To find the time stamp when a capacitor in an RC circuit reaches a certain fraction (like 15/16) of its total charge, we can make use of the formula for charging a capacitor through a resistor. The voltage across the capacitor \( V(t) \) at time \( t \) is given by:
\[ V(t) = V_0 \left(1 - e^{-\frac{t}{RC}}\right) \]
where:
- \( V_0 \) is the maximum voltage (or total charge voltage) the capacitor will reach,
- \( R \) is the resistance in ohms,
- \( C \) is the capacitance in farads,
- \( e \) is the base of the natural logarithm,
- \( t \) is the time in seconds.
To find the time when the capacitor reaches \( \frac{15}{16} \) of its total voltage \( V_0 \), set \( V(t) = \frac{15}{16} V_0 \):
\[ \frac{15}{16} V_0 = V_0 \left(1 - e^{-\frac{t}{RC}}\right) \]
Dividing both sides by \( V_0 \):
\[ \frac{15}{16} = 1 - e^{-\frac{t}{RC}} \]
Rearranging gives:
\[ e^{-\frac{t}{RC}} = 1 - \frac{15}{16} = \frac{1}{16} \]
Taking the natural logarithm of both sides:
\[ -\frac{t}{RC} = \ln\left(\frac{1}{16}\right) \]
Which can be rewritten as:
\[ t = -RC \ln\left(\frac{1}{16}\right) \]
Using the property of logarithms, we have:
\[ \ln\left(\frac{1}{16}\right) = -\ln(16) \]
So the equation simplifies to:
\[ t = RC \ln(16) \]
Given that:
\[ \ln(16) = 4 \ln(2) \quad \text{(since } 16=2^4\text{)} \]
If you have specific values for \( R \) and \( C \), you can plug them into this equation to calculate the time \( t \) when the capacitor reaches \( \frac{15}{16} \) of its total charge.
Example Calculation:
If \( R = 1 , \text{k}\Omega = 1000 , \Omega \) and \( C = 10 , \mu F = 10 \times 10^{-6} , F \):
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Calculate \( RC \): \[ RC = 1000 \times 10 \times 10^{-6} = 0.01 , \text{seconds} , (or , 10 , ms) \]
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Now find \( t \): \[ t = 0.01 \ln(16) = 0.01 \times 4 \ln(2) \approx 0.01 \times 4 \times 0.693 \approx 0.01 \times 2.772 \approx 0.02772 , \text{s} , (or , 27.72 , ms) \]
This means the capacitor reaches \( \frac{15}{16} \) of its charge after approximately 27.72 ms.