To solve this problem, we need to find the time it takes for a capacitor to reach 60% of its maximum charge in an RC circuit. The time constant \(\tau\) for a resistor and capacitor in series is given by:
\[ \tau = R \cdot C \]
Where:
- \(R\) is the resistance (in ohms)
- \(C\) is the total capacitance (in farads)
- Calculate the total capacitance \(C\): The two capacitors are in parallel, so their total capacitance is the sum of their individual capacitances:
\[ C = C_1 + C_2 = 2,\text{nF} + 2,\text{nF} = 4,\text{nF} = 4 \times 10^{-9},\text{F} \]
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Identify resistance \(R\): We have a resistance of \(2 \times 10^6, \Omega\).
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Calculate the time constant \(\tau\):
\[ \tau = R \cdot C = (2 \times 10^6, \Omega)(4 \times 10^{-9},\text{F}) = 8 \times 10^{-3},\text{s} = 0.008,\text{s} \]
- Determine the time to reach 60% of maximum charge: The voltage \(V(t)\) across the capacitor at time \(t\) is given by the equation:
\[ V(t) = V_0 (1 - e^{-t/\tau}) \]
Where \(V_0\) is the maximum voltage across the capacitor. We are interested in finding the time \(t\) when \(V(t) = 0.6 V_0\):
\[ 0.6 V_0 = V_0 (1 - e^{-t/\tau}) \]
Dividing both sides by \(V_0\) (assuming \(V_0 \neq 0\)):
\[ 0.6 = 1 - e^{-t/\tau} \]
Rearranging gives:
\[ e^{-t/\tau} = 1 - 0.6 = 0.4 \]
Taking the natural logarithm of both sides:
\[ -t/\tau = \ln(0.4) \]
Thus:
\[ t = -\tau \ln(0.4) \]
- Substituting \(\tau\):
\[ t = -0.008 , \text{s} \cdot \ln(0.4) \]
Calculating \(\ln(0.4)\):
\[ \ln(0.4) \approx -0.9163 \]
So now substituting this value in:
\[ t = -0.008 , \text{s} \cdot (-0.9163) \approx 0.00737 , \text{s} \]
Thus, the time it takes for 60% of the maximum charge to be deposited on the capacitor is approximately:
\[ \boxed{0.00737 , \text{s}} \text{ or } 7.37 , \text{ms} \]