🔋 DC Circuit Variables and Ohm's Law

Learning objectives

By the end of this lesson, you should be able to:

  • Define electric charge, current, voltage, resistance, and conductance with their units.
  • State Ohm's law and use it to solve for voltage, current, or resistance.
  • Calculate DC power with $P = VI$, $P = I^2R$, and $P = V^2/R$.
  • Convert common prefixes such as milli and kilo before substituting values.
  • Apply the passive sign convention to decide whether a component absorbs or delivers power.

Theory

Before you touch a simulator, get comfortable with four quantities: charge, current, voltage, and resistance. Most beginner mistakes are unit mistakes, not algebra mistakes.

Charge, current, voltage, and resistance

Quantity Symbol Unit Working definition
Charge $Q$ coulomb (C) Quantity of electric charge
Current $I$ ampere (A) Rate of charge flow, $I = Q/t$
Voltage $V$ volt (V) Energy per unit charge between two points, $V = W/Q$
Resistance $R$ ohm (Ω) Opposition to current, $R = V/I$
Conductance $G$ siemens (S) Reciprocal of resistance, $G = 1/R$

One ampere is one coulomb per second. One volt is one joule per coulomb. One ohm is one volt per ampere.

Conventional current is drawn as flowing from the positive terminal, through the external circuit, to the negative terminal. Electron flow is opposite, but circuit analysis normally uses conventional current.

Ohm's law

For a resistor, Ohm's law states:

$$ V = IR $$

Rearranged, $I = V/R$ and $R = V/I$.

A material is ohmic when its resistance stays roughly constant over the operating range, so its voltage-current graph is a straight line through the origin. A lamp filament is not ohmic: its resistance rises as it heats up. Ohm's law still defines the instantaneous ratio $V/I$, but $R$ is not constant.

Power in a DC circuit

Power is the rate of energy transfer:

$$ P = VI = I^2R = \frac{V^2}{R} $$

The unit is the watt (W), equal to one joule per second. Use $P = VI$ when you know voltage and current. Use $P = I^2R$ when current is known. Use $P = V^2/R$ when voltage is known.

[!TIP] Convert prefixes before you calculate. 8 mA = 0.008 A, and 2.2 kΩ = 2200 Ω. Mixing them is the most common source of wrong answers in early circuit problems.

Key facts to recall

Ohm's law for a resistor is {{c1::V = IR}}. Power can be written as {{c2::P = VI = I²R = V²/R}}. Resistance is measured in {{c3::ohms (Ω)}}, and conductance is {{c4::G = 1/R}}. In a fixed resistor, doubling the voltage causes the current to {{c5::double}}.

Worked examples

Example 1: Find current and power

A 12 V source is connected across a 240 Ω resistor. Find the current and the power dissipated.

  1. Knowns: $V = 12\ \text{V}$, $R = 240\ \Omega$.
  2. Use Ohm's law: $I = V/R = 12/240 = 0.05\ \text{A}$.
  3. Convert to milliamperes: $0.05\ \text{A} = 50\ \text{mA}$.
  4. Find power: $P = VI = 12(0.05) = 0.6\ \text{W}$.
  5. Check with another form: $P = I^2R = (0.05)^2(240) = 0.6\ \text{W}$.

Answer: $I = 50\ \text{mA}$, $P = 0.6\ \text{W}$.

Example 2: Find resistance and power

A resistor has 6 V across it and carries 20 mA. Find its resistance and power.

  1. Convert current: $20\ \text{mA} = 0.02\ \text{A}$.
  2. Use Ohm's law: $R = V/I = 6/0.02 = 300\ \Omega$.
  3. Find power: $P = VI = 6(0.02) = 0.12\ \text{W}$.
  4. Convert to milliwatts: $0.12\ \text{W} = 120\ \text{mW}$.

Answer: $R = 300\ \Omega$, $P = 120\ \text{mW}$.

Example 3: Work backward from power

A 0.25 W resistor has 10 V across it. Find the current and resistance.

  1. Use $P = VI$: $I = P/V = 0.25/10 = 0.025\ \text{A} = 25\ \text{mA}$.
  2. Use Ohm's law: $R = V/I = 10/0.025 = 400\ \Omega$.
  3. Check: $P = V^2/R = 100/400 = 0.25\ \text{W}$.

Answer: $I = 25\ \text{mA}$, $R = 400\ \Omega$.

Guided practice

Use this four-step habit for every problem:

  1. List knowns with units.
  2. Convert prefixes to base units.
  3. Choose the equation that contains your knowns and unknown.
  4. Substitute, then check the result with a second power formula when possible.

Try these before opening the answers.

  1. A 9 V battery is connected across a 470 Ω resistor. Find the current and power.
  2. A 2.2 kΩ resistor carries 8 mA. Find the voltage across it and the power dissipated.
  3. A resistor rated at 0.5 W has 10 V across it. Find the current and resistance. Is the resistor operating at its rating?
  4. A 12 V supply is connected to a resistor, and the measured current is 100 mA. What is the actual resistance? What current would you expect from a 100 Ω resistor?
  5. A 60 W lamp operates at 120 V. Find the current and the hot resistance of the lamp.
Click to reveal answers and guidance
  1. $I = 9/470 \approx 0.0191\ \text{A} = 19.1\ \text{mA}$. $P = 9(0.0191) \approx 0.172\ \text{W} = 172\ \text{mW}$.
  2. $V = IR = 0.008(2200) = 17.6\ \text{V}$. $P = VI = 17.6(0.008) = 0.1408\ \text{W} \approx 141\ \text{mW}$.
  3. $I = P/V = 0.5/10 = 0.05\ \text{A} = 50\ \text{mA}$. $R = 10/0.05 = 200\ \Omega$. Yes, it is operating exactly at its 0.5 W rating, which is not a good design margin in practice.
  4. $R = V/I = 12/0.1 = 120\ \Omega$. A 100 Ω resistor would draw $I = 12/100 = 0.12\ \text{A} = 120\ \text{mA}$.
  5. $I = P/V = 60/120 = 0.5\ \text{A}$. $R = V/I = 120/0.5 = 240\ \Omega$. This is the hot resistance, not the cold resistance you would measure with a multimeter on an unpowered lamp.

Expert corner

  • ==Convert every prefix before you substitute numbers.== The shortcut mA × kΩ = V works because $10^{-3} \times 10^3 = 1$, but only if you apply it deliberately.
  • Use the passive sign convention: when current enters the positive voltage terminal of a component, that component absorbs power. If current leaves the positive terminal of a source, the source delivers power.
  • Do not treat Ohm's law as a universal definition of every device. It is the model for an ohmic resistor. Diodes and lamps are nonlinear.
  • An ideal open circuit has zero current. An ideal short circuit has zero voltage across it. Neither is a resistor, so do not plug zero or infinity into Ohm's law and expect a useful number.
  • Measure voltage across a component and current through it. Reversing that habit creates confusing sign errors.
  • In exams, write the formula, substitute with units, then give the answer with the correct prefix. Graders look for the unit as much as the number.
[
  { "front": "What is electric current?", "back": "The net rate of charge flow, I = Q/t, measured in amperes (A)." },
  { "front": "What is voltage?", "back": "Energy transferred per unit charge between two points, measured in volts (V)." },
  { "front": "What is resistance?", "back": "The opposition to current flow, measured in ohms (Ω)." },
  { "front": "State Ohm's law for a resistor.", "back": "V = IR, where V is the voltage across the resistor and I is the current through it." },
  { "front": "Give the three common DC power formulas.", "back": "P = VI = I²R = V²/R." },
  { "front": "What is the unit of power?", "back": "The watt (W), equal to one joule per second." },
  { "front": "Which direction does conventional current flow?", "back": "From the positive terminal, through the external circuit, to the negative terminal." },
  { "front": "What does an ohmic device mean?", "back": "Its resistance stays constant, so its V-I graph is a straight line through the origin." },
  { "front": "If voltage across a fixed resistor doubles, what happens to current?", "back": "It doubles, because I = V/R." },
  { "front": "If voltage across a fixed resistor doubles, what happens to power?", "back": "It quadruples, because P = V²/R." },
  { "front": "Convert 1 kΩ to ohms.", "back": "1 kΩ = 1000 Ω." },
  { "front": "Convert 1 mA to amperes.", "back": "1 mA = 0.001 A." },
  { "front": "What is conductance and its unit?", "back": "G = 1/R, measured in siemens (S)." },
  { "front": "What is the current in an ideal open circuit?", "back": "Zero amperes." },
  { "front": "What is the voltage across an ideal short circuit?", "back": "Zero volts." }
]

Summary and next steps

The core of this lesson is short: $V = IR$ relates voltage, current, and resistance, and $P = VI = I^2R = V^2/R$ relates power to those same quantities. Keep your units consistent, convert prefixes early, and check every answer with a second formula when you can.

Next, you will use these relationships inside full networks. Kirchhoff's laws and series-parallel reduction let you move from one resistor to many, and Ohm's law remains the local rule you apply at every branch and element.