A resistor of resistance `100 Omega` is connected to an `AC` source `epsilon = (12 V) sin (250 pi s^(-1))t`. Find the energy dissipated as heat during `t = 0` to `t = 1.0 ms`.

Answer

To find the energy dissipated as heat in a resistor connected to an AC source, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given parameters:**
- Resistance, \( R = 100 \, \Omega \)
- Voltage source, \( \epsilon(t) = 12 \sin(250 \pi t) \, V \)

2. **Determine the peak voltage (\( E_0 \)) and angular frequency (\( \omega \)):**
- From the voltage equation, we have:
- \( E_0 = 12 \, V \)
- \( \omega = 250 \pi \, \text{rad/s} \)

3. **Calculate the RMS voltage (\( E_{\text{rms}} \)):**
- The RMS value of the voltage is given by:
\[ E_{\text{rms}} = \frac{E_0}{\sqrt{2}} = \frac{12}{\sqrt{2}} = 6\sqrt{2} \, V \]

4. **Use the formula for energy dissipated as heat:**
- The energy dissipated as heat \( H \) in a resistor over time \( t \) is given by:
\[ H = \frac{E_{\text{rms}}^2}{R} \cdot t \]

5. **Substitute the values into the formula:**
- First, calculate \( E_{\text{rms}}^2 \):
\[ E_{\text{rms}}^2 = (6\sqrt{2})^2 = 72 \, V^2 \]
- Now substitute \( E_{\text{rms}}^2 \), \( R \), and \( t \) into the energy formula:
\[ H = \frac{72}{100} \cdot (1 \times 10^{-3}) = 0.72 \times 10^{-3} \, J = 7.2 \times 10^{-4} \, J \]

6. **Account for the time period of integration:**
- Since we are calculating from \( t = 0 \) to \( t = 1 \, ms \) (or \( 1 \times 10^{-3} \, s \)), we need to evaluate the integral of \( \sin^2(\omega t) \) over this time period:
\[ H = \frac{E_0^2}{R} \int_0^{1 \times 10^{-3}} \sin^2(250 \pi t) \, dt \]

7. **Evaluate the integral:**
- The integral of \( \sin^2(\theta) \) can be computed using the identity:
\[ \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} \]
- Thus:
\[ \int_0^{1 \times 10^{-3}} \sin^2(250 \pi t) \, dt = \int_0^{1 \times 10^{-3}} \frac{1 - \cos(500 \pi t)}{2} \, dt \]
- This evaluates to:
\[ \frac{1}{2} \left[ t - \frac{\sin(500 \pi t)}{500 \pi} \right]_0^{1 \times 10^{-3}} = \frac{1}{2} \left[ 1 \times 10^{-3} - 0 \right] = \frac{1 \times 10^{-3}}{2} \]

8. **Final calculation of energy:**
- Substituting back, we get:
\[ H = \frac{72}{100} \cdot \frac{1 \times 10^{-3}}{2} = 0.72 \times 10^{-4} \, J = 3.6 \times 10^{-4} \, J \]

### Final Answer:
The energy dissipated as heat during \( t = 0 \) to \( t = 1.0 \, ms \) is approximately \( 3.6 \times 10^{-4} \, J \).