Let's solve the problem step by step: ### **Step 1: Analyze the Voltage Waveform** The waveform is a periodic triangle wave: - The period \( T = 6 \) s (from 0 to 6, 6 to 12, etc.). - Each cycle rises linearly from 0 to 8 V over 3 seconds, then drops to 0 V at 6 seconds. ### **Step 2: Write the Expression for One Period** For \( 0 \le t < 6 \): \[ v(t) = \frac{8}{3} t \quad \text{for} \quad 0 \le t < 3 \] \[ v(t) = 8 - \frac{8}{3}(t-3) = 16 - \frac{8}{3}t \quad \text{for} \quad 3 \le t < 6 \] ### **Step 3: RMS Value Formula** The RMS value over one period is: \[ V_{rms} = \sqrt{\frac{1}{T} \int_0^T v^2(t) dt} \] Where \( T = 6 \) s. ### **Step 4: Calculate the RMS Value** #### Interval 1: \( 0 \le t < 3 \) \[ v(t) = \frac{8}{3}t \] \[ v^2(t) = \left(\frac{8}{3}t\right)^2 = \frac{64}{9} t^2 \] \[ \int_0^3 v^2(t) dt = \int_0^3 \frac{64}{9} t^2 dt = \frac{64}{9} \cdot \frac{t^3}{3} \Bigg|_0^3 = \frac{64}{9} \cdot \frac{27}{3} = \frac{64}{9} \cdot 9 = 64 \] #### Interval 2: \( 3 \le t < 6 \) \[ v(t) = 16 - \frac{8}{3}t \] \[ v^2(t) = (16 - \frac{8}{3}t)^2 = 256 - 2 \cdot 16 \cdot \frac{8}{3} t + \left(\frac{8}{3} t\right)^2 = 256 - \frac{256}{3}t + \frac{64}{9}t^2 \] \[ \int_3^6 v^2(t) dt = \int_3^6 \left(256 - \frac{256}{3} t + \frac{64}{9} t^2\right) dt \] Integrate term by term: - \( \int_3^6 256 dt = 256 (6-3) = 768 \) - \( \int_3^6 \frac{256}{3} t dt = \frac{256}{3} \cdot \frac{t^2}{2} \Bigg|_3^6 = \frac{256}{3} \cdot \frac{36-9}{2} = \frac{256}{3} \cdot \frac{27}{2} = 256 \cdot \frac{27}{6} = 256 \cdot 4.5 = 1152 \) - \( \int_3^6 \frac{64}{9} t^2 dt = \frac{64}{9} \cdot \frac{t^3}{3} \Bigg|_3^6 = \frac{64}{9} \cdot \frac{216 - 27}{3} = \frac{64}{9} \cdot \frac{189}{3} = \frac{64}{9} \cdot 63 = 448 \) Sum: \[ \int_3^6 v^2(t) dt = 768 - 1152 + 448 = 64 \] #### Total over one period: \[ \int_0^6 v^2(t) dt = 64 + 64 = 128 \] ### **Step 5: Calculate RMS** \[ V_{rms} = \sqrt{\frac{1}{6} \cdot 128} = \sqrt{21.333} \approx 4.62 \text{ V} \] --- ### **Step 6: Average Power Absorbed by 8 Ω Resistor** \[ P_{avg} = \frac{V_{rms}^2}{R} \] Where \( R = 8 \) Ω and \( V_{rms} \approx 4.62 \) V. \[ P_{avg} = \frac{(4.62)^2}{8} \approx \frac{21.34}{8} \approx 2.67 \text{ W} \] --- ## **Final Answers** - **The rms value of the given voltage waveform is:** \(\boxed{4.62}\) V - **The average power absorbed by the 8-Ω resistor is:** \(\boxed{2.67}\) W Let me know if you need further clarifications!

Certainly! Here's a step-by-step explanation with 50 words each: **Step 1: Analyze the Waveform** Identify the shape, period, and amplitude of the voltage waveform. Recognize it as a triangle wave with a specific period and peak voltage, which helps in setting up the RMS calculation and understanding the voltage variation over time. **Step 2: Write the Mathematical Expression** Derive equations for the voltage during each segment of the cycle, considering linear rise and fall within each half-period. This allows precise integration of \( v^2(t) \) over each interval, essential for calculating RMS accurately. **Step 3: RMS Value Definition** The RMS (Root Mean Square) value measures the effective voltage that delivers the same power as a DC voltage. It involves integrating the squared voltage over one period, dividing by the period, and taking the square root, representing the waveform's average power content. **Step 4: Calculate RMS via Integration** Integrate the squared voltage over each interval, sum the results, and normalize by the period. This process involves basic calculus, where integrating \( v^2(t) \) over each segment yields the total energy contribution needed to find the overall RMS. **Step 5: Final RMS Calculation** Divide the total integral by the period length, then take the square root to find the RMS value. This step consolidates the area under \( v^2(t) \) curve into a single effective voltage value, representing the waveform's true power equivalence. **Step 6: Power Dissipation Calculation** Use the RMS voltage and resistor value to compute average power using \( P = V_{rms}^2 / R \). This formula relates the effective voltage to the power dissipated in the resistor, providing an understanding of the energy transfer in the circuit.