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newSchoolWorks.com › VoTechWorks › Building & Construction › Solar / PV Installation › Electrical Fundamentals for PV Systems › Lesson 1
BLD-SOLA09-03 · Lesson 1 · 12 clock hours · theory

DC Circuit Theory: Voltage, Current, Resistance, and Power

Objective. Given a DC circuit description, the student will correctly apply Ohm's Law and Watt's Law to calculate voltage, current, resistance, and power with 100% accuracy on a 10-problem set.

⚠ Pending professional review

This lesson, its worked examples and its practice questions has not yet been verified by a licensed professional engineer or master tradesperson or by a certification or approval body. It is a study draft and may contain errors. Do not use it as instructions for work on a job site or on live equipment. Always follow your instructor, your supervising professional, and current official references and regulations.

VoTechWorks program approval and accreditation are still in progress. Credentials are issued only by the certifying bodies named in each program, after their own exams.

DC Circuit Theory: Voltage, Current, Resistance, and Power

Every photovoltaic system is, at its core, a DC electrical circuit. Before you can safely size conductors, select overcurrent protection, or troubleshoot a string that is underperforming, you must be able to calculate voltage, current, resistance, and power with confidence. This lesson covers the two laws you will use on every job: Ohm's Law and Watt's Law.

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The Four Fundamental Quantities

Voltage (V) is electrical pressure — the force that pushes electrons through a conductor. It is measured in volts (V). On a PV module spec sheet you will see two voltage ratings: open-circuit voltage (Voc) and maximum-power-point voltage (Vmp).

Current (I) is the flow rate of electrons through a conductor. It is measured in amperes, or amps (A). The spec sheet lists short-circuit current (Isc) and maximum-power-point current (Imp).

Resistance (R) opposes the flow of current. It is measured in ohms (Ω). Every conductor, connection, and component adds some resistance to the circuit.

Power (P) is the rate at which electrical energy is converted to another form (heat, light, mechanical work, or stored energy). It is measured in watts (W) or kilowatts (kW).

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Ohm's Law

Ohm's Law states that in a DC resistive circuit, voltage equals current multiplied by resistance:

> V = I × R

Rearranged to solve for the other quantities:

A useful memory tool is the Ohm's Law triangle: write V at the top, I and R at the bottom. Cover the quantity you want to find and the remaining symbols show the calculation.

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Watt's Law

Watt's Law relates power to voltage and current:

> P = V × I

Combined with Ohm's Law, two additional forms are commonly used:

Rearranged:

On a 300 W module operating at its maximum power point, if Vmp = 32.5 V, then Imp = 300 ÷ 32.5 ≈ 9.23 A. You will use this relationship constantly when verifying spec-sheet values.

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Applying These Laws to PV Circuits

Series Strings

When modules are wired in series, voltages add and current stays the same throughout the string. If four 300 W modules each have a Vmp of 32.5 V and an Imp of 9.23 A:

Parallel Arrays

When strings are wired in parallel, currents add and voltage stays the same. Two of the above strings in parallel:

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Units and Decimal Precision

Always write the unit with every number. A naked "32.5" means nothing on an inspection document; "32.5 V" is unambiguous. Round calculated answers to two decimal places unless a spec sheet or problem states otherwise. Carry full precision through intermediate steps to avoid rounding error.

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Safety Notes

PV modules generate voltage the moment light hits them — you cannot switch off the sun. Even a single 300 W module can produce lethal DC voltage and arc-flash energy in full sunlight.

Understanding Ohm's Law and Watt's Law helps you recognize when a measured value is dangerously outside the expected range — an essential troubleshooting and safety skill.

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Summary

| Law | Primary Formula | Common Rearrangements |

|---|---|---|

| Ohm's Law | V = I × R | I = V ÷ R; R = V ÷ I |

| Watt's Law | P = V × I | V = P ÷ I; I = P ÷ V; P = I² × R |

Master these two laws and you have the mathematical foundation for every electrical calculation in this program.

Key terms10

Voltage (V)The electrical potential difference between two points in a circuit, measured in volts (V); the 'pressure' that drives current through a conductor.
Current (I)The rate of flow of electric charge through a conductor, measured in amperes (A).
Resistance (R)The opposition a material offers to the flow of electric current, measured in ohms (Ω).
Power (P)The rate at which electrical energy is transferred or converted, measured in watts (W); equal to voltage multiplied by current in a DC resistive circuit.
Ohm's LawThe principle that voltage in a DC resistive circuit equals current multiplied by resistance (V = I × R).
Watt's LawThe principle that power equals voltage multiplied by current (P = V × I), with derived forms P = I² × R and P = V² ÷ R.
Open-Circuit Voltage (Voc)The maximum voltage a PV module produces when no load is connected and no current is flowing; used for string voltage calculations and equipment ratings.
Maximum Power Point (Vmp / Imp)The operating voltage and current at which a PV module delivers its rated peak power output under standard test conditions.
Series StringA configuration in which PV modules are connected positive terminal to negative terminal so that voltages add while current remains the same throughout the string.
Parallel ArrayA configuration in which two or more series strings are connected so that currents add while voltage remains the same across all strings.

Worked examples4

1. A PV module has a Vmp of 32.5 V and an Imp of 9.23 A. What is the module's power output at the maximum power point?
  1. Identify the known values: V = 32.5 V, I = 9.23 A.
  2. Select the correct formula: P = V × I.
  3. Substitute: P = 32.5 V × 9.23 A.
  4. Calculate: P = 299.975 W, rounded to 300.00 W.
Answer: P ≈ 300.00 W — consistent with the module's 300 W nameplate rating.
2. A DC combiner output circuit carries 18.46 A through a conductor with a total resistance of 0.75 Ω. What is the voltage drop across that conductor?
  1. Identify the known values: I = 18.46 A, R = 0.75 Ω.
  2. Select the correct formula: V = I × R.
  3. Substitute: V = 18.46 A × 0.75 Ω.
  4. Calculate: V = 13.845 V, rounded to 13.85 V.
Answer: Voltage drop = 13.85 V. This would represent a significant loss and would prompt the installer to use a larger conductor to reduce resistance.
3. An inverter input circuit operates at 320 V DC and consumes 3,200 W. What current does the circuit carry?
  1. Identify the known values: P = 3,200 W, V = 320 V.
  2. Select the correct formula: I = P ÷ V.
  3. Substitute: I = 3,200 W ÷ 320 V.
  4. Calculate: I = 10.00 A.
Answer: I = 10.00 A. This value would be used to select the minimum ampacity of the inverter input conductors and overcurrent protection.
4. Four identical 300 W modules, each with Vmp = 32.5 V and Imp = 9.23 A, are wired in a series string. What are the string Vmp, string Imp, and total string power?
  1. For a series string, voltages add and current is unchanged: String Vmp = 32.5 V × 4 = 130.00 V.
  2. String Imp = 9.23 A (current is the same throughout a series string).
  3. Calculate string power: P = V × I = 130.00 V × 9.23 A = 1,199.90 W, rounded to 1,199.90 W (≈ 1,200 W).
  4. Verify with nameplate: 4 modules × 300 W = 1,200 W — results are consistent.
Answer: String Vmp = 130.00 V; String Imp = 9.23 A; String power ≈ 1,199.90 W (≈ 1,200 W).

The math in this lesson2

Learners compute series string voltage, current and power from module values.
A-SSE.1A-CED.4
Brush up in the math halls at newSchoolWorks.com:
In class: Instructor-led lecture using a PV module and battery bank as real-world references; students label V, I, R, and P on diagrams of a series string and a parallel array. · Paired problem-solving: students work through 20 Ohm's Law and Watt's Law problems drawn from typical PV module and string specifications, checking each other's arithmetic and units. · Whiteboard relay game where teams race to write the correct formula and solve a randomly selected circuit problem, with instructor correction of errors in real time.
Materials: Whiteboard, sample PV module spec sheets (e.g., 300 W module with Voc, Isc, Vmp, Imp listed), scientific calculators, student workbook Chapter 1, Dunlop 'Photovoltaic Systems' textbook (NJATC).
Check: 10-question written quiz covering Ohm's Law and Watt's Law applied to DC PV circuits; pass mark 80% (8/10 correct); instructor reviews each incorrect answer with the student before moving on.

Practice check8

Answer each question, then check it. Nothing is recorded; this is your own practice.

Electrical Fundamentals / PV System Design
1. A PV module produces Vmp = 36.0 V and Imp = 8.33 A. Using Watt's Law, what is its rated power output?
Electrical Fundamentals
2. Which formula correctly rearranges Ohm's Law to solve for resistance?
PV System Design and Installation
3. Six identical modules, each with Voc = 40.5 V, are wired in series. What is the string Voc?
Electrical Fundamentals
4. A DC circuit has a total resistance of 4.0 Ω and carries a current of 5.0 A. What power is dissipated as heat in the resistance?
PV System Design and Installation
5. Two identical series strings, each with Vmp = 120.0 V and Imp = 9.0 A, are connected in parallel. What is the combined array Imp?
Electrical Fundamentals
6. A conductor in a PV system has a voltage drop of 8.0 V when 4.0 A flows through it. What is the conductor's resistance?
Safety and Safe Work Practices
7. Before measuring voltage on an energized DC PV circuit, a technician should FIRST:
PV System Design and Installation
8. An array produces 5,400 W at a DC bus voltage of 270 V. What current does the inverter receive from the array?
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