The heat produced in a given resistor in a given time by the sinusoidal current `I_(0)` sin `omegat` will be the same as that by a steady current of magnitude .

Options

A. `I_(0)/(sqrt2`
B. `I_(0)`
C. `I_(0)sqrt2`
D. `I_(0)/2`

Explanation

To solve the problem, we need to find the steady current (DC) that produces the same amount of heat in a resistor as the sinusoidal alternating current (AC) given by \( I_0 \sin(\omega t) \).

### Step-by-Step Solution:

1. **Understanding Heat Produced in a Resistor:**
The heat produced in a resistor due to a current is given by the formula:
\[ H = I^2 R t \]
where \( H \) is the heat produced, \( I \) is the current, \( R \) is the resistance, and \( t \) is the time.

2. **Heat Produced by AC Current:**
For the sinusoidal current \( I(t) = I_0 \sin(\omega t) \), we need to calculate the average current over a complete cycle. The effective (RMS) value of the sinusoidal current is given by:
\[ I_{\text{rms}} = \frac{I_0}{\sqrt{2}} \]

3. **Calculating Heat Produced by AC:**
The heat produced by the AC current over time \( t \) can be expressed as:
\[ H_{\text{AC}} = (I_{\text{rms}})^2 R t = \left(\frac{I_0}{\sqrt{2}}\right)^2 R t = \frac{I_0^2}{2} R t \]

4. **Heat Produced by Steady Current (DC):**
Let the steady current (DC) that produces the same heat be \( I \). The heat produced by this steady current over the same time \( t \) is:
\[ H_{\text{DC}} = I^2 R t \]

5. **Equating Heat Produced by AC and DC:**
Since the heat produced by both currents is the same, we can set the equations equal to each other:
\[ H_{\text{AC}} = H_{\text{DC}} \]
This gives us:
\[ \frac{I_0^2}{2} R t = I^2 R t \]

6. **Simplifying the Equation:**
We can cancel \( R \) and \( t \) from both sides (assuming they are non-zero):
\[ \frac{I_0^2}{2} = I^2 \]

7. **Solving for Steady Current \( I \):**
Taking the square root of both sides, we find:
\[ I = \frac{I_0}{\sqrt{2}} \]

### Final Answer:
The magnitude of the steady current \( I \) that produces the same heat as the sinusoidal current \( I_0 \sin(\omega t) \) is:
\[ I = \frac{I_0}{\sqrt{2}} \]