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.
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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:
- I = V ÷ R (current equals voltage divided by resistance)
- R = V ÷ I (resistance equals voltage divided by current)
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:
- P = I² × R (useful when you know current and resistance)
- P = V² ÷ R (useful when you know voltage and resistance)
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:
- String Vmp = 32.5 × 4 = 130 V
- String Imp = 9.23 A (unchanged)
- String power = 130 × 9.23 ≈ 1,200 W
Parallel Arrays
When strings are wired in parallel, currents add and voltage stays the same. Two of the above strings in parallel:
- Array Vmp = 130 V (unchanged)
- Array Imp = 9.23 × 2 = 18.46 A
- Array power = 130 × 18.46 ≈ 2,400 W
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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.
- Lockout/tagout (LOTO): Follow your site's written LOTO procedure before working on any part of the circuit. Cover modules with an opaque tarp when you must de-energize the DC side.
- PPE: Wear voltage-rated insulated gloves appropriate for the system voltage, safety glasses, and arc-flash-rated clothing when the hazard assessment requires it.
- Never assume a circuit is de-energized. Use a calibrated meter to verify voltage is absent before touching conductors.
- DC arc flash: DC arcs are harder to extinguish than AC arcs. Treat every DC conductor as energized until proven otherwise.
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?
- Identify the known values: V = 32.5 V, I = 9.23 A.
- Select the correct formula: P = V × I.
- Substitute: P = 32.5 V × 9.23 A.
- 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?
- Identify the known values: I = 18.46 A, R = 0.75 Ω.
- Select the correct formula: V = I × R.
- Substitute: V = 18.46 A × 0.75 Ω.
- 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?
- Identify the known values: P = 3,200 W, V = 320 V.
- Select the correct formula: I = P ÷ V.
- Substitute: I = 3,200 W ÷ 320 V.
- 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?
- For a series string, voltages add and current is unchanged: String Vmp = 32.5 V × 4 = 130.00 V.
- String Imp = 9.23 A (current is the same throughout a series string).
- 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).
- 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
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.
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