Current, Resistance & Power

You’ll be able to

Define current as dQ/dt and relate it to charge transported
Compute resistance from resistivity and geometry, and apply Ohm’s law
Calculate electrical power with P = IV and its equivalent forms

Current is the rate of charge flow

Electric current is the rate at which charge passes a cross-section of a conductor: I = dQ/dt, measured in amperes (1 A = 1 C/s). By convention current points the way positive charge would move — in a metal the electrons actually drift the other way, slowly (∼10⁻⁴ m/s), while the field that drives them propagates at nearly light speed. Because I is a derivative, the charge delivered over an interval is the integral Q = ∫ I dt: the area under the I–t graph.

Resistance comes from geometry and material

A conductor’s resistance depends on what it is made of and how it is shaped: R = ρL/A, where resistivity ρ (Ω·m) is a material property, L is the length, and A is the cross-sectional area. A longer wire resists more (charges suffer more collisions); a fatter wire resists less (more parallel paths). Ohm’s law, V = IR, then relates the potential difference across a resistor to the current through it. Ohmic materials keep R constant over a wide range of V; devices like diodes and bulb filaments do not.

Current, resistance & Ohm’s law

I = dQ/dt ⇔ Q = ∫ I dt; R = ρL/A; V = IR

Integrate a varying current for charge — never multiply a changing current by total time. Doubling length doubles R; doubling radius quarters R, because A = πr² quadruples.

Power: energy delivered per second

A charge dq falling through potential difference V loses energy dU = V·dq, so the rate of energy transfer is P = dU/dt = V·dq/dt = IV. For a resistor, substitute Ohm’s law to get the two equivalent forms P = I²R and P = V²/R. In a resistor this power appears as heat (Joule heating); in a motor it becomes mechanical work; in a charging battery it becomes stored chemical energy.

Electrical power

P = IV = I²R = V²/R

All three are the same statement. Pick the form whose two variables you actually know for the element in question.

Worked example

The current in a wire varies as I(t) = 4.0 − 2.0t (I in amperes, t in seconds). How much charge passes a cross-section between t = 0 and t = 2.0 s, and at what time is the current zero?

1.Charge is the integral of current: Q = ∫₀² (4.0 − 2.0t) dt.
2.Antidifferentiate: Q = [4.0t − t²]₀².
3.Evaluate: Q = (8.0 − 4.0) − 0 = 4.0 C.
4.Current is zero when 4.0 − 2.0t = 0, i.e. t = 2.0 s — the I–t graph is a straight line hitting the axis exactly at the end of the interval.

Answer: Q = 4.0 C of charge passes; I = 0 at t = 2.0 s

Worked example

A copper wire (ρ = 1.7 × 10⁻⁸ Ω·m) is 10 m long with cross-sectional area 1.7 × 10⁻⁶ m². Find its resistance, and the power it dissipates when carrying 2.0 A.

1.Resistance from geometry: R = ρL/A = (1.7 × 10⁻⁸)(10) / (1.7 × 10⁻⁶).
2.Numerator: 1.7 × 10⁻⁷ Ω·m². Divide by 1.7 × 10⁻⁶ m²: R = 0.10 Ω.
3.With the current known, use P = I²R = (2.0)²(0.10) = 0.40 W.
4.Check with the other forms: V = IR = 0.20 V, so P = IV = (2.0)(0.20) = 0.40 W. Consistent.

Answer: R = 0.10 Ω, P = 0.40 W

Tip

When comparing bulbs or resistors, first decide what they share. Elements in series share the same I, so compare with P = I²R (bigger R wins). Elements in parallel share the same V, so compare with P = V²/R (smaller R wins). Using the wrong form is how the “brighter bulb” questions trap you.

Checkpoint

The charge through a device is Q(t) = 3t² (Q in coulombs, t in seconds). What is the instantaneous current at t = 2.0 s?

Checkpoint

A bulb is rated 60 W at 120 V. What is its operating resistance?

Answer the 2 checkpoints as you read.