ENGG 1015 Tutorial - University of Hong Kong

ENGG 1015 Tutorial - University of Hong Kong

ENGG 1015 Tutorial Circuit Analysis 5 Nov Learning Objectives News Analysis circuits through circuit laws (Ohms Law, KCL and KVL) HW1 deadline (5 Nov 23:55) Ack.: HKU ELEC1008 and MIT OCW 6.01 1 Quick Checking

NOT Always always True true R2 If R4 R3 , then i6 0 R5 i2 i3 i4 i5 i2 i6 i3 e1 R4

R2 R4 If i6 0, then R2 R2 R4 R3 V0 R3 R5 2 What is a Circuit? Circuits are connects of components Through which currents flow Across which voltages develop

3 Rules Governing Flow and Voltages Rule 1: Currents flow in loops Rule 2: Like the flow of water, the flow of electrical current (charged particles) is incompressible The same amount of current flows into the bulb (top path) and out of the bulb (bottom path) Kirchoffs Current Law (KCL): the sum of the currents into a node is zero Rule 3: Voltages accumulate in loops

Kirchoffs Voltage Law (KVL): the sum of the voltages around a closed loop is zero 4 Rules Governing Components Each component is represented by a relationship between the voltage (V) across the component to the current (I) through the component Ohms Law (V = IR) R: Resistance 5 Question 1: Current and Voltage

If R = 0 ohm, I1 = If R = 1 ohm, V1 = 6 Solution 1 If R = 0 ohm, I1 = 6V/3 ohm = 2A If R = 1 ohm, V1 6 V1 V1 5 0 V 3V 1 3 1 1 7 Parallel/Series Combinations

To simplify the circuit for analysis v R1i R2i Series v Rs i Rs R1 R2 Rp 1 1 R1 R1 // R2 1 R2 R1 R2 R1 R2 Parallel

8 Voltage/Current Divider Voltage Divider I V R1 R2 V1 R1 I R1 V R1 R2 V2 R2 I R2 V R1 R2 Current Divider

V R1 // R2 I I1 R2 V R1 // R2 I I R1 R1 R1 R2 I2 R1 I R1 R2 9 Question 2a: Voltage Calculation Find V2 using single loop analysis

Without simplifying the circuit Simplifying the circuit Vs1 2v, Vs 2 2v, Vs 3 2v, R1 1, R2 2, R3 4 Vs2 R1 Vs1 Vs3 R2 + R3 10 Solution 2a Choose loop current

Vs2 R1 Vs1 Vs3 Apply KVL R2 + R3 Replace V2 by R2I Vs 2 R1 I R2 I R3 I Vs 3 Vs1 0 2 I A 7

Find V2 4 V2 R2 I v 7 11 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. 12 Question 3: Potential Difference Assume all resistors have the same resistance, R. Determine the voltage vAB. 13 Solution 3 Determine VAB

We assign VG=0 R2 VA 5 2.5V R1 R2 R4 VB 3 1.5V R3 R4 VAB VA VB 2.5 1.5 4V 14 Question 4: Current Calculation using Parallel/Series Combinations For the circuit in the figure, determine i1 to i5. 15 Solution 4 We apply:

(i) 3 4 40V 1 V = IR Series / Parallel Combinations Current Divider 2 (ii) 2 1 // 2

3 (iii) 3 40V 4/7 4 25 3 // 7 7 3 4 40V 2/3 2 4 4 // 3

7 (iv) 40V 25/7 16 Solution 4 i3 i1 (vi) (v) 40V V IR 40 i1 i2 i3

4 2 25/7 (vii) i4 i5 4 2 i1 i2 i3 25 2 i1 11.2 A 1 3 7 i2 i1 11.2 1.6 A

42 7 3 i3 11.2 1.6 9.6 A i3 i4 i5 i4 2 i5 3 9.6 6.4 A 1 9.6 3.2 A 3 17 Question 5: Resistance Calculation using Parallel/Series Combinations Find Req and io in the circuit of the figure. 18 Solution 5 (i)

60 i0 12 5 6 40V 15 20 80 12 // 6 4 20 // 80 16 Req (ii) 60

i0 5 40V 4 15 16 4 16 20 Req 19 Solution 5 (iii) i0 5 40V 15

20 60 Req Req 15 // 20 // 60 7.5 V IR 40 i0 7.5 5 i0 3.2 A 20 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 21 Question 6: Circuit Analysis (I) R1 = 80, R2 = 10, R3 = 20, R4 = 90, R5 = 100 Battery: V1 = 12V, V2 = 24V, V3 = 36V

Resistor: I1, I2, , I5 = ? P1, P2, , P5 = ? 22 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

23 Solution 6a VM VB = R5I1 + V1 + R1I1 I1 = (VM VB V1)/(R5 + R1) = (24 VB)/180 Step 3 24 Solution 6a VN VB = R2I2 + R3I2 I2 = (VN VB)/(R2 + R3) = VB/30 Step 3 25

Solution 6a VM VB = V2 + R4I4 I4 = (VM VB V2)/R4 = (12 VB)/90 We get three relationships now (I1, I2, I4) Step 3 26 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 27 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 = 0.178A Step 5 28 Solution 6a

P = I2R = P1 = (0.104)2 80 = 0.86528W P4 = (0.074)2 90 = 0.49284W = VR42 / R (6.66V, 90) 29 Solution 6b Lets try another reference ground VM = 0 I1: B R1 V1 R5 M I2: B R2 R3 V3 M

I4: B R4 V2 M 30 Quick Checking I1: B R1 V1 R5 M I2: B R2 R3 V3 M I4: B R4 V2 M Different direction, different result? 31

Solution 6b KCL of Node B: I1 + I2 + I4 = 0 VB VM = R1I1 V1 + R5I1 I1 = (VB VM + V1)/(R1 + R5) = (VB + 12)/180 32 Solution 6b VB VM = R2I2 + R3I2 V3 I2 = (VB VM + V3)/(R2 + R3) = (VB + 36)/30

33 Solution 6b VB VM = R4I4 V2 I4 = (VB VM + V2)/R4 = (VB + 24)/90 34 Solution 6b KCL of Node B: I1 + I2 + I4 = 0 (VB + 12)/180 + (VB + 36)/30 + (VB + 24)/90 = 0 VB = 92/3 V

35 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 36 Question 7: Circuit Analysis (II) Find vo in the circuit of the figure.

37 Solution 7 Step 1: Define the node voltage (v1,v2,v3) Step 2: Define the current direction 5A v2 v1 2 1 4 v3 8 + v0

-- 40V 20V 38 Solution 7 Apply: 1) V = IR 2) KCL Step 3: Consider node 1 v1 v2 40 v1 5 3v1 v2 70 2 1 1 5A

5A v2 v1 2 v1 (v1-v2)/2 (40-v1)/1 1 4 v3 8 + v0 -- 40V 20V

39 Solution 7 Step 3: Consider node 2 v1 v2 v2 v2 v3 5 4v1 7v2 20 2 4 8 Step 4, 5: From (1) and (2), v1 = 30V, v2 = 20V, v0 = v2 = 20V 2 5A v2

v1 2 5A v2 (v2-v0)/8 (v1-v2)/2 1 4 v3 8 + v0 -- 40V 20V

(v1-0)/4 40 Quick Checking NOT Always always True true R2 If R4 R3 , then i6 0 R5 i2 i3 i4 i5 i2 i6 i3

e1 R4 R2 R4 If i6 0, then R2 R2 R4 R3 V0 R3 R5 41 Quick Checking NOT Always always True true R2

If R4 R3 , then i6 0 R5 i2 i3 i4 i5 i2 i6 i3 e1 R4 R2 R4 If i6 0, then R2

R2 R4 R3 V0 R3 R5 42

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