To determine how long it takes for the energy stored in a capacitor to reach one-half its maximum (equilibrium) value, we can use the formula for the energy stored in a capacitor:
\[ U = \frac{1}{2} C V^2 \]
where:
- \( U \) is the energy stored,
- \( C \) is the capacitance, and
- \( V \) is the voltage across the capacitor.
The voltage across a charging capacitor as a function of time \( t \) is given by:
\[ V(t) = V_0 (1 - e^{-t/RC}) \]
where:
- \( V_0 \) is the maximum voltage,
- \( R \) is the resistance, and
- \( C \) is the capacitance.
The energy stored in the capacitor at time \( t \) can then be expressed as:
\[ U(t) = \frac{1}{2} C (V(t))^2 = \frac{1}{2} C (V_0 (1 - e^{-t/RC}))^2 \]
To find when the energy stored is half of the equilibrium value, we can set up the equation:
\[ U(t) = \frac{1}{2} U_{max} \]
Substituting for \( U(t) \) gives:
\[ \frac{1}{2} C (V_0 (1 - e^{-t/RC}))^2 = \frac{1}{2} \left( \frac{1}{2} C V_0^2 \right) \]
Simplifying this gives:
\[ (V_0 (1 - e^{-t/RC}))^2 = \frac{1}{2} V_0^2 \]
Now dividing both sides by \( V_0^2 \) (assuming \( V_0 \neq 0 \)) gives:
\[ (1 - e^{-t/RC})^2 = \frac{1}{2} \]
Taking the square root of both sides provides:
\[ 1 - e^{-t/RC} = \frac{1}{\sqrt{2}} \]
Rearranging this expression to solve for \( e^{-t/RC} \):
\[ e^{-t/RC} = 1 - \frac{1}{\sqrt{2}} = \frac{\sqrt{2} - 1}{\sqrt{2}} \]
Taking the natural logarithm of both sides yields:
\[ -\frac{t}{RC} = \ln \left( \frac{\sqrt{2} - 1}{\sqrt{2}} \right) \]
Thus,
\[ t = -RC \ln \left( \frac{\sqrt{2} - 1}{\sqrt{2}} \right) \]
Now let's calculate this value. Using the approximation \( \ln(1) \approx 0.693147 \):
\[ \ln \left( \frac{\sqrt{2} - 1}{\sqrt{2}} \right) \approx \ln(0.4142) \approx -0.8853 \]
So,
\[ t \approx RC \cdot 0.8853 \]
Now converting this result into a ratio involving \( RC \):
\[ t \approx 0.885 , RC \]
From the options given, the closest is approximately 0.8 RC.
Therefore, the answer is:
0.8 RC
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