Question

A $10\mu F$capacitor initially charged to $20\mu C$is discharged through a $1.0k\Omega$ resistor. How long does it take to reduce the capacitor’s charge to $10\mu C$?

Short answer

$t=6.9ms$ time will be taken to reduce the cpacitor's charge to$10\mu C$.

Step-by-step solution

Given Information  

We need to find that how long does it take to reduce the capacitor’s charge to $10\mu C$.

Simplify

The switch is off, the capacitor discharge and the current flows through the resistor. To discharge the capacitor, the charges flow through the circuit in time $r$. The initial charge $Q_o$is related to the final charge $Q$by equation in the form

$Q=Q_o{e}^{\raisebox{1ex}{$-t$}\!\left/ \!\raisebox{-1ex}{$RC$}\right.}$ (1)

Here, $R$is the resistance and $C$is the capacitance. Let us solve equation (1) to get it by

$t=-RCIn\left(\frac{Q}{Q_o}\right)$ (2)

Now, putting the values for$R,C,Q$and $Q_o$into equation (2) to get $t$

$$\begin{gathered} t=-RCIn\left(\frac{Q}{Q_o}\right) \\ =-\left(1000\right)(10\times {10}^{-6}F)In\left(\frac{10\mu C}{20\mu C}\right) \\ =6.9\times {10}^{-3}s \\ =6.9ms \end{gathered}$$