The time constant (τ) of an RC circuit is given by the product of the resistance (R) and the capacitance (C):

τ = RC

In this case, R = 1 MΩ = \(10^6\) Ω and C = 10 µF = \(10 \times 10^{-6}\) F. Therefore,

τ = \(10^6 \times 10 \times 10^{-6}\) = 10 seconds

The charging equation for a capacitor in an RC circuit is:

\(V(t) = V_f (1 - e^{-t/τ})\)

where:

• \(V(t)\) is the voltage across the capacitor at time t
• \(V_f\) is the final voltage across the capacitor
• \(τ\) is the time constant

We want to find the time it takes for the capacitor to reach 90% of its final charge, meaning \(V(t) = 0.9 V_f\). Substituting this into the equation:

\(0.9 V_f = V_f (1 - e^{-t/τ})\)

Dividing both sides by \(V_f\):

\(0.9 = 1 - e^{-t/τ}\)

\(e^{-t/τ} = 1 - 0.9 = 0.1\)

Taking the natural logarithm of both sides:

\(-t/τ = \ln(0.1)\)

Solving for t:

\(t = -τ \ln(0.1)\)

Substituting τ = 10 seconds:

\(t = -10 \times \ln(0.1) \approx -10 \times (-2.303) \approx 23.03\) seconds

Therefore, it will take approximately 23.03 seconds for the capacitor to reach 90% of its final charge.