Dimensional analysis of the time constant If you’ve been working through the Purves textbook, you may be surprised that if we multiply two electrical measurements, resistance and capacitance, we get an answer in units of time (milliseconds). Let’s see how it happens. First, a reminder of the time constant equation: τ = RmCm Resistance is measured in ohms, and from Ohm’s law: V = IR therefore: R =V I where V is voltage, I is current and R is resistance. The units will be: Ω=V A Current is charge per unit time, i.e. I =Q or in terms of units: t A=C s So if we insert our new definition of current into the expression for resistance, we get: R= V Q = Vt Q = Vs C t or in terms of units: Ω= V C s Capacitance (C) is measured in farads (F), and the farad is a derived SI unit. As you may remember, the amount of charge (Q) a capacitor can hold, in coulombs (C), is related to its capacitance and the voltage (V) applied across it: Q = CV therefore: C =Q V The units will be: F=C V (In case you’re getting confused between quantities and their units, remember that the quantities - voltage V, charge Q, capacitance C and resistance R - are in italics, while the units - volt V, coulomb C, farad F and ohm Ω - are in ordinary type). So when we multiply resistance and capacitance, we get: RC = V • Q = Vt • Q = t I V Q V or in terms of their units: Ω•F= V A •C V = Vs • C = s C V So one ohm multiplied by one farad gives one second. This may be a little difficult to grasp at first, but follow it through thoroughly and slowly, and you’ll get the idea after a few hundred ohm-farads.