7.2 Ohm's Law & Power Calculations

Ohm's Law & Power Calculations

Quick Answer: Ohm's Law ($V = IR$) defines the linear relationship between voltage, current, and resistance. Electrical power ($P = VI = I^2 R = V^2 / R$) measures the rate of energy conversion. When evaluating circuits on the DAA, use resistor shortcuts—such as Product-Over-Sum for two parallel resistors or $R/n$ for $n$ identical resistors—and remember that series batteries add voltage while parallel batteries add amp-hour capacity.

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1. Georg Simon Ohm and the Physics of $V = IR$

In 1827, German physicist Georg Simon Ohm published his experimental findings establishing the direct mathematical relationship governing electrical circuits. Ohm's Law states that the current passing through an ideal conductor is directly proportional to the voltage applied across it and inversely proportional to its electrical resistance.

The standard algebraic formula is: [ V = I \times R ]

Where:

• $V$ = Voltage across the conductor, measured in Volts (V)
• $I$ = Current flowing through the conductor, measured in Amperes (A)
• $R$ = Resistance of the conductor, measured in Ohms ((\Omega))

Proportionality Rules:

1. Direct Proportionality ($V \propto I$): If resistance remains constant, doubling the supply voltage will exactly double the current flowing through the circuit.
2. Inverse Proportionality ($I \propto 1/R$): If voltage remains constant, doubling the resistance will reduce the current to exactly half its original value.

Ohmic vs. Non-Ohmic Conductors:

• Ohmic Components: Materials and devices that maintain a constant resistance regardless of applied voltage or current (e.g., carbon film resistors). Their current-voltage ($I-V$) graph is a straight line passing through the origin.
• Non-Ohmic Components: Devices whose resistance changes based on operating conditions such as voltage, current, or temperature (e.g., diodes, incandescent lightbulbs, thermistors). An incandescent bulb's filament increases resistance as it gets hot, causing its current to level off non-linearly.

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2. Practical Tool: The Ohm's Law Memory Triangle

During timed aptitude testing like the DAA, mental speed and accuracy are vital. Sketching the Ohm's Law Triangle on your scratch paper provides an instant visual tool to isolate any missing variable.

To solve for an unknown quantity:

1. To solve for Voltage ($V$): Cover $V$. The remaining variables are $I$ and $R$ side-by-side. [ V = I \times R ]
2. To solve for Current ($I$): Cover $I$. The remaining format shows $V$ over $R$. [ I = \frac{V}{R} ]
3. To solve for Resistance ($R$): Cover $R$. The remaining format shows $V$ over $I$. [ R = \frac{V}{I} ]

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3. Electrical Power Calculations ($P = VI$, $P = I^2 R$, $P = V^2 / R$)

While voltage drives current through a resistance, Electrical Power ($P$) quantifies how rapidly electrical energy is being converted into work, heat, light, or mechanical motion. Power is measured in Watts (W), where one Watt equals one Joule of energy per second ($1\text{ W} = 1\text{ J/s}$).

The primary power formula is: [ P = V \times I ]

Where:

• $P$ = Power in Watts (W)
• $V$ = Potential difference in Volts (V)
• $I$ = Current in Amperes (A)

Combining Ohm's Law with Power (The Derived Equations)

By substituting Ohm's Law ($V = IR$ or $I = V/R$) into the base power equation ($P = VI$), we derive two extremely powerful alternative formulas essential for exam calculations.

Derivation 1: Current and Resistance ($P = I^2 R$)

When you know the current flowing through a component and its resistance, but do not know the voltage drop across it: [ P = I^2 \times R ] Application: This formula is known as the Joule Heating or $I^2 R$ Loss equation. It shows that power dissipated as heat in a wire scales with the square of current. Doubling the current through a cable quadruples thermal line loss!

Derivation 2: Voltage and Resistance ($P = V^2 / R$)

When you know the voltage applied across a component and its resistance, but do not know the current: [ P = \frac{V^2}{R} ] Application: This formula is ideal for parallel appliances connected to a fixed supply voltage (such as 12V vehicle battery systems or 230V mains). Lower resistance heating elements draw more power and generate higher thermal output.

Electrical Energy ($E$)

Power consumed over time equals total electrical energy expended: [ E = P \times t ] Where energy ($E$) is measured in Joules (J) or Watt-hours (Wh) / Kilowatt-hours (kWh) (with time $t$ in seconds or hours respectively).

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4. Resistor Combinations & Calculation Shortcuts

To compute total circuit current or power, you must first calculate the equivalent resistance ($R_{\text{eq}}$) of the network.

Resistors in Series

Series resistances add directly together: [ R_{\text{eq}} = R_1 + R_2 + R_3 + \dots + R_n ]

Resistors in Parallel: Standard Reciprocal Formula

The general parallel resistance equation is: [ \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots + \frac{1}{R_n} ]

Time-Saving Parallel Shortcuts for the DAA:

Shortcut 1: Product-Over-Sum (For Exactly Two Resistors)

When calculating equivalent resistance for two parallel resistors, avoid complex fraction additions by using the Product-Over-Sum rule: [ R_{\text{eq}} = \frac{R_1 \times R_2}{R_1 + R_2} ]

Example: Calculate the equivalent resistance of a $60,\Omega$ resistor and a $30,\Omega$ resistor in parallel. [ R_{\text{eq}} = \frac{60 \times 30}{60 + 30} = \frac{1800}{90} = 20,\Omega ]

Shortcut 2: Identical Resistors in Parallel ($R / n$)

When $n$ parallel branches contain identical resistors of value $R$, the equivalent resistance is simply the single resistance value divided by the total number of branches: [ R_{\text{eq}} = \frac{R}{n} ]

Example: Four $100,\Omega$ resistors connected in parallel. [ R_{\text{eq}} = \frac{100,\Omega}{4} = 25,\Omega ]

Shortcut 3: Boundaries Sanity Check

The total equivalent resistance of any parallel combination is always strictly less than the single smallest resistor in that parallel group. If you connect a $1000,\Omega$ resistor in parallel with a $5,\Omega$ resistor, $R_{\text{eq}}$ must be less than $5,\Omega$.

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5. Battery Configurations: Voltage vs. Amp-Hour Capacity

Power sources (cells and batteries) can also be wired in series or parallel to tailor voltage and operating duration.

Batteries in Series

• Wiring: Connect the positive terminal of the first battery to the negative terminal of the second.
• Voltage: Voltages add together ($V_{\text{total}} = V_1 + V_2 + V_3$).
• Capacity: Total Amp-hour (Ah) runtime capacity remains equal to a single battery.
• Example: Four 1.5V, 2,000 mAh AA batteries connected in series provide 6.0 Volts with a capacity of 2,000 mAh.

Batteries in Parallel

• Wiring: Connect positive terminals together, and negative terminals together.
• Voltage: Output voltage remains equal to a single battery ($V_{\text{total}} = V_{\text{single}}$).
• Capacity: Amp-hour (Ah) runtime capacities add together ($ ext{Capacity}_{\text{total}} = C_1 + C_2 + C_3$).
• Example: Two 12V, 100 Ah tactical batteries connected in parallel provide 12 Volts with an expanded capacity of 200 Ah.

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6. Circuit Troubleshooting & Diagnostic Logic

When evaluating circuit faults on the DAA, use Ohm's Law to predict multimeter readings across components:

1. Measuring Voltage Across an Open Circuit:
   • Because no current flows ($I = 0$), there is zero voltage drop ($V = I \times R = 0$) across intact resistors.
   • The full supply voltage appears directly across the open gap (e.g., open switch or blown fuse) because the open break represents infinite resistance ($R = \infty$).
2. Measuring Voltage Across a Short Circuit:
   • A short circuit has zero resistance ($R \approx 0$).
   • Therefore, the voltage drop across a shorted component is zero ($V = I \times 0 = 0\text{ V}$), while line current spikes violently.

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7. Step-by-Step Worked Calculation Examples

Example 1: Series Circuit Analysis

Problem: A 24-Volt battery is connected in series with three resistors: $R_1 = 10,\Omega$, $R_2 = 20,\Omega$, and $R_3 = 30,\Omega$. Find (a) total resistance, (b) circuit current, and (c) the voltage drop across $R_2$.

• Step 1: Total Resistance ($R_{\text{total}}$) [ R_{\text{total}} = R_1 + R_2 + R_3 = 10 + 20 + 30 = 60,\Omega ]
• Step 2: Total Current ($I$) [ I = \frac{V_{\text{total}}}{R_{\text{total}}} = \frac{24\text{ V}}{60,\Omega} = 0.4\text{ Amps} ]
• Step 3: Voltage Drop across $R_2$ ($V_2$) [ V_2 = I \times R_2 = 0.4\text{ A} \times 20,\Omega = 8.0\text{ Volts} ]

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Example 2: Parallel Circuit Analysis & Power Consumption

Problem: A 120-Volt supply feeds two parallel load heaters: $R_1 = 30,\Omega$ and $R_2 = 60,\Omega$. Calculate (a) equivalent resistance, (b) total current, and (c) total power dissipated.

• Step 1: Equivalent Resistance ($R_{\text{eq}}$) Using Product-Over-Sum: [ R_{\text{eq}} = \frac{30 \times 60}{30 + 60} = \frac{1800}{90} = 20,\Omega ]
• Step 2: Total Supply Current ($I_{\text{total}}$) [ I_{\text{total}} = \frac{V}{R_{\text{eq}}} = \frac{120\text{ V}}{20,\Omega} = 6.0\text{ Amps} ]
• Step 3: Total Power ($P_{\text{total}}$) Using $P = V \times I$: [ P_{\text{total}} = 120\text{ V} \times 6.0\text{ A} = 720\text{ Watts} ] (Verification using $P = V^2 / R_{\text{eq}}$: $120^2 / 20 = 14,400 / 20 = 720\text{ W}$. Result is verified!)

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Example 3: Combination Circuit Master Problem

Problem: A 36-Volt supply is connected to a circuit with resistor $R_1 = 4,\Omega$ in series with a parallel pair consisting of $R_2 = 12,\Omega$ and $R_3 = 6,\Omega$. Determine total current leaving the source.

• Step 1: Collapse Parallel Pair ($R_{2,3}$) [ R_{2,3} = \frac{12 \times 6}{12 + 6} = \frac{72}{18} = 4,\Omega ]
• Step 2: Add Series Element ($R_1$) for Total Resistance [ R_{\text{total}} = R_1 + R_{2,3} = 4,\Omega + 4,\Omega = 8,\Omega ]
• Step 3: Calculate Total Source Current [ I_{\text{total}} = \frac{36\text{ V}}{8,\Omega} = 4.5\text{ Amps} ]

Mastering these derivations, visual shortcuts, and calculation steps guarantees top-tier performance on the quantitative sections of the Defence Aptitude Assessment.