Solution:

The charging of a capacitor through a resistor is described by the following equation:

$$Q(t) = Q_{\text{max}} \left( 1 - e^{-t/\tau} \right)$$

Where:

• 
  $Q(t)$ 

  is the charge on the capacitor at time $t$ 

  ,

• 
  $Q_{\text{max}}$ 

  is the maximum (steady-state) charge the capacitor can hold,

• 
  $\tau$ 

  is the time constant, $\tau = R \cdot C$ 

  , where $R$ 

  is the resistance and $C$ 

  is the capacitance,

• 
  $t$ 

  is the time.

Step 1: Given condition (10% of steady charge)

We are told that at time

$t$

, the capacitor has collected 10% of the steady charge, so:

$$Q(t) = 0.1 \cdot Q_{\text{max}}$$

Step 2: Substitute into the charging equation

Substitute

$Q(t) = 0.1 \cdot Q_{\text{max}}$

into the charging formula:

$$0.1 \cdot Q_{\text{max}} = Q_{\text{max}} \left( 1 - e^{-t/\tau} \right)$$

Cancel

$Q_{\text{max}}$

from both sides:

$$0.1 = 1 - e^{-t/\tau}$$

Step 3: Solve for $t$

Rearrange the equation to solve for

$e^{-t/\tau}$

:

$$e^{-t/\tau} = 1 - 0.1 = 0.9$$

Take the natural logarithm of both sides:

$$-\frac{t}{\tau} = \ln(0.9)$$

$$t = -\tau \ln(0.9)$$

Using the fact that

$\ln(0.9) = -\ln(10/9)$

:

$$t = \tau \ln\left(\frac{10}{9}\right)$$

Final Answer:

The time at which the capacitor has collected 10% of the steady charge is

$t = \tau \ln\left(\frac{10}{9}\right)$

.