AC source and load modeling in Simulink - Step-by-Step Execution

Concept Flow - AC source and load modeling

Define AC Source Parameters

↓

Generate AC Voltage Signal

↓

Connect Load Model

↓

Calculate Current and Power

↓

Analyze Output Waveforms

↓

End

This flow shows how an AC source is defined, connected to a load, and how current and power are calculated and analyzed.

Execution Sample

Simulink

ac_voltage = Vm * sin(2 * pi * f * t + phase)
current = ac_voltage / load_impedance
power = ac_voltage * current
plot(t, ac_voltage, t, current)

This code models an AC voltage source, calculates current through a load, computes power, and plots voltage and current over time.

Execution Table

Step	Time (t)	AC Voltage (V)	Load Impedance (Ohm)	Current (A)	Power (W)
1	0	0.0	10	0.0	0.0
2	0.001	6.28	10	0.628	3.94
3	0.002	12.35	10	1.235	15.25
4	0.003	17.32	10	1.732	30.0
5	0.004	20.0	10	2.0	40.0
6	0.005	20.0	10	2.0	40.0
7	0.006	17.32	10	1.732	30.0
8	0.007	12.35	10	1.235	15.25
9	0.008	6.28	10	0.628	3.94
10	0.009	0.0	10	0.0	0.0
11	0.01	-6.28	10	-0.628	3.94
12	0.011	-12.35	10	-1.235	15.25
13	0.012	-17.32	10	-1.732	30.0
14	0.013	-20.0	10	-2.0	40.0
15	0.014	-20.0	10	-2.0	40.0
16	0.015	-17.32	10	-1.732	30.0
17	0.016	-12.35	10	-1.235	15.25
18	0.017	-6.28	10	-0.628	3.94
19	0.018	0.0	10	0.0	0.0
20	0.019	6.28	10	0.628	3.94
Exit	0.02	0.0	10	0.0	0.0

💡 Simulation time reached 0.02 seconds, completing one full AC cycle.

Variable Tracker

Variable	Start	After 1	After 2	After 3	After 4	After 5	After 6	After 7	After 8	After 9	After 10	Final
ac_voltage (V)	0.0	6.28	12.35	17.32	20.0	20.0	17.32	12.35	6.28	0.0	0.0	0.0
current (A)	0.0	0.628	1.235	1.732	2.0	2.0	1.732	1.235	0.628	0.0	0.0	0.0
power (W)	0.0	3.94	15.25	30.0	40.0	40.0	30.0	15.25	3.94	0.0	0.0	0.0

Key Moments - 2 Insights

Why does the current have the same shape as the voltage but scaled down?

Because current is voltage divided by load impedance (Ohm's law), so it follows the voltage waveform shape but with smaller amplitude as shown in execution_table rows 2-10.

Why is power always positive even when voltage and current are negative?

Power is voltage times current; negative times negative gives positive power, indicating energy flow direction, as seen in execution_table rows 11-18.

Visual Quiz - 3 Questions

Look at the execution table at step 5, what is the current value?

A 2.0 A

B 1.732 A

C 0.628 A

D 0.0 A

💡 Hint

Check the 'Current (A)' column at step 5 in the execution_table.

At which step does the AC voltage first become negative?

A Step 10

B Step 11

C Step 9

D Step 12

💡 Hint

Look at the 'AC Voltage (V)' column in execution_table rows 10-12.

If the load impedance doubles, how does the current at step 5 change?

A It doubles

B It stays the same

C It halves

D It becomes zero

💡 Hint

Current = Voltage / Load Impedance; see variable_tracker for current changes.

Concept Snapshot

AC Source and Load Modeling:
- AC voltage: V = Vm * sin(2πft + phase)
- Current: I = V / Load Impedance
- Power: P = V * I
- Load affects current amplitude
- Power positive when voltage and current signs match

Full Transcript

This visual execution traces AC source and load modeling in Simulink. We start by defining AC source parameters, generating the voltage waveform, connecting a load, calculating current and power, and analyzing waveforms. The execution table shows voltage, current, and power values at each time step over one AC cycle. Current follows voltage shape scaled by load impedance. Power remains positive even when voltage and current are negative due to multiplication of two negatives. Key moments clarify why current waveform matches voltage shape and why power is always positive. The quiz tests understanding of current values, voltage sign changes, and effect of load impedance on current. The snapshot summarizes formulas and key points for AC source and load modeling.

Practice

1. What is the main purpose of using an AC Voltage Source block in Simulink when modeling electrical circuits?

easy

A. To convert AC to DC voltage

B. To simulate an alternating current power supply with specific voltage and frequency

C. To measure the resistance of a load in the circuit

D. To store electrical energy like a battery

Solution

1. Step 1: Understand the role of AC Voltage Source block

   The AC Voltage Source block provides an alternating voltage signal with set amplitude and frequency, simulating real AC power supplies.

2. Step 2: Compare with other options

   Measuring resistance, converting AC to DC, or storing energy are not functions of this block.

3. Final Answer:

   To simulate an alternating current power supply with specific voltage and frequency -> Option B

4. Quick Check:

   AC Voltage Source = simulate AC power [OK]

Hint: AC Voltage Source creates AC power signals [OK]

Common Mistakes:

• Confusing AC source with DC source
• Thinking it measures resistance
• Assuming it stores energy

2. Which of the following is the correct way to connect an RLC Load block to an AC Voltage Source block in Simulink for proper simulation?

easy

A. Connect the positive terminal of AC Voltage Source to the RLC Load input and the negative terminal to the RLC Load output

B. Connect the AC Voltage Source output directly to ground without the RLC Load

C. Connect the RLC Load output to the AC Voltage Source input terminal

D. Connect the AC Voltage Source output to the RLC Load input terminal and the other terminal of RLC Load to ground

Solution

1. Step 1: Understand connection basics in Simulink

   The AC Voltage Source output should connect to the RLC Load input terminal to supply voltage, and the other terminal of the load should connect to ground to complete the circuit.

2. Step 2: Eliminate incorrect options

   Connect the positive terminal of AC Voltage Source to the RLC Load input and the negative terminal to the RLC Load output incorrectly connects terminals; Connect the AC Voltage Source output directly to ground without the RLC Load misses the load; Connect the RLC Load output to the AC Voltage Source input terminal reverses input/output roles.

3. Final Answer:

   Connect the AC Voltage Source output to the RLC Load input terminal and the other terminal of RLC Load to ground -> Option D

4. Quick Check:

   Source output to load input, load to ground [OK]

Hint: Connect source output to load input, load other end to ground [OK]

Common Mistakes:

• Reversing input and output terminals
• Omitting ground connection
• Connecting load directly to ground without source

3. Consider this Simulink setup: An AC Voltage Source block set to 120 V RMS at 60 Hz connected to an RLC Load with R=10 Ω, L=0.1 H, and C=100 μF. What is the expected behavior of the current waveform?

medium

A. The current waveform will be a sine wave with the same frequency but phase shifted due to the RLC load

B. The current waveform will be a constant DC value because of the capacitor

C. The current waveform will be zero because the load blocks AC

D. The current waveform will be a square wave due to the inductance

Solution

1. Step 1: Analyze the AC source and RLC load interaction

   The AC source provides a sine voltage at 60 Hz. The RLC load causes the current to also be sinusoidal but shifted in phase depending on resistance, inductance, and capacitance.

2. Step 2: Evaluate waveform types

   Capacitors and inductors cause phase shifts but do not convert sine waves to DC or square waves. The current will not be zero unless the load is open.

3. Final Answer:

   The current waveform will be a sine wave with the same frequency but phase shifted due to the RLC load -> Option A

4. Quick Check:

   RLC load causes phase shift, current sine wave [OK]

Hint: RLC load shifts phase, current stays sine wave [OK]

Common Mistakes:

• Thinking capacitor creates DC current
• Assuming inductance creates square wave
• Believing load blocks all AC current

4. You built a Simulink model with an AC Voltage Source and an RLC Load, but the simulation shows zero current. What is the most likely error?

medium

A. The RLC Load block is not connected properly to the AC Voltage Source

B. The AC Voltage Source frequency is set to zero

C. The simulation time is too short to see current

D. The RLC Load parameters are set to zero resistance, inductance, and capacitance

Solution

1. Step 1: Check connection between source and load

   If the load is not connected properly to the source, no current can flow, resulting in zero current in simulation.

2. Step 2: Evaluate other options

   Frequency zero means DC but current can still flow; short simulation time shows initial values; zero parameters mean a short circuit, current would be high, not zero.

3. Final Answer:

   The RLC Load block is not connected properly to the AC Voltage Source -> Option A

4. Quick Check:

   Improper connection causes zero current [OK]

Hint: Check connections first if current is zero [OK]

Common Mistakes:

• Ignoring wiring errors
• Assuming zero frequency means no current
• Thinking zero parameters mean no current

5. You want to model an AC circuit in Simulink with a 230 V RMS, 50 Hz source feeding a load that has a resistance of 20 Ω and an inductance of 0.05 H. How do you calculate the load's total impedance magnitude to analyze current amplitude?

hard

A. Calculate impedance as \( Z = R + L \) directly added

B. Calculate impedance as \( Z = R \times L \) multiplied

C. Calculate impedance as \( Z = \sqrt{R^2 + (2\pi f L)^2} \) where \( f \) is frequency

D. Calculate impedance as \( Z = R - 2\pi f L \) subtracting reactance

Solution

1. Step 1: Recall impedance formula for R-L circuit

   The total impedance magnitude \( Z \) is the square root of the sum of squares of resistance \( R \) and inductive reactance \( X_L = 2\pi f L \).

2. Step 2: Apply formula with given values

   Use \( Z = \sqrt{R^2 + (2\pi f L)^2} \) to find magnitude, not simple addition or subtraction.

3. Final Answer:

   Calculate impedance as \( Z = \sqrt{R^2 + (2\pi f L)^2} \) where \( f \) is frequency -> Option C

4. Quick Check:

   Impedance magnitude = sqrt(R² + (2πfL)²) [OK]

Hint: Use Pythagorean formula for R and inductive reactance [OK]

Common Mistakes:

• Adding resistance and inductance directly
• Multiplying resistance by inductance
• Subtracting reactance from resistance