6.3 Ohm's Law, Power & Direct Current Calculations

Ohm's Law, Power & Direct Current Calculations

The fundamental relationship linking potential difference, electric current, and resistance was formulated in 1827 by Bavarian physicist Georg Simon Ohm. Known universally as Ohm's Law, it forms the mathematical cornerstone of electrical engineering science and installation practice.

Alongside Ohm's Law, understanding how electrical potential and current transfer energy over time—electrical power ($P$) and electrical energy ($E$)—is vital for selecting cable sizes, rating protective switchgear, evaluating thermal dissipation, and calculating commercial electrical running costs.

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Ohm's Law: Statement and Mathematical Formulations

Ohm's Law states that:

The electric current flowing through a conductor is directly proportional to the potential difference across its ends, provided that all physical conditions (particularly temperature) remain strictly constant.

Mathematically, this direct proportionality is expressed as:

$V \propto I \implies \frac{V}{I} = \text{Constant} = R$

Where:

• $V$ = Potential difference (voltage) across the conductor in Volts ($\text{V}$)
• $I$ = Electric current flowing through the conductor in Amperes ($\text{A}$)
• $R$ = Electrical resistance of the conductor in Ohms ($\Omega$)

The Three Mathematical Formulations

Depending on which circuit variable is unknown, Ohm's Law is transposed into three forms:

$V = I \times R, \qquad I = \frac{V}{R}, \qquad R = \frac{V}{I}$

• Calculating Voltage ($V = IR$): If a current of $4.0\text{ A}$ passes through a circuit resistance of $25\ \Omega$, the potential difference is $V = 4.0 \times 25 = 100\text{ V}$.
• Calculating Current ($I = \frac{V}{R}$): If a $230\text{ V}$ supply is connected across an element having a resistance of $46\ \Omega$, the current is $I = \frac{230}{46} = 5.0\text{ A}$.
• Calculating Resistance ($R = \frac{V}{I}$): If a potential difference of $24\text{ V}$ causes a current of $1.5\text{ A}$ to flow, the resistance is $R = \frac{24}{1.5} = 16\ \Omega$.

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Ohmic versus Non-Ohmic Components

Ohm's Law holds true only when physical conditions—most critically conductor temperature—remain constant. Components that comply with this linear relationship are termed ohmic conductors, whereas components whose resistance changes under varying electrical conditions are termed non-ohmic.

1. Ohmic Conductors

• Standard Carbon or Metal Film Resistors: When operated within their rated power dissipation limits so that temperature rise is negligible, these components exhibit a constant resistance regardless of applied voltage or current direction.
• Metallic Conductor Wires at Constant Temperature: Test circuits maintained in oil baths or subjected to brief, low-current measurement pulses exhibit linear $V$-$I$ behaviour.

2. Non-Ohmic Devices

• Tungsten Filament Lamps: A filament lamp is a prime example of a non-ohmic device. When unenergized at room temperature ($20^\circ\text{C}$), the cold resistance of the fine tungsten wire is very low (typically less than $10\%$ of its rated operating resistance). When switched on, a large initial inrush current surge flows. Within milliseconds, resistive heating elevates the filament temperature to over $2,500^\circ\text{C}$. Because tungsten has a strong Positive Temperature Coefficient (PTC), its resistance increases by a factor of 10 to 12. As a result, the $V$-$I$ curve flattens out at higher voltages—current increases much more slowly than voltage.
• Semiconductor Diodes: A p-n junction diode conducts current predominantly in one direction. In forward bias, virtually zero current flows until the applied voltage overcomes the internal barrier potential (approximately $0.6\text{ V}$ to $0.7\text{ V}$ for silicon). Once this threshold is exceeded, current increases exponentially with negligible additional voltage rise. In reverse bias, only microamperes of leakage current flow until catastrophic reverse breakdown occurs. The $V$-$I$ curve is completely non-linear.
• Gas Discharge Lamps (Fluorescent, Sodium, Metal Halide): Once an electric arc is struck through the ionized gas, the dynamic resistance of the lamp drops sharply. In fact, discharge lamps display negative incremental resistance—as current increases, more gas atoms ionize, lowering the resistance further. Without a series inductive choke (ballast) or electronic control gear to limit current, the lamp would draw destructive overcurrent and self-destruct instantly.

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Electrical Power ($P$) in Direct Current Circuits

Electrical power ($P$) is defined as the rate at which electrical energy is converted into other forms of energy (such as heat, light, mechanical motion, or chemical change). The SI unit of electrical power is the Watt (symbol $\text{W}$), named after Scottish engineer James Watt. One Watt equals one Joule of energy converted per second ($1\text{ W} = 1\text{ J/s}$).

Derivation of the Three Fundamental Power Formulas

1. The Primary Power Formula: $P = V \times I$ By definition, electric potential difference ($V$) represents work done ($W$) per unit charge ($Q$), so $V = \frac{W}{Q} \implies W = V \times Q$. Current ($I$) represents the rate of charge flow, so $I = \frac{Q}{t} \implies Q = I \times t$. Substituting this into work done yields $W = V \times I \times t$. Power is rate of work done ($P = \frac{W}{t}$): $P = \frac{V \times I \times t}{t} = V \times I$ Power (Watts) = Potential Difference (Volts) $\times$ Current (Amperes).

2. The Joule Heating Formula: $P = I^2 R$ Substituting Ohm's Law ($V = I \times R$) into $P = V \times I$: $P = (I \times R) \times I = I^2 R$ This equation reveals that power dissipated as heat in a conductor is proportional to the square of the current. Doubling the current flowing through a distribution cable quadruples the heat dissipation ($2^2 = 4$). This formula is essential for calculating cable thermal ratings, volt drop energy losses, and protective device clearing capacities.

3. The Constant-Resistance Voltage Formula: $P = \frac{V^2}{R}$ Substituting Ohm's Law transposed for current ($I = \frac{V}{R}$) into $P = V \times I$: $P = V \times \left(\frac{V}{R}\right) = \frac{V^2}{R}$ This equation is used to calculate the power output of fixed-resistance electrical appliances (such as immersion heaters, panel heaters, and shower elements) when subjected to variations in supply voltage.

The Ohm's Law and Power Summary Circle

The combined relationships yield 12 fundamental equations connecting $V$, $I$, $R$, and $P$:

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Electrical Energy: Joules vs Kilowatt-Hours (kWh)

Electrical energy ($E$ or $W$) represents the total quantity of work performed by an electrical system over a period of time:

$\text{Energy} = \text{Power} \times \text{Time}$

1. Base SI Unit: The Joule ($\text{J}$)

In physical science, energy is measured in Joules (symbol $\text{J}$), where one Joule is the energy transferred by a power of one Watt operating for one second ($1\text{ J} = 1\text{ W}\cdot\text{s}$):

$E(\text{Joules}) = P(\text{Watts}) \times t(\text{seconds})$

While the Joule is the base SI unit, it is impractical for commercial electricity metering because it is too small. For example, a standard domestic kettle consuming $3,000\text{ W}$ for just 5 minutes converts: $E = 3,000\text{ W} \times (5 \times 60\text{ s}) = 3,000 \times 300 = 900,000\text{ Joules}$ Recording residential or industrial consumption in Joules would require multi-digit numbers running into hundreds of millions.

2. Commercial Metering Unit: The Kilowatt-Hour ($\text{kWh}$)

To provide a practical commercial unit, utility providers meter electricity in kilowatt-hours (symbol $\text{kWh}$), historically designated the "Board of Trade Unit" or simply the "Unit" of electricity.

One kilowatt-hour is defined as the total electrical energy converted by a device consuming one kilowatt ($1,000\text{ W}$) continuously for one hour ($3,600\text{ seconds}$):

$1\text{ kWh} = 1,000\text{ W} \times 3,600\text{ s} = 3,600,000\text{ Joules} = 3.6 \times 10^6\text{ J} = 3.6\text{ MJ}$

To calculate commercial energy consumption in $\text{kWh}$:

$E(\text{kWh}) = \frac{P(\text{Watts}) \times t(\text{hours})}{1,000} = P(\text{kW}) \times t(\text{hours})$

3. Utility Tariff Calculations

The financial cost of operating electrical installations is calculated directly from energy consumed in $\text{kWh}$:

$\text{Total Operating Cost} = \text{Energy Consumed (kWh)} \times \text{Tariff Rate (pence or \pounds per kWh)}$

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Multi-Step Worked Calculation Examples

Worked Example 1: Domestic Immersion Heater Analysis

Problem: A domestic hot water immersion heater is connected to a $230\text{ V}$ single-phase AC supply and is rated at $3.0\text{ kW}$. Calculate:

1. The continuous load current drawn by the heater.
2. The electrical resistance of the heating element.
3. The total electrical energy consumed in Joules and in $\text{kWh}$ if the heater runs continuously for $4\text{ hours and } 30\text{ minutes}$.
4. The financial running cost of this heating cycle at an electricity tariff of $26\text{p per kWh}$.

Solution:

1. Load Current ($I$): $I = \frac{P}{V} = \frac{3,000\text{ W}}{230\text{ V}} \approx 13.04\text{ A}$
2. Element Resistance ($R$): $R = \frac{V^2}{P} = \frac{230^2}{3,000} = \frac{52,900}{3,000} \approx 17.63\ \Omega$ (Verification via Ohm's Law: $R = \frac{V}{I} = \frac{230}{13.04} = 17.64\ \Omega$.)
3. Energy Consumed ($E$):
   • In $\text{kWh}$: Time $t = 4.5\text{ hours}$. $E(\text{kWh}) = 3.0\text{ kW} \times 4.5\text{ h} = 13.5\text{ kWh}$
   • In Joules: Time $t = 4.5 \times 3,600\text{ s} = 16,200\text{ s}$. $E(\text{Joules}) = 3,000\text{ W} \times 16,200\text{ s} = 48,600,000\text{ J} = 48.6\text{ MJ}$ (Check: $13.5\text{ kWh} \times 3.6\text{ MJ/kWh} = 48.6\text{ MJ}$.)
4. Running Cost: $\text{Cost} = 13.5\text{ kWh} \times \pounds 0.26 = \pounds 3.51$

Worked Example 2: Voltage Sag Impact on an Electric Shower

Problem: An instantaneous electric shower has an element rated at $9.6\text{ kW}$ when supplied at a nominal voltage of $240\text{ V}$. During peak winter evening demand, network loading causes the supply voltage at the shower terminals to drop by $10\%$ to $216\text{ V}$. Assuming the heating element resistance remains constant, calculate:

1. The resistance of the shower element.
2. The actual heat power output at $216\text{ V}$.
3. The percentage reduction in heating performance.

Solution:

1. Element Resistance ($R$): $R = \frac{V^2}{P} = \frac{240^2}{9,600} = \frac{57,600}{9,600} = 6.0\ \Omega$
2. Actual Power at Reduced Voltage ($216\text{ V}$): $P_{\text{actual}} = \frac{V^2}{R} = \frac{216^2}{6.0} = \frac{46,656}{6.0} = 7,776\text{ W} = 7.776\text{ kW}$
3. Percentage Reduction in Heat Output: $\text{Reduction} = \frac{9.600 - 7.776}{9.600} \times 100\% = \frac{1.824}{9.600} \times 100\% = 19.0\%$

Worked Example 3: Commercial LED Lighting Installation Costing

Problem: An industrial warehouse installation contains 80 commercial LED luminaire fittings. Each fitting consumes $60\text{ W}$. The lighting installation operates $12\text{ hours per day}$, $5\text{ days per week}$, for $50\text{ weeks per year}$. Calculate:

1. The total connected lighting load in kilowatts.
2. The annual energy consumption in $\text{kWh}$.
3. The annual electricity running cost at a commercial tariff of $22\text{p per kWh}$.

Solution:

1. Total Connected Load ($P$): $P = 80 \times 60\text{ W} = 4,800\text{ W} = 4.8\text{ kW}$
2. Annual Operating Hours ($t$): $t = 12\text{ hours/day} \times 5\text{ days/week} \times 50\text{ weeks/year} = 3,000\text{ hours/year}$
3. Annual Energy Consumption ($E$): $E = 4.8\text{ kW} \times 3,000\text{ hours} = 14,400\text{ kWh}$
4. Annual Running Cost: $\text{Cost} = 14,400\text{ kWh} \times \pounds 0.22 = \pounds 3,168.00$