Question

Calculating time stamp for 15/16 total charge in RC circuit with given voltage measurements.

Answer 1

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 \( C = 10 \, \mu F = 10 \times 10^{-6} \, F \):

1. Calculate \( RC \):
\[
RC = 1000 \times 10 \times 10^{-6} = 0.01 \, \text{seconds} \, (or \, 10 \, ms)
\]

2. 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.