# Why do farads multiplied by ohms produce a result that has a unit of seconds?

Why is the time constant (RC) measured in seconds even though the units are farads x ohms?

This is to fulfill my own curiosity as I haven't had much luck at finding the answer. I would be most grateful if someone could give me a solid answer or send me in the right direction.

• Because hours and minutes would be silly. What option is there besides seconds? Nov 12 '14 at 23:05
• The second is the SI unit of time. What do you propose as an alternative? Nov 12 '14 at 23:09
• Sorry I was asking why Farads multiplied by ohms gives seconds? Nov 12 '14 at 23:37
• Unit cancellation is weird: what-if.xkcd.com/11 Nov 14 '14 at 0:39
• It's math. You just gotta take it on faith. Jun 18 '17 at 23:18

It's the way the units work out.

Broken down to its form in SI units, a volt is

$$\mathrm{V = \frac{kg \cdot m^2}{A \cdot s^3}}$$

where A is amperes. So, when you divide by current to get ohms, you see that

$$\Omega = \mathrm{\frac{kg \cdot m^2}{A^2\cdot s^3}}$$

$$\mathrm{F=\frac{s^4 \cdot A^2}{m^2 \cdot kg}}$$

So when you multiply Ohms by Farads, you're left with seconds:

$$\Omega \cdot \mathrm{F = \frac{kg \cdot m^2}{A^2\cdot s^3} \cdot \frac{s^4 \cdot A^2}{m^2 \cdot kg} = s}$$

• Great answer, but IMO too far. I think it would have been enough to say $$\Omega \cdot \mathrm{F = \frac{V}{A} \cdot \frac{As}{V} = s}$$. Nov 13 '14 at 18:45

If you charge the capacitor to some level and then connect it in parallel with the resistor, a current will begin to flow.

In reality this current will become smaller as the capacitor discharges (and the voltage across it therefore drops), but if we imagine that we somehow forced the current to stay at the initial magnitude through the resistor until the capacitor was fully discharged then it would take a certain amount of time until the capacitor was discharged to 0 V.

It turns out that this "certain amount of time" is the same no matter how much or little you charged the capacitor originally. (If you charge it more, there will be more charge to discharge, but the current will be proportionally higher because higher charge produces more voltage). This time is the product of the capacitance with the resistance -- or in other words your time constant.

And that is, intuitively, why the time constant has units of time.

(Alternatively, the time constant is how long it will take for the voltage to drop to $\frac1e$ of its original value, in the more realistic situation where we leave the system alone and let the current drop with the voltage as per Ohm's Law).

Because seconds (s) is the fundamental unit of measurement of time. The other fundamental SI units are:
1. Meters (m) for distance
2. Kilograms (kg) for mass
3. Ampere (A) for current
4. Kelvin (K) for temperature
5. Mole (mol) for quantity
6. Candela (cd) for light intensity

All other units of measurement relevant to physics are derived from those seven fundamental units.

Resistance, represented in fundamental form, is $\frac{kg*m^2}{A^2*s^3}$
Likewise, Capacitance in fundamental form is $\frac{s^4*A^2}{m^2*kg}$
Therefore, $R*C = \frac{kg*m^2}{A^2*s^3}*\frac{s^4*A^2}{m^2*kg} = s$

\begin{align} v&=IR \\ R&=\frac{v}{I} \\ &=\frac{1V}{1A} \\ I&=\frac{q}{t} \\ 1A &=\frac{1C}{1s} \\ C&=\frac{Q}{v} \\ &=\frac{1C}{1V} \end{align}

\begin{align} \text{.'. unit of } RC &=\frac{1V}{1A} \cdot \frac{1C}{1V} \\ &=\frac{1C}{1A} \\ &=\frac{1C}{1C/1s} \\ &=1s \\ \end{align}

Why is the time constant (RC) measured in seconds even though the units are farads x ohms?

Because farad is defined as charges held per unit of volt across the capacitor.

Charges are current times time. So farad is current x time over volt, or time over ohm.