1. **Problem Statement:** We are given an AC voltage applied across a 6 ohm resistor, with a stepped waveform voltage described at specific times. 2. **Understanding the waveform:** The voltage steps are: - At $-10t$, voltage is $-1005$ V (negative step mirrored) - At $-5t$, voltage is $-505$ V (negative step mirrored) - At $5t$, voltage is $505$ V (positive step) - At $10t$, voltage is $1005$ V (positive step) 3. **Power dissipation formula:** Power dissipated by a resistor is given by $$P = \frac{V_{rms}^2}{R}$$ where $V_{rms}$ is the root mean square voltage across the resistor and $R$ is the resistance. 4. **Calculate $V_{rms}$ for the stepped waveform:** Assuming the waveform is periodic and symmetric, the RMS voltage is calculated by averaging the square of voltage values over one period and then taking the square root. 5. **Calculate the mean of squared voltages:** The voltage levels are $\pm 505$ V and $\pm 1005$ V. Assuming equal time intervals for each step, the mean square voltage is: $$V_{rms}^2 = \frac{505^2 + 1005^2 + 505^2 + 1005^2}{4} = \frac{2 \times 505^2 + 2 \times 1005^2}{4} = \frac{2(505^2 + 1005^2)}{4} = \frac{505^2 + 1005^2}{2}$$ Calculate the squares: $$505^2 = 255025$$ $$1005^2 = 1010025$$ So, $$V_{rms}^2 = \frac{255025 + 1010025}{2} = \frac{1265050}{2} = 632525$$ 6. **Calculate $V_{rms}$:** $$V_{rms} = \sqrt{632525} \approx 795.3 \text{ volts}$$ 7. **Calculate power dissipation:** Given $R = 6$ ohms, $$P = \frac{V_{rms}^2}{R} = \frac{632525}{6} \approx 105420.8 \text{ watts}$$ **Final answer:** The power dissipated by the 6 ohm resistor is approximately **105420.8 watts**.