How many time constants will elapse before a capacitor gains 99% of its steady state charge?

Options

A. 1
B. 9.2
C. 4.6
D. 18.4

Explanation

To determine how many time constants will elapse before a capacitor gains 99% of its steady state charge, we can follow these steps:

### Step 1: Understand the charging equation of a capacitor
The charge \( Q \) on a capacitor at any time \( t \) during charging can be expressed using the formula:
\[ Q = Q_0 \left(1 - e^{-\frac{t}{\tau}}\right) \]
where:
- \( Q_0 \) is the maximum charge (steady state charge),
- \( \tau \) is the time constant of the circuit.

### Step 2: Set up the equation for 99% charge
We want to find the time \( t \) when the capacitor reaches 99% of its steady state charge. Therefore, we set:
\[ Q = 0.99 Q_0 \]

### Step 3: Substitute into the charging equation
Substituting \( Q \) into the charging equation gives:
\[ 0.99 Q_0 = Q_0 \left(1 - e^{-\frac{t}{\tau}}\right) \]

### Step 4: Simplify the equation
We can cancel \( Q_0 \) from both sides (assuming \( Q_0 \neq 0 \)):
\[ 0.99 = 1 - e^{-\frac{t}{\tau}} \]
Rearranging this gives:
\[ e^{-\frac{t}{\tau}} = 1 - 0.99 = 0.01 \]

### Step 5: Take the natural logarithm of both sides
Taking the natural logarithm (ln) of both sides results in:
\[ -\frac{t}{\tau} = \ln(0.01) \]

### Step 6: Solve for \( t \)
We know that \( \ln(0.01) = \ln\left(\frac{1}{100}\right) = -\ln(100) \). Thus:
\[ -\frac{t}{\tau} = -\ln(100) \]
This simplifies to:
\[ \frac{t}{\tau} = \ln(100) \]

### Step 7: Calculate \( \ln(100) \)
Using the property of logarithms:
\[ \ln(100) = \ln(10^2) = 2 \ln(10) \approx 2 \times 2.3 = 4.6 \]
Thus, we have:
\[ \frac{t}{\tau} = 4.6 \]

### Step 8: Conclusion
This means that the time \( t \) taken to reach 99% of the steady state charge is approximately \( 4.6 \tau \). Therefore, the answer is:
\[ \text{4.6 time constants} \]