Electrical circuits

Growth and decay

When the rate of change of a quantity is proportional to the quantity itself, dN/dt = kN, and the solution is N = N₀e^(kt). If k > 0 it is exponential growth (populations, compound interest); if k < 0 it is decay (radioactivity, drug in the blood). The half-life is t½ = ln 2 ÷ |k|.

N = 100e^(−0.3t). The half-life is ln 2 ÷ 0.3 ≈ 2.3.

Cooling of bodies

Newton’s law of cooling: the rate of loss of heat of a body is proportional to the difference between its temperature T and the temperature Tₛ of its surroundings: dT/dt = −k(T − Tₛ). The solution is T = Tₛ + (T₀ − Tₛ)e^(−kt).

Worked exampleA cup of tea

Tea at 90 °C cools in a room at 20 °C. After 10 minutes it is 60 °C. Find k.

Substitute.
Take logs.

Answer: k ≈ 0.056 per minute.

Falling bodies

A body falling under gravity with air resistance proportional to its speed has m dv/dt = mg − kv. The solution is v = (mg/k)(1 − e^(−kt/m)): the speed increases and approaches the terminal velocity mg/k.

In an RC circuit the charge on a capacitor obeys R dq/dt + q/C = V. Charging gives q = CV(1 − e^(−t/RC)), and discharging gives q = q₀e^(−t/RC). The time constant is RC. In an RL circuit the current rises as I = (V/R)(1 − e^(−Rt/L)).

v = 50(1 − e^(−0.2t)) approaches 50 m/s.

Try itNewton’s law of cooling

Choose the starting temperature, the room temperature, k and the time.

Check yourself The solution of dN/dt = −kN is

Check yourself As t → ∞, a cooling body’s temperature approaches

Check yourself The terminal velocity in m dv/dt = mg − kv is

Key terms

Half-life: The time for a quantity to halve.
Newton’s law of cooling: Cooling rate proportional to temperature difference.
Terminal velocity: The constant speed reached when drag balances weight.
Time constant: RC in a charging capacitor.

Summary

dN/dt = kN gives N = N₀e^(kt).
Newton’s cooling: T = Tₛ + (T₀ − Tₛ)e^(−kt).
Falling bodies approach a terminal velocity.
RC circuits charge and discharge exponentially.