A charged capacitor of capacitance C is discharging through a resistor of resistance R. At what time the charge on the capacitor is equal to one half of its initial value?

Options

A. `(RC ) /(2)`
B. `(RC)/(log_(e)2 )`
C. RC `"log"_(e) 2`
D. `(1)/(RC) "log"_(e) 2 `

Explanation

To solve the problem step by step, we will use the formula for the charge on a discharging capacitor and find the time at which the charge is half of its initial value.

### Step 1: Understand the formula for discharging capacitor
The charge \( q(t) \) on a discharging capacitor at time \( t \) is given by the formula:
\[ q(t) = q_0 e^{-\frac{t}{RC}} \]
where:
- \( q_0 \) is the initial charge,
- \( R \) is the resistance,
- \( C \) is the capacitance.

### Step 2: Set up the equation for half charge
We want to find the time \( t \) when the charge \( q(t) \) is half of the initial charge \( q_0 \). Therefore, we set up the equation:
\[ q(t) = \frac{q_0}{2} \]
Substituting this into the equation gives:
\[ \frac{q_0}{2} = q_0 e^{-\frac{t}{RC}} \]

### Step 3: Simplify the equation
We can divide both sides of the equation by \( q_0 \) (assuming \( q_0 \neq 0 \)):
\[ \frac{1}{2} = e^{-\frac{t}{RC}} \]

### Step 4: Take the natural logarithm of both sides
To solve for \( t \), we take the natural logarithm of both sides:
\[ \ln\left(\frac{1}{2}\right) = -\frac{t}{RC} \]

### Step 5: Solve for \( t \)
Now, we can isolate \( t \):
\[ t = -RC \ln\left(\frac{1}{2}\right) \]

### Step 6: Simplify the logarithm
Using the property of logarithms, we know that:
\[ \ln\left(\frac{1}{2}\right) = -\ln(2) \]
Thus, we can rewrite \( t \):
\[ t = RC \ln(2) \]

### Final Answer
The time at which the charge on the capacitor is equal to one half of its initial value is:
\[ t = RC \ln(2) \]