = 0

VB = – 92/3 V

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Solution 6b

I1 = (VB + 12)/180 = –14/135 A = – 0.104A

I2 = (VB + 36)/30 = 8/45 A = 0.178A

I4 = (VB + 24)/90 = –2/27 A = – 0.074A

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Question 7: Circuit Analysis (II) Find vo in the circuit of the figure.

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Solution 7

Step 1: Define the node voltage (v1,v2,v3) Step 2: Define the current direction

1Ω

40V

2Ω

4Ω

8Ω

20V

v1v2 v3

5A

+v0

--

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Solution 7

Apply: 1) V = IR 2) KCL Step 3: Consider node 1

1 2 11 2

405 3 70 1

2 1

v v vv v

v1

5A

(40-v1)/1(v1-v2)/2

1Ω

40V

2Ω

4Ω

8Ω

20V

v1v2 v3

5A

+v0

--

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Solution 7

Step 3: Consider node 2

Step 4, 5: From (1) and (2),v1 = 30V, v2 = 20V, v0 = v2 = 20V

2 31 2 21 25 4 7 20 2

2 4 8

v vv v vv v

v2

5A

(v1-0)/4

(v2-v0)/8

(v1-v2)/2

1Ω

40V

2Ω

4Ω

8Ω

20V

v1v2 v3

5A

+v0

--

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Quick CheckingNOT always true

Always True

If , then

If , then

32

4 5

RRR R

6 0i

2 3 4 5i i i i

2 6 3i i i

4

1 02 4

Re VR R

6 0i

32

2 4 3 5

RRR R R R

42

Quick CheckingNOT always true

Always True

If , then

If , then

32

4 5

RRR R

6 0i

2 3 4 5i i i i

2 6 3i i i

4

1 02 4

Re VR R

6 0i

32

2 4 3 5

RRR R R R

√

√

√

√

√