The peak value of the a.c. current flowing throw a resistor is given by

Options

A. `I_(0)=e_(0)//R`
B. `I=e//R`
C. `I_(0)=e_(0)`
D. `I_(0)=R//e_(0)`

Answer

To determine the peak value of the alternating current (AC) flowing through a resistor, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the AC Voltage**:
We start with the AC voltage applied across the resistor, which can be expressed as:
\[ V(t) = V_0 \sin(\omega t) \]
where \( V_0 \) is the peak voltage (maximum voltage), and \( \omega \) is the angular frequency.

2. **Ohm's Law**:
According to Ohm's Law, the current \( I(t) \) flowing through the resistor \( R \) is given by:
\[ I(t) = \frac{V(t)}{R} \]
Substituting the expression for \( V(t) \):
\[ I(t) = \frac{V_0 \sin(\omega t)}{R} \]

3. **Peak Current**:
The peak value of the current \( I_0 \) can be found by taking the maximum value of \( I(t) \):
\[ I_0 = \frac{V_0}{R} \]
This is because the maximum value of \( \sin(\omega t) \) is 1.

4. **Final Expression**:
Thus, the peak value of the AC current flowing through the resistor is:
\[ I_0 = \frac{V_0}{R} \]

### Conclusion:
The peak value of the AC current flowing through a resistor is given by \( I_0 = \frac{V_0}{R} \).