In an `AC` circuit, the rms value of the current `I_("rms")` is related to the peak current `I_(0)` as

Options

A. 1) `i_(rms)=sqrt2i_(0)`
B. 2) `i_(rms)=pii_(0)`
C. 3) `i_(rms)=(i_(0))/(pi)`
D. 4) `i_(rms)=(1)/(sqrt2)i_(0)`

Answer

To find the relationship between the RMS value of the current \( I_{\text{rms}} \) and the peak current \( I_0 \) in an AC circuit, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding RMS Current**:
The RMS (Root Mean Square) value of the current is defined as the value of the direct current (DC) that would deliver the same power to a resistor as the alternating current (AC) does over one complete cycle.

2. **Power Equivalence**:
The power consumed by the AC circuit can be expressed as:
\[ P = I_{\text{rms}}^2 R \]
where \( R \) is the resistance.

3. **Expression for AC Current**:
The instantaneous current in an AC circuit can be expressed as:
\[ I(t) = I_0 \sin(\omega t) \]
where \( I_0 \) is the peak current and \( \omega \) is the angular frequency.

4. **Calculating Power in AC Circuit**:
The average power over one complete cycle can also be expressed as:
\[ P = \frac{1}{T} \int_0^T I(t)^2 R \, dt \]
where \( T \) is the time period of the AC signal.

5. **Substituting for \( I(t) \)**:
Substitute \( I(t) \) into the power equation:
\[ P = \frac{1}{T} \int_0^T (I_0 \sin(\omega t))^2 R \, dt \]
Simplifying gives:
\[ P = R I_0^2 \frac{1}{T} \int_0^T \sin^2(\omega t) \, dt \]

6. **Using Trigonometric Identity**:
We can use the identity \( \sin^2(x) = \frac{1 - \cos(2x)}{2} \) to simplify the integral:
\[ \int_0^T \sin^2(\omega t) \, dt = \frac{T}{2} \]
Thus, the average power becomes:
\[ P = R I_0^2 \frac{1}{T} \cdot \frac{T}{2} = \frac{R I_0^2}{2} \]

7. **Equating Powers**:
Now, equate the two expressions for power:
\[ I_{\text{rms}}^2 R = \frac{R I_0^2}{2} \]
Cancel \( R \) from both sides (assuming \( R \neq 0 \)):
\[ I_{\text{rms}}^2 = \frac{I_0^2}{2} \]

8. **Taking the Square Root**:
Finally, taking the square root of both sides gives:
\[ I_{\text{rms}} = \frac{I_0}{\sqrt{2}} \]

### Final Answer:
The relationship between the RMS value of the current \( I_{\text{rms}} \) and the peak current \( I_0 \) is:
\[ I_{\text{rms}} = \frac{I_0}{\sqrt{2}} \]