A capacitor of capacitance C is allowed to discharge through a resistance R. The net charge flown through resistance during one time constant is (`I_(0)` is the maximum current)

Options

A. `CRl_(0)((1)/(e)+1)`
B. `CRl_(0)(1-(1)/(e))`
C. `CRl_(0)`
D. `(CRl_(0))/(e)`

Answer

To solve the problem of finding the net charge that flows through a resistance \( R \) during one time constant while a capacitor of capacitance \( C \) discharges, we can follow these steps:

### Step 1: Understand the Initial Conditions
At \( t = 0 \), the initial charge \( Q_0 \) on the capacitor is given by:
\[ Q_0 = C E_0 \]
where \( E_0 \) is the initial voltage across the capacitor.

### Step 2: Define the Maximum Current
The maximum current \( I_0 \) in the circuit can be expressed as:
\[ I_0 = \frac{E_0}{R} \]
This is derived from Ohm's law, where \( E_0 \) is the voltage and \( R \) is the resistance.

### Step 3: Determine the Charge at One Time Constant
The charge \( Q(t) \) on the capacitor at any time \( t \) during the discharge can be expressed as:
\[ Q(t) = Q_0 e^{-\frac{t}{\tau}} \]
where \( \tau = R C \) is the time constant of the circuit.

At \( t = \tau \):
\[ Q(\tau) = Q_0 e^{-1} = C E_0 e^{-1} \]

### Step 4: Calculate the Charge Flowed Through the Resistance
The charge that has flowed through the resistance during one time constant can be calculated as the difference between the initial charge and the charge at \( t = \tau \):
\[ \Delta Q = Q_0 - Q(\tau) \]
Substituting the expressions we derived:
\[ \Delta Q = C E_0 - C E_0 e^{-1} \]
Factoring out \( C E_0 \):
\[ \Delta Q = C E_0 (1 - e^{-1}) \]

### Step 5: Substitute \( E_0 \) in Terms of \( I_0 \)
Since \( E_0 = I_0 R \), we can substitute this into our equation:
\[ \Delta Q = C (I_0 R) (1 - e^{-1}) \]
Thus, we can simplify this to:
\[ \Delta Q = C I_0 R (1 - \frac{1}{e}) \]

### Final Answer
The net charge that has flowed through the resistance during one time constant is:
\[ \Delta Q = C I_0 R \left(1 - \frac{1}{e}\right) \]