Exam Revision Notes
Revision Week - Questions and Topics
Revision week, focusing on addressing questions and topics.
Friday will cover questions from class; filter circuits were previously discussed with Kane.
Inquiries about assignments, labs, or course content are welcome.
Delta Star Questions: Three-Phase Circuits
Basic level questions on Delta Star configurations.
Problem: Determine three line currents in a delta-connected load.
Circuit Configuration:
Delta connection with points labeled A, B, C (capital letters) and a, b, c (small letters).
Impedances: Z{ac}, Z{ba}, Z_{cb}.
Lines: A, B, C.
Given Impedances:
Z_{ac} = 60 \Omega \angle 0^{\circ}
Z_{cb} = 30 \Omega \angle -30^{\circ}
Z_{ba} = 30 \Omega \angle 45^{\circ}
Voltages (Phase):
E_{ab} = 120 V \angle 0^{\circ}
E_{bc} = 120 V \angle -120^{\circ}
E_{ca} = 120 V \angle 120^{\circ}
Delta Connection Relationship:
V{line} = V{phase}
I{line} = \sqrt{3} \times I{phase}
Calculating Phase Currents:
I{ac} = \frac{V{ac}}{Z_{ac}}
I{cb} = \frac{V{cb}}{Z_{cb}}
I{ba} = \frac{V{ba}}{Z_{ba}}
Calculations:
I_{ac} = \frac{120 \angle 0^{\circ}}{60 \angle 0^{\circ}} = 2 A \angle 0^{\circ}
I_{cb} = \frac{120 \angle 120^{\circ}}{30 \angle -30^{\circ}} = 4 A \angle 150^{\circ}
I_{ba} = \frac{120 \angle -120^{\circ}}{30 \angle 45^{\circ}} = 4 A \angle -165^{\circ}
Line Currents and Phasor Diagram
Line currents are different from phase currents.
Ia = Iac - Iba
Ib = Iba - Icb
Ic = Icb - Iac
Convert polar to rectangular, perform subtraction, then convert back to polar form.
Example of calculating Ia:
I_a = 2.0 \angle 0^{\circ} - 4 \angle -165^{\circ}
Convert to rectangular: 2 + j0 - (-3.864 - j1.035) = 5.864 + j1.035
Convert back to polar: 6 A \angle 10^{\circ}
Ib = 3.1 A at -98 degrees.
Ic = 6 A at 160 degrees.
Unbalanced Load: Different loads result in different currents.
Wye (Y) Connection Running a Delta-Connected Load
Balanced load with impedance Z = 52 \Omega \angle 45^{\circ} in each arm of the delta.
Y-connected source:
Voltage: 120V (RMS) or 208V (peak).
Objective: Find the magnitude of the current in each coil of the source.
Calculations:
I{ac} = I{cb} = I_{ba} = \frac{208 V}{52 \Omega} = 4 A
I{line} = \sqrt{3} \times I{peak} = 1.732 \times 4 A = 6.93 A
Total Power Calculation
Find the total power input to the load.
Power factor: \cos(\theta) = \cos(45^{\circ}) = 0.707 lagging (inductive).
Total power: Pt = \sqrt{3} \times V{line} \times I_{line} \times \cos(\phi)
P_t = 1.732 \times 208 V \times 6.93 A \times 0.707 = 1.8 kW
Efficiency Example
550V three-phase motor delivers 15 horsepower to a mechanical load.
Efficiency: 80% (\frac{P{out}}{P{in}} \times 100).
Power factor: 0.9 lagging.
Objective: Find the line current.
Calculations:
Convert horsepower to watts: 1 HP = 746 W
Total power input: P_{total} = \frac{15 HP \times 746 W/HP}{0.8} = 14 kW
Line current: I{line} = \frac{P{total}}{\sqrt{3} \times V_L \times power factor} = \frac{14000 W}{\sqrt{3} \times 550 V \times 0.9} = 16 A
Delta Load Example with Different Impedances
Impedances:
Z_{ab} = 52 \Omega \angle -30^{\circ}
Z_{bc} = 52 \Omega \angle 45^{\circ}
Z_{ca} = 104 \Omega \angle 0^{\circ}
Line voltage: 208V (peak).
Objective: Find the magnitude of the current in each line for two phase sequences:
ABC
CBA
Sequence ABC
Step 1: Calculate phase currents.
I{ab} = \frac{V{ab}}{Z_{ab}} = \frac{208 V \angle 0^{\circ}}{52 \Omega \angle -30^{\circ}} = 4 A \angle 30^{\circ}
I{bc} = \frac{V{bc}}{Z_{bc}} = \frac{208 V \angle -120^{\circ}}{52 \Omega \angle 45^{\circ}} = 4 A \angle -165^{\circ}
I{ca} = \frac{V{ca}}{Z_{ca}} = \frac{208 V \angle 120^{\circ}}{104 \Omega \angle 0^{\circ}} = 2 A \angle 120^{\circ}
Step 2: Calculate line currents.
Ia = I{ab} - I_{ca} = 4 \angle 30^{\circ} - 2 \angle 120^{\circ} (convert to rectangular, subtract, convert back to polar: 4.64 + j0.268 A
Ib = I{bc} - I_{ab} = 4 \angle -165^{\circ} - 4 \angle 30^{\circ}
Ic = I{ca} - I_{bc} = 2 \angle 120^{\circ} - 4 \angle -165^{\circ} = 4 A \angle 0^{\circ}
Sequence CBA
I{ab} = \frac{V{ab}}{Z_{ab}} = \frac{208 V \angle 0^{\circ}}{52 \Omega \angle -30^{\circ}} = 4 A \angle 30^{\circ}
I{bc} = \frac{V{bc}}{Z_{bc}} = \frac{208 V \angle 0^{\circ}}{52 \Omega \angle 45^{\circ}} = 4 A \angle 75^{\circ}
I{ca} = \frac{V{ca}}{Z_{ca}} = \frac{208 V \angle -120^{\circ}}{104 \Omega \angle 0^{\circ}} = 2 A \angle -120^{\circ}
Calculate Line Currents:
Ia = I{ab} - I_{ca}
Ib = I{bc} - I_{ab}
Ic = I{ca} - I_{bc}
Star (Y) Connection: Resistor and Capacitor
Components: Resistors (12 ohms) and capacitors (-j9 ohms).
Objective: Compute Ia, Ib, and I_c.
Given: V_{an} = 120 V \angle 0^{\circ} (RMS, phase voltage).
Calculations:
Ia = \frac{V{an}}{Z_{an}} = \frac{120 V \angle 0^{\circ}}{12 - j9} = \frac{120 V \angle 0^{\circ}}{15 \angle -36.87^{\circ}} = 8 A \angle 36.87^{\circ}
Ib = \frac{V{bn}}{Z_{bn}}
Ic = \frac{V{cn}}{Z_{cn}}
Additional Problems
Problem 1: For the same Y-connected circuit:
If V{ab} = 600 V \angle 0^{\circ}, what are Ia, Ib, and Ic?
Problem 2: For the same Y-connected circuit:
If V{bc} = 600 V \angle -120^{\circ}, what are Ia, Ib, and Ic?
Answers:
Problem 1:
I_a = 23.1 A \angle 6.9^{\circ}
I_b = 23.1 A \angle -113.1^{\circ}
I_c = 23.1 A \angle 126.9^{\circ}
Problem 2:
I_a = 23.1 A \angle 36.9
I_b = 23.1 A \angle -83.1^{\circ}
I_c = 23.1 A \angle 156.9^{\circ}
Delta-Connected Circuit Problems
Circuit: Delta connection with resistors and inductors.
Values: 10 ohms and j3 ohms.
Given: V_{ab} = 240 V \angle 15^{\circ}
Objectives:
Find phase current (I_{phase}).
Find line current (I_{line}).
Sketch the phasor diagram.
Note: In delta, V{phase} = V{line}, and I{line} = \sqrt{3} \times I{phase}.
Hint: For inductive circuits, the line current lags the phase current by 30 degrees.
Ia lags I{ab} by 30 degrees.
Ib lags I{bc} by 30 degrees.
Ic lags I{ca} by 30 degrees.
More Delta-Connected Circuit Problems
Problem 1: If Ia = 17.32 A \angle 20^{\circ}, determine I{ab} and V_{ab}.
Problem 2: If I{bc} = 5 A \angle -140^{\circ}, what is V{ab}?
Answers:
Problem 1:
I_{ab} = 10 A \angle 50^{\circ}
V_{ab} = 240 V \angle 66.7^{\circ}
Problem 2:
V_{ab} = 52.2 V \angle -3.3^{\circ}