1.1 Ohm's Law, Watt's Law & Circuit Mathematics

1.1 Ohm's Law, Watt's Law & Circuit Mathematics

Quick Answer: Direct-current (DC) circuit mathematics rests on two foundational relationships: Ohm's Law ($E = I \times R$) and Watt's Law ($P = E \times I$). In series circuits, current remains constant throughout the loop while voltage drops divide across individual loads. In parallel circuits, voltage across all branches remains identical while total current divides among the branches according to their individual resistances. Mastering circuit reductions and power calculations is essential for passing the Massachusetts Journeyman Electrician examination.

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Core Electrical Quantities and Physical Units

Every electrical circuit consists of an electromotive force acting upon charge carriers moving through an opposing medium to perform work. Understanding the formal definitions and international units (SI) of these quantities is fundamental:

• Electromotive Force / Voltage ($E$ or $V$): The electrical potential difference between two points, representing the work required to move a unit of electric charge. One volt ($1\text{ V}$) equals one joule of energy per coulomb of charge ($1\text{ V} = 1\text{ J/C}$).
• Current Intensity ($I$): The rate of electrical charge flow past a specific cross-section of a conductor. One ampere ($1\text{ A}$) represents the flow of one coulomb of charge per second ($1\text{ A} = 1\text{ C/s} = 6.242 \times 10^{18}\text{ electrons/s}$).
• Resistance ($R$): The opposition offered by a physical material to the flow of electric current. One ohm ($1\ \Omega$) is the resistance that permits one ampere of current to flow when an electromotive force of one volt is applied ($1\ \Omega = 1\text{ V/A}$).
• Power ($P$): The rate at which electrical energy is converted into heat, light, or mechanical work. One watt ($1\text{ W}$) equals one joule of energy converted per second ($1\text{ W} = 1\text{ J/s} = 1\text{ V} \times 1\text{ A}$).

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Ohm's Law Formulas and Algebraic Manipulations

Formulated by Georg Simon Ohm in 1827, Ohm's Law states that current through a conductor between two points is directly proportional to the potential difference across the two points and inversely proportional to the resistance between them.

$$E = I \times R$$ $$I = \frac{E}{R}$$ $$R = \frac{E}{I}$$

Where:

• $E$ = Electromotive force in Volts (V)
• $I$ = Current intensity in Amperes (A)
• $R$ = Resistance in Ohms ($\Omega$)

Mathematical Principles

1. Direct Proportionality ($I \propto E$): If the circuit resistance remains constant, doubling the applied voltage doubles the resulting current flow.
2. Inverse Proportionality ($I \propto 1/R$): If the source voltage remains constant, doubling the circuit resistance reduces the resulting current flow by half.

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Watt's Power Law and Energy Calculations

Watt's Law defines the relationship between electrical power, voltage, and current. Combining Watt's Law ($P = E \times I$) with Ohm's Law allows power to be calculated using any two known variables:

$$P = E \times I$$ $$P = I^2 \times R \quad \text{(substituting } E = I \times R \text{ into Watt's Law)}$$ $$P = \frac{E^2}{R} \quad \text{(substituting } I = \frac{E}{R} \text{ into Watt's Law)}$$

The 12 Formulas of the Ohm's / Watt's Law Wheel

Electrical Work vs. Electrical Power

Power is instantaneous; work (energy consumed) accumulates over time. Electric utilities bill customers for total electrical work measured in kilowatt-hours (kWh):

$$\text{Energy (kWh)} = \frac{\text{Power (Watts)} \times \text{Operating Time (Hours)}}{1,000}$$ $$\text{Cost} = \text{Energy (kWh)} \times \text{Utility Rate (\$/kWh)}$$

Practical Calculation: Water Heater Energy Consumption

An electric storage water heater with a $4,500\text{ W}$ ($4.5\text{ kW}$) resistive element operates an average of $3.5\text{ hours}$ per day. Calculate the energy consumed in a $30\text{-day}$ billing cycle and the total cost at an electric rate of $\text{\textdollar}0.26\text{ per kWh}$:

1. Daily energy consumption: $4.5\text{ kW} \times 3.5\text{ hours} = 15.75\text{ kWh/day}$
2. Monthly energy consumption: $15.75\text{ kWh/day} \times 30\text{ days} = 472.5\text{ kWh}$
3. Billing cost: $472.5\text{ kWh} \times \text{\textdollar}0.26\text{/kWh} = \text{\textdollar}122.85$

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Series Circuits and Kirchhoff's Voltage Law (KVL)

A series circuit provides only a single, continuous conductive path for current flow. Every electron leaving the negative terminal must pass through every connected component before returning to the positive terminal.

The Fundamental Laws of Series Circuits

1. Current is Uniform: Current is identical at every point in the circuit: $$I_T = I_1 = I_2 = I_3 = \dots = I_n$$
2. Resistance is Additive: Total equivalent circuit resistance equals the arithmetic sum of individual 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 circuit loop equals zero. Stated another way, the source voltage equals the sum of the individual voltage drops across the series loads: $$E_T = V_1 + V_2 + V_3 + \dots + V_n$$
4. Total Power is Additive: Total power dissipated equals the sum of powers dissipated by each individual component: $$P_T = P_1 + P_2 + P_3 + \dots + P_n$$

Worked Example: Three Resistors in Series

A $120\text{-volt}$ DC circuit powers three series-connected resistive heating elements with resistances $R_1 = 15\ \Omega$, $R_2 = 25\ \Omega$, 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 ($I_T$): $$I_T = \frac{E_T}{R_T} = \frac{120\text{ V}}{60\ \Omega} = 2.0\text{ A}$$
3. Calculate the individual voltage drops across each resistor: $$V_1 = I_T \times R_1 = 2.0\text{ A} \times 15\ \Omega = 30.0\text{ V}$$ $$V_2 = I_T \times R_2 = 2.0\text{ A} \times 25\ \Omega = 50.0\text{ V}$$ $$V_3 = I_T \times R_3 = 2.0\text{ A} \times 20\ \Omega = 40.0\text{ V}$$ KVL Verification: $30.0\text{ V} + 50.0\text{ V} + 40.0\text{ V} = 120.0\text{ V}$.
4. Calculate power dissipation per resistor: $$P_1 = I_T^2 \times R_1 = (2.0)^2 \times 15 = 4 \times 15 = 60\text{ W}$$ $$P_2 = I_T^2 \times R_2 = (2.0)^2 \times 25 = 4 \times 25 = 100\text{ W}$$ $$P_3 = I_T^2 \times R_3 = (2.0)^2 \times 20 = 4 \times 20 = 80\text{ W}$$ Total Power: $P_T = 60\text{ W} + 100\text{ W} + 80\text{ W} = 240\text{ W}$. Check: $E_T \times I_T = 120\text{ V} \times 2.0\text{ A} = 240\text{ W}$.

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Parallel Circuits and Kirchhoff's Current Law (KCL)

A parallel circuit connects components across the same two electrical common nodes, providing multiple independent paths for current flow.

The Fundamental Laws of Parallel Circuits

1. Voltage is Identical Across All Branches: The full potential difference of the source appears across each parallel branch: $$E_T = V_1 = V_2 = V_3 = \dots = V_n$$
2. Kirchhoff's Current Law (KCL): The algebraic sum of currents entering any electrical junction (node) must equal the sum of currents leaving that junction. In a parallel circuit, total supply current equals the sum of the individual branch currents: $$I_T = I_1 + I_2 + I_3 + \dots + I_n$$
3. Total Equivalent Resistance Decreases: Adding parallel paths always reduces total equivalent circuit resistance. Total resistance is always less than the lowest individual branch resistance: $$\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots + \frac{1}{R_n}$$ $$R_T = \frac{1}{\frac{1}{R_1} + \frac{1}{R_2} + \dots + \frac{1}{R_n}}$$
4. Product-Over-Sum Rule (Two Resistors Only): When exactly two resistors are in parallel, equivalent resistance simplifies to: $$R_T = \frac{R_1 \times R_2}{R_1 + R_2}$$
5. Equal Resistors in Parallel: When $N$ identical resistors of resistance $R$ are connected in parallel: $$R_T = \frac{R}{N}$$

Worked Example: Three Resistors in Parallel

A $240\text{-volt}$ circuit supplies three parallel branch loads: Heater A ($R_1 = 20\ \Omega$), Heater B ($R_2 = 30\ \Omega$), and Heater C ($R_3 = 60\ \Omega$).

1. Calculate total equivalent resistance ($R_T$): $$\frac{1}{R_T} = \frac{1}{20} + \frac{1}{30} + \frac{1}{60} = \frac{3}{60} + \frac{2}{60} + \frac{1}{60} = \frac{6}{60} = \frac{1}{10\ \Omega}$$ $$R_T = 10\ \Omega$$ (Notice that $10\ \Omega$ is smaller than the smallest branch resistance of $20\ \Omega$.)
2. Calculate individual branch currents: $$I_1 = \frac{E_T}{R_1} = \frac{240\text{ V}}{20\ \Omega} = 12.0\text{ A}$$ $$I_2 = \frac{E_T}{R_2} = \frac{240\text{ V}}{30\ \Omega} = 8.0\text{ A}$$ $$I_3 = \frac{E_T}{R_3} = \frac{240\text{ V}}{60\ \Omega} = 4.0\text{ A}$$
3. Calculate total circuit current ($I_T$) via KCL: $$I_T = 12.0\text{ A} + 8.0\text{ A} + 4.0\text{ A} = 24.0\text{ A}$$ Ohm's Law Verification: $I_T = \frac{E_T}{R_T} = \frac{240\text{ V}}{10\ \Omega} = 24.0\text{ A}$.
4. Calculate total power dissipated: $$P_T = E_T \times I_T = 240\text{ V} \times 24.0\text{ A} = 5,760\text{ W} = 5.76\text{ kW}$$

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Comparison of Series vs. Parallel Circuit Characteristics

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Combination (Series-Parallel) Circuits: Step-by-Step Circuit Reduction

Most practical electrical installations represent combination circuits containing both series and parallel sections. To analyze a combination circuit, use the network reduction method:

1. Identify sub-branches connected strictly in parallel or series.
2. Calculate the equivalent resistance of each isolated sub-group.
3. Redraw the circuit replacing those sub-groups with single equivalent resistors.
4. Repeat the process until the entire circuit simplifies to a single total resistance ($R_T$).
5. Calculate total circuit current ($I_T$).
6. Work backward from the source to determine intermediate node voltages and branch currents.

Step-by-Step Reduction Problem

A $120\text{-volt}$ DC circuit features resistor $R_1 = 10\ \Omega$ connected in series with the positive line. This series resistor feeds a parallel bank consisting of $R_2 = 30\ \Omega$ and $R_3 = 60\ \Omega$. The return path from the parallel bank passes through another series resistor $R_4 = 10\ \Omega$ back to the negative source terminal.

(+) ----- [ R1: 10 ohms ] -----+----- [ R2: 30 ohms ] -----+----- [ R4: 10 ohms ] ----- (-)
                               |                           |
                               +----- [ R3: 60 ohms ] -----+

Step 1: Reduce the Parallel Bank ($R_2 \parallel R_3$)

Use the product-over-sum formula for $R_2$ and $R_3$:

$$R_p = \frac{R_2 \times R_3}{R_2 + R_3} = \frac{30 \times 60}{30 + 60} = \frac{1,800}{90} = 20.0\ \Omega$$

Step 2: Calculate Total Equivalent Resistance ($R_T$)

The circuit is now reduced to three series resistances: $R_1$, the parallel equivalent $R_p$, and $R_4$:

$$R_T = R_1 + R_p + R_4 = 10.0\ \Omega + 20.0\ \Omega + 10.0\ \Omega = 40.0\ \Omega$$

Step 3: Calculate Total Circuit Current ($I_T$)

$$I_T = \frac{E_T}{R_T} = \frac{120.0\text{ V}}{40.0\ \Omega} = 3.0\text{ A}$$

Step 4: Calculate Voltage Drops Across Series Sections and Parallel Bank

• Voltage drop across $R_1$: $V_{R1} = I_T \times R_1 = 3.0\text{ A} \times 10.0\ \Omega = 30.0\text{ V}$
• Voltage drop across parallel bank: $V_p = I_T \times R_p = 3.0\text{ A} \times 20.0\ \Omega = 60.0\text{ V}$
• Voltage drop across $R_4$: $V_{R4} = I_T \times R_4 = 3.0\text{ A} \times 10.0\ \Omega = 30.0\text{ V}$

KVL Check: $30.0\text{ V} + 60.0\text{ V} + 30.0\text{ V} = 120.0\text{ V}$ (matches source voltage).

Step 5: Calculate Individual Branch Currents Through $R_2$ and $R_3$

The full $60.0\text{ V}$ parallel bank drop appears across both $R_2$ and $R_3$:

$$I_2 = \frac{V_p}{R_2} = \frac{60.0\text{ V}}{30.0\ \Omega} = 2.0\text{ A}$$ $$I_3 = \frac{V_p}{R_3} = \frac{60.0\text{ V}}{60.0\ \Omega} = 1.0\text{ A}$$

KCL Check: $I_2 + I_3 = 2.0\text{ A} + 1.0\text{ A} = 3.0\text{ A} = I_T$.

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Common Massachusetts Exam Traps: Circuit Mathematics

• The Voltage-Squared Power Trap ($P \propto E^2$): If a heating element rated at $240\text{ V}$ and $2,400\text{ W}$ is connected to a $120\text{ V}$ circuit, candidates frequently guess the power drops in half to $1,200\text{ W}$. Because resistance is fixed ($R = E^2 / P = 240^2 / 2400 = 24\ \Omega$), operating at $120\text{ V}$ yields $P = 120^2 / 24 = 14,400 / 24 = 600\text{ W}$. Cutting voltage in half cuts power to one-quarter (25%)!
• Direct Addition of Parallel Resistors: Candidates under time pressure sometimes add parallel resistances directly ($20 + 30 + 60 = 110\ \Omega$) instead of using reciprocals. Remember: total parallel resistance must always be lower than the smallest branch resistor.
• Opening a Branch in a Parallel Circuit: Opening a switch in one branch of a parallel circuit does not increase the current in the other branches (assuming a stiff voltage source). The other branch currents remain identical, but total line current drops.
• Multi-Wire Branch Circuit (MWBC) Open Neutral: An open neutral on a $120/240\text{V}$ multiwire branch circuit converts two parallel $120\text{V}$ branch circuits into a single $240\text{V}$ series circuit. The higher-resistance load (lower wattage device) experiences excessive overvoltage and burns out, while the lower-resistance load experiences undervoltage.