2.1 Exam Math, Ohm's Law, Watt's Law & Series/Parallel Circuits

2.1 Exam Math, Ohm's Law, Watt's Law & Series/Parallel Circuits

Every electrical installation governed by the National Electrical Code rests on fundamental physical laws that describe how electrical charges behave under potential difference. For the journeyworker electrician, mastery of direct current (DC) theory is not merely academic—it forms the mathematical foundation for sizing conductors, calculating overcurrent protection, assessing equipment heat dissipation, and troubleshooting field wiring faults.

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Fundamentals of Direct Current Theory

Direct current (DC) represents the unidirectional flow of electric charge. Four basic physical quantities govern every DC circuit:

1. Electromotive Force (Voltage, $V$ or $E$): Measured in volts ($\text{V}$), voltage represents the electrical potential difference between two points, providing the pressure that forces free electrons through a conductive medium.
2. Current ($I$): Measured in amperes ($\text{A}$), current represents the rate of charge flow through a conductor, where one ampere equals one coulomb of charge moving past a specific cross-section per second ($1\text{ A} = 1\text{ C/s} = 6.242 \times 10^{18}\text{ electrons/s}$).
3. Resistance ($R$): Measured in ohms ($\Omega$), resistance represents the opposition a material offers to the movement of free electrons, converting electrical energy into heat.
4. Power ($P$): Measured in watts ($\text{W}$), power represents the rate at which electrical energy is transformed into another form of work or heat, where one watt equals one joule per second ($1\text{ W} = 1\text{ J/s}$).

The Ohm's Law and Watt's Law Formula Wheel

In 1827, German physicist Georg Simon Ohm verified experimentally that current through a conductor between two points is directly proportional to voltage across the points and inversely proportional to resistance. James Watt established the relationship between potential, current, and mechanical work. Combining Ohm's Law ($V = I \times R$) and Watt's Law ($P = V \times I$) yields 12 fundamental algebraic equations.

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DC Series Circuit Analysis

A series circuit provides exactly one continuous, uninterrupted conductive path through which current can flow. Because there are no junction points or alternate paths, every electron leaving the negative power terminal must pass sequentially through every connected load before returning to the source.

The Three Cardinal Laws of Series Circuits

1. Current is Uniform: Current flow is identical at every point in the series string: $I_T = I_1 = I_2 = I_3 = \dots = I_n$
2. Resistance is Additive: Total circuit resistance equals the arithmetic sum of individual component resistances: $R_T = R_1 + R_2 + R_3 + \dots + R_n$
3. Kirchhoff's Voltage Law (KVL): The algebraic sum of all voltages around any closed loop must equal zero. In practice, the total source voltage equals the sum of the individual component voltage drops: $V_T = V_1 + V_2 + V_3 + \dots + V_n$

The Voltage Divider Principle

Because current is identical through all series elements, the voltage dropped across any single resistor is directly proportional to its share of the total resistance:

$V_x = V_T \times \left(\frac{R_x}{R_T}\right)$

Step-by-Step Series Circuit Calculation

Consider an industrial control circuit powered by a $120\text{ V}$ DC supply connected to three series resistors: $R_1 = 15\ \Omega$, $R_2 = 25\ \Omega$, and $R_3 = 20\ \Omega$.

1. Calculate Total Resistance ($R_T$): $R_T = 15\ \Omega + 25\ \Omega + 20\ \Omega = 60\ \Omega$
2. Calculate Total Circuit Current ($I_T$): $I_T = \frac{V_T}{R_T} = \frac{120\text{ V}}{60\ \Omega} = 2.0\text{ A}$
3. Calculate Individual Voltage Drops ($V_1, V_2, V_3$):
   • $V_1 = I_T \times R_1 = 2.0\text{ A} \times 15\ \Omega = 30\text{ V}$
   • $V_2 = I_T \times R_2 = 2.0\text{ A} \times 25\ \Omega = 50\text{ V}$
   • $V_3 = I_T \times R_3 = 2.0\text{ A} \times 20\ \Omega = 40\text{ V}$
   • Verification: $30\text{ V} + 50\text{ V} + 40\text{ V} = 120\text{ V}$ (KVL confirmed).
4. Calculate Power Dissipation:
   • $P_1 = I^2 \times R_1 = (2.0)^2 \times 15 = 60\text{ W}$
   • $P_2 = I^2 \times R_2 = (2.0)^2 \times 25 = 100\text{ W}$
   • $P_3 = I^2 \times R_3 = (2.0)^2 \times 20 = 80\text{ W}$
   • Total Power: $P_T = 60 + 100 + 80 = 240\text{ W}$ (Or $P_T = V_T \times I_T = 120 \times 2.0 = 240\text{ W}$).

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DC Parallel Circuit Analysis

A parallel circuit connects two or more electrical loads between the same pair of common electrically common nodes. Parallel wiring is the standard distribution topology for premises wiring because every branch load operates at full system voltage regardless of the presence, operation, or failure of other connected loads.

The Three Cardinal Laws of Parallel Circuits

1. Voltage is Uniform: Every parallel branch experiences the exact same potential difference: $V_T = V_1 = V_2 = V_3 = \dots = V_n$
2. Kirchhoff's Current Law (KCL): The total current entering a junction equals the total current leaving the junction. Total circuit current equals the sum of branch currents: $I_T = I_1 + I_2 + I_3 + \dots + I_n$
3. Reciprocal Equivalent Resistance: Adding parallel paths increases the cross-sectional conductive area available to current, meaning total equivalent resistance is always strictly less than the smallest individual branch resistance: $\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots + \frac{1}{R_n}$

Specialized Parallel Resistance Shortcuts

• Two Resistors in Parallel (Product Over Sum): $R_T = \frac{R_1 \times R_2}{R_1 + R_2}$
• $n$ Identical Resistors in Parallel: $R_T = \frac{R}{n}$

The Current Divider Principle

When current divides between two parallel branches, branch current is inversely proportional to branch resistance:

$I_1 = I_T \times \left(\frac{R_2}{R_1 + R_2}\right)$

Step-by-Step Parallel Circuit Calculation

A $240\text{ V}$ DC supply feeds three parallel heating elements: $R_1 = 20\ \Omega$, $R_2 = 30\ \Omega$, and $R_3 = 60\ \Omega$.

1. Calculate Total Equivalent Resistance ($R_T$): $\frac{1}{R_T} = \frac{1}{20} + \frac{1}{30} + \frac{1}{60} = \frac{3 + 2 + 1}{60} = \frac{6}{60} = \frac{1}{10\ \Omega}$ $R_T = 10\ \Omega$
2. Calculate Branch Currents ($I_1, I_2, I_3$):
   • $I_1 = \frac{240\text{ V}}{20\ \Omega} = 12\text{ A}$
   • $I_2 = \frac{240\text{ V}}{30\ \Omega} = 8\text{ A}$
   • $I_3 = \frac{240\text{ V}}{60\ \Omega} = 4\text{ A}$
3. Calculate Total Circuit Current ($I_T$): $I_T = 12\text{ A} + 8\text{ A} + 4\text{ A} = 24\text{ A}$
   • Verification using Ohm's Law: $I_T = \frac{240\text{ V}}{10\ \Omega} = 24\text{ A}$ (KCL confirmed).
4. Calculate Power Dissipation:
   • $P_1 = 240\text{ V} \times 12\text{ A} = 2,880\text{ W}$
   • $P_2 = 240\text{ V} \times 8\text{ A} = 1,920\text{ W}$
   • $P_3 = 240\text{ V} \times 4\text{ A} = 960\text{ W}$
   • $P_T = 2,880 + 1,920 + 960 = 5,760\text{ W} = 5.76\text{ kW}$.

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Combination (Series-Parallel) Resistor Networks

Practical field circuits often present combination series-parallel configurations. A classic example is a long branch circuit where the conductor lead resistance sits in series with parallel-connected lighting or receptacle loads.

The Step-by-Step Reduction Method

To solve a complex combination network:

1. Identify Pure Parallel Clusters: Locate sub-networks of resistors connected across identical nodes and reduce them to single equivalent resistances ($R_p$).
2. Identify Pure Series Elements: Locate resistors connected in series with the equivalent blocks and sum them.
3. Redraw the Simplified Circuit: Repeat reduction until a single equivalent total resistance ($R_T$) remains.
4. Calculate Total Circuit Current: Apply Ohm's Law ($I_T = \frac{V_T}{R_T}$).
5. Traverse Backward: Apply Ohm's Law, KVL, and KCL to calculate intermediate node voltages and branch currents.

Worked Example: Combination Circuit Analysis

A $120\text{ V}$ DC power source supplies a series resistor $R_1 = 10\ \Omega$ connected in line with a parallel bank containing $R_2 = 30\ \Omega$ and $R_3 = 60\ \Omega$.

1. Reduce the Parallel Bank ($R_{2,3}$): $R_{2,3} = \frac{R_2 \times R_3}{R_2 + R_3} = \frac{30 \times 60}{30 + 60} = \frac{1800}{90} = 20\ \Omega$
2. Calculate Total Resistance ($R_T$): $R_T = R_1 + R_{2,3} = 10\ \Omega + 20\ \Omega = 30\ \Omega$
3. Calculate Total Current ($I_T$): $I_T = \frac{V_T}{R_T} = \frac{120\text{ V}}{30\ \Omega} = 4.0\text{ A}$
4. Calculate Voltage Drop Across Series Resistor $R_1$: $V_{R1} = I_T \times R_1 = 4.0\text{ A} \times 10\ \Omega = 40\text{ V}$
5. Calculate Voltage Across the Parallel Bank ($V_{2,3}$): $V_{2,3} = V_T - V_{R1} = 120\text{ V} - 40\text{ V} = 80\text{ V}$
6. Calculate Individual Branch Currents ($I_2, I_3$):
   • $I_2 = \frac{V_{2,3}}{R_2} = \frac{80\text{ V}}{30\ \Omega} \approx 2.67\text{ A}$
   • $I_3 = \frac{V_{2,3}}{R_3} = \frac{80\text{ V}}{60\ \Omega} \approx 1.33\text{ A}$
   • Verification: $I_2 + I_3 = 2.67\text{ A} + 1.33\text{ A} = 4.0\text{ A} = I_T$.

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The Math DLI Expects You to Do by Hand

DLI's exam guide lists general mathematics for every license: transposing equations and calculating area, volume, and percentages. Its own worked examples show the level expected:

• Percent of a number: 70% of 140 is $140 \times 0.70 = 98$.
• Increasing by a percent: increasing 120 by 25% is $120 \times 1.25 = 150$. The same move gives 125% of a continuous load.
• Ratios: a 480/120 V transformer has a primary-to-secondary voltage ratio of $480 \div 120 = 4$, written 4:1.
• Transposing: if $A \times B = C$, dividing both sides by $B$ gives $A = C \div B$. Ohm's law and Watt's law can each be rearranged this way into the other forms in the table above.
• Area and volume: box fill uses cubic inches (a 60 cu in box holds $60 \div 2.25 = 26.67$, so 26 conductors of 12 AWG, because you never round up), and conduit fill uses square inches from Chapter 9.

DLI writes Ohm's law as $E = I \times R$, with E for electromotive force in volts. Treat E and V as the same quantity. Unless a question says otherwise, DLI assumes unity power factor, so $P = E \times I$ for single-phase loads.

DLI's own parallel-resistance examples

For a series circuit, DLI's example uses 20 Ω, 40 Ω, and 60 Ω carrying 2 A: $R_T = 120$ Ω and $E = 2 \times 120 = 240$ V.

Comparison Table: Series vs. Parallel Circuits