Ohm's Law

Learning Objectives

By the end of this module you will be able to:

• State Ohm's Law and its algebraic rearrangements.
• Compute voltage, current, or resistance from two known quantities.
• Calculate power dissipated in resistive elements using three equivalent formulas.
• Apply Ohm's Law to series, parallel, and series-parallel circuits.
• Use Ohm's Law correctly with measurement instruments (multimeter).
• Recognize Ohmic vs non-ohmic behavior and limitations of Ohm's Law.
• Solve step-by-step circuit problems and check units for correctness.

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1. What is Ohm's Law?

Ohm's Law describes a linear relationship between voltage, current, and resistance for many electrical components (called ohmic devices).

The law is most commonly written as:

$$\boxed{V = I \, R}$$

Where:

• $V$ is voltage (potential difference) in volts (V),
• $I$ is current in amperes (A),
• $R$ is resistance in ohms ($\Omega$).

You can rearrange Ohm's Law to solve for any one variable:

$$\boxed{I = \dfrac{V}{R}} \qquad\text{and}\qquad \boxed{R = \dfrac{V}{I}}$$

Ohm's Law is empirical

Ohm's Law was formulated based on observation: many materials (like metal wires) show a linear $V$ vs $I$ relationship over a useful operating range. Devices that do not show a constant $R$ (e.g., diodes, transistors, incandescent bulbs at varying temperature) are called non-ohmic.

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2. Units and Symbols — quick reference

• Voltage: $V$ — volts (V)
• Current: $I$ — amperes (A)
• Resistance: $R$ — ohms ($\Omega$)

SI prefixes you will commonly see: milli (m, $10^{-3}$), kilo (k, $10^3$), mega (M, $10^6$).

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3. Power and Ohm's Law

Power dissipated by an electrical element (usually as heat in a resistor) is:

$$P = V \, I$$

Using Ohm's Law this can be written in alternative forms:

$$\boxed{P = V I = I^2 R = \dfrac{V^2}{R}}$$

Units: $P$ in watts (W).

Why three formulas? Depending on which two quantities you know (e.g., $V$ and $R$ or $I$ and $R$), you pick the appropriate form.

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4. Measurement tips (multimeter)

• Measure voltage across (in parallel with) the component. Set meter to volts.
• Measure current in series with the circuit path. Break the circuit and insert the meter in series; set meter to amps.
• Measure resistance only with the circuit power off (or isolate the resistor), because Ohm's Law $(R = V/I)$ requires measured voltage and current under the same powered condition, while resistance from a cold, isolated resistor uses an ohmmeter.

Passive sign convention

Using the passive sign convention, if current enters the positive-voltage terminal of an element, power $P = VI$ is absorbed (positive). If current leaves the positive terminal, the element is delivering power (negative $P$).

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5. Worked examples — step-by-step (digit-by-digit arithmetic)

Per good practice, arithmetic is shown in explicit steps to avoid mistakes.

Example 1 — Simple: find current

Problem: A resistor of $R = 4\ \Omega$ has $V = 12\ \text{V}$ across it. Find $I$.

Use $I = \dfrac{V}{R}$.

Step arithmetic:

1. $V = 12$, $R = 4$.

2. Compute $12 \div 4$.

   • $4 \times 3 = 12$.
   • Therefore $12 \div 4 = 3$.

3. $I = 3\ \text{A}$.

Answer: $I = 3\ \text{A}$.

Example 2 — Power from current and resistance

Problem: A resistor $R = 10\ \Omega$ carries $I = 2\ \text{A}$. Find power $P$.

Use $P = I^2 R$.

Step arithmetic:

1. $I^2 = 2^2 = 4$.

2. Multiply by $R$: $4 \times 10$.

   • $4 \times 10 = 40$.

3. $P = 40\ \text{W}$.

Answer: $P = 40\ \text{W}$.

Example 3 — Two resistors in series

Problem: Three resistors $R_1=10\ \Omega$, $R_2=20\ \Omega$, $R_3=30\ \Omega$ are in series across a $V_s = 12\ \text{V}$ battery. Find the current and voltage across each resistor.

Steps:

1. Series total $R_T = R_1 + R_2 + R_3$.

   • $R_1 + R_2 = 10 + 20 = 30$.
   • $30 + R_3 = 30 + 30 = 60$.
   • So $R_T = 60\ \Omega$.

2. Total current $I = V_s / R_T$.

   • Compute $12 \div 60$.
   • $12 \div 60 = 0.2$ because $60 \times 0.2 = 12$.
   • So $I = 0.2\ \text{A}$.

3. Voltage across each resistor $V_n = I \, R_n$.

   • $V_1 = 0.2 \times 10 = 2.0\ \text{V}$. (since $0.2 \times 10 = 2$)
   • $V_2 = 0.2 \times 20 = 4.0\ \text{V}$. (since $0.2 \times 20 = 4$)
   • $V_3 = 0.2 \times 30 = 6.0\ \text{V}$. (since $0.2 \times 30 = 6$)

Check: $2.0 + 4.0 + 6.0 = 12.0\ \text{V}$ matches supply.

Example 4 — Two resistors in parallel

Problem: $R_1 = 100\ \Omega$ and $R_2 = 200\ \Omega$ in parallel connected to $V_s = 12\ \text{V}$. Find equivalent resistance, total current, and branch currents.

Steps:

1. Compute reciprocals:

   • $1/R_1 = 1/100 = 0.01$. (since $1 \div 100 = 0.01$)
   • $1/R_2 = 1/200 = 0.005$. (since $1 \div 200 = 0.005$)

2. Sum reciprocals:

   • $1/R_T = 0.01 + 0.005 = 0.015$.

3. Equivalent resistance $R_T = 1 / 0.015$.

   • Note $0.015 = 15/1000$.
   • Reciprocal $= 1000 / 15$.
   • $1000 \div 15 = 66.666\ldots$ (repeating). We'll keep three decimals: $66.667\ \Omega$.

4. Total current $I_T = V_s / R_T$.

   • Use the exact reciprocal method: $I_T = V_s \times (1/R_T) = 12 \times 0.015$.
   • $12 \times 0.015 = 12 \times (15/1000) = (12 \times 15) / 1000 = 180 / 1000 = 0.180$.
   • So $I_T = 0.180\ \text{A}$.

5. Branch currents (use $I = V/R$):

   • $I_1 = 12 \div 100 = 0.12\ \text{A}$. (because $100 \times 0.12 = 12$)
   • $I_2 = 12 \div 200 = 0.06\ \text{A}$. (because $200 \times 0.06 = 12$)

6. Check: $I_1 + I_2 = 0.12 + 0.06 = 0.18\ \text{A} = I_T$.

Answers: $R_T \approx 66.667\ \Omega$, $I_T = 0.180\ \text{A}$, $I_1 = 0.12\ \text{A}$, $I_2 = 0.06\ \text{A}$.

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6. Using Ohm's Law in series-parallel circuits

Strategy (stepwise simplification):

1. Identify simple series or parallel groups.
2. Replace each group with its equivalent resistance.
3. Recompute until you have a single equivalent resistance for the whole network.
4. Use $I = V / R_T$ to find total current, then work backwards (voltage division, current division) to find branch quantities.

Tip: Keep units consistent (volts, ohms, amps) and track significant figures sensibly.

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7. Common pitfalls and troubleshooting

• Forgetting units: e.g., using $k\Omega$ vs $\Omega$ without converting causes large errors. Always convert to base units (V, A, Ω) before arithmetic.
• Measuring current incorrectly: never measure current by placing the meter across a voltage source (that will short the source).
• Applying Ohm's Law to non-ohmic elements: for diodes, transistors, many lamps, and some semiconductor devices, $R$ is not constant — treat those with their I–V curve or small-signal resistance.
• Temperature effects: resistor values can change with temperature (see the Resistance module). For high accuracy, use temperature coefficients or measure under operating conditions.

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8. Quick reference — formulas

• Ohm's Law: $\;V = I R$
• Current: $\;I = \dfrac{V}{R}$
• Resistance: $\;R = \dfrac{V}{I}$
• Power: $\;P = V I = I^2 R = \dfrac{V^2}{R}$
• Series resistors: $\;R_T = \displaystyle\sum R_n$
• Parallel resistors: $\;\dfrac{1}{R_T} = \displaystyle\sum \dfrac{1}{R_n}$

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9. Practice problems (with answers)

1. Find $R$: A lamp draws $I=0.5\ \text{A}$ from a $120\ \text{V}$ source. What is the lamp's resistance?

   • $R = V / I = 120 \div 0.5$. $0.5 \times 240 = 120$ so $120 \div 0.5 = 240$.
   • Answer: $R = 240\ \Omega$.

2. Power check: Using the lamp above ($R=240\ \Omega$, $V=120\ \text{V}$), what is power $P$?

   • Use $P = V^2 / R$.
   • $V^2 = 120^2 = 14400$.
   • $P = 14400 \div 240$. Since $240 \times 60 = 14400$, $P = 60\ \text{W}$.
   • Answer: $P = 60\ \text{W}$.

3. Mixed network: $R_1 = 50\ \Omega$ in series with a parallel pair $R_2 = 100\ \Omega$ and $R_3 = 100\ \Omega$. Supply $V_s = 24\ \text{V}$. Find total current.

   • Parallel $R_{23} = (100 \times 100) / (100 + 100) = 10000 / 200 = 50\ \Omega$.
   • Total $R_T = R_1 + R_{23} = 50 + 50 = 100\ \Omega$.
   • $I_T = 24 \div 100 = 0.24\ \text{A}$ because $100 \times 0.24 = 24$.
   • Answer: $I_T = 0.24\ \text{A}$.

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10. Limitations & final notes

• Ohm's Law is local and linear — it holds for components whose voltage–current relationship is linear over the operating range. For non-linear devices, use their I–V curves or linearize for small signals (small-signal resistance).
• For high-precision work, account for temperature coefficients of resistors and wiring resistance.
• Combining Ohm's Law with Kirchhoff's Voltage and Current Laws lets you analyze arbitrarily complex linear circuits.

Key takeaway

Ohm's Law connects voltage, current, and resistance in a simple algebraic way: $V=IR$. Use the rearranged forms, paired with series/parallel simplification and the power formulas, to solve most basic circuit problems. Always check units and the device's linearity before applying it.