91Ó°ÊÓ

We have to prove that the product $RC$has units of s.

The resistance is $R$and the capacitance of the capacitor is $C$. The capacitor is a device that accumulates the charges $Q$on its plates and due to this accumulation, a potential difference across the capacitor is created $â–³V_C$. This potential difference of capacitor is related to the capacitance $C$by

localid="1648663675252" $C=\frac{Q}{â–³V_C}....\left(1\right)$

Using Ohm's law to get the resistance $R$in the circuit where Ohm's law states that the current $I$flows through a resistance due to the potential difference $â–³V$between the resistance.

localid="1648663717970" $R=\frac{â–³V_R}{I}....\left(2\right)$

From both equations, we can get an expression in the term of$RC$by

localid="1648663740584" $RC=\left(\frac{â–³V_R}{I}\right)\left(\frac{Q}{â–³V_R}\right)=\frac{Q}{I}....\left(3\right)$

connect the two devices, the potentials at an instant are equal $â–³V_C=â–³V_R$. The current is the rate flow of the charge $I=\raisebox{1ex}{$Q$}\!\left/ \!\raisebox{-1ex}{$t$}\right.$, the charge is $Q=It$where $t$is the time. Use the expression of $Q$into equation (2)

$RC=\frac{It}{I}=t$

$t$is the time in seconds, so conclude that the term $RC$ has the unit of time(seconds)