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