Let's solve the step by step: ## Given Data- **Source:** \( V = -8 16\cos(\omega t) \) V - **RS (series resistor):** 2kΩ - **RL (load resistor):** 8kΩ - **D1:** Ideal diode--- ## 1. **Analyzing the Circuit** ### **Source Voltage** \( V_S = -8 + 16\cos(\omega t) \) - This is a cosine wave with an offset of \(-8\) V and amplitude \(16\) V. ### **Operation of the Diode** - The diode conducts only when its anode (left side) is more positive than its cathode (right side, connected to RL). - When \( V_S > \), the diode conducts. - When \( V_S < \), the diode is off. --- ## 2. **Finding Conduction Interval** Set \( V_S = \) to find when the diode starts/stops conducting: \[ -8 + 16\cos(\omega t) = \implies \cos(\omega t) = .5 \] \[ \omega t = \pm\cos^{-1}(.5) = \pm \frac{\pi}{3} \] So, the diode conducts from \( \omega t = -\frac{\pi}{3} \) to \( +\frac{\pi}{3} \) in each cycle. --- ## 3. **a. Average Value of \( V_{RL} \)** When the diode conducts, \( V_{RL} \) is the voltage across RL in the voltage divider: \[ V_{RL} = V_S \cdot \frac{R_L}{R_S + R_L} \] Plug in values: \[ \frac{R_L}{R_S + R_L} = \frac{8k}{2k + 8k} = \frac{8}{10} = .8 \] So, \[ V_{RL} = .8 V_S = .8(-8 + 16\cos(\omega t)) = -6.4 + 12.8\cos(\omega t) \] But this is valid **only when the diode conducts** (\( -\frac{\pi}{3} < \omega t < \frac{\pi}{3} \)). ### **Average Value Calculation** The average over one period \( T \) is: \[ V_{RL,avg} = \frac{1}{T} \int_{-\frac{\pi}{3\omega}}^{+\frac{\pi}{3\omega}} (-6.4 + 12.8\cos(\omega t))\,dt \] But it's easier to use the symmetry and period \( T = 2\pi/\omega \): \[ V_{RL,avg} = \frac{1}{2\pi} \int_{-\frac{\pi}{3}}^{+\frac{\pi}{3}} (-6.4 + 12.8\cos(\theta))\,d\theta \] Evaluate: \[ = \frac{1}{2\pi} \left[ -6.4\theta + 12.8\sin(\theta) \right]_{-\frac{\pi}{3}}^{+\frac{\pi}{3}} \] \[ = \frac{1}{2\pi} \left[ -6.4\left(\frac{\pi}{3} - \left(-\frac{\pi}{3}\right)\right) + 12.8\left(\sin\left(\frac{\pi}{3}\right) - \sin\left(-\frac{\pi}{3}\right)\right) \right] \] \[ = \frac{1}{2\pi} \left[ -6.4\left(\frac{2\pi}{3}\right) + 12.8\left(2\sin\left(\frac{\pi}{3}\right)\right) \right] \] \[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \] \[ = \frac{1}{2\pi} \left[ -6.4 \cdot \frac{2\pi}{3} + 12.8 \cdot 2 \cdot \frac{\sqrt{3}}{2} \right] \] \[ = \frac{1}{2\pi} \left[ -\frac{12.8\pi}{3} + 12.8\sqrt{3} \right] \] \[ = \frac{1}{2\pi} \left[ 12.8\sqrt{3} - \frac{12.8\pi}{3} \right] \] \[ = \frac{12.8}{2\pi} \left[ \sqrt{3} - \frac{\pi}{3} \right] \] \[ = \frac{6.4}{\pi} \left[ \sqrt{3} - \frac{\pi}{3} \right] \quad \boxed{\text{(final answer in symbolic form)}} \] If you want a **numerical value**: - \(\sqrt{3} \approx 1.732\) - \(\pi \approx 3.142\) \[ V_{RL,avg} = \frac{6.4}{3.142} \left(1.732 - \frac{3.142}{3}\right) \] \[ = 2.037 \cdot (1.732 - 1.047) \] \[ = 2.037 \cdot .685 = 1.396 \ \text{V (rounded)} \] --- ## 4. **b. Peak Value of \( V_{RS} \)** \( V_{RS} = V_S - V_{RL} \), but only when diode conducts. \[ V_{RS} = V_S - V_{RL} = V_S - .8V_S = .2V_S \] \[ V_{RS} = .2(-8 + 16\cos(\omega t)) = -1.6 + 3.2\cos(\omega t) \] Peak value of \( V_{RS} \) **when diode conducts** (\( \omega t = \)): \[ V_{RS,peak} = -1.6 + 3.2 \cdot 1 = 1.6 \ \text{V} \] --- ## 5. **c. Peak Value of \( V_{RL} \)** As before: \[ V_{RL} = -6.4 + 12.8\cos(\omega t) \] Peak value (\( \cos(\omega t) = 1 \)): \[ V_{RL,peak} = -6.4 + 12.8 = 6.4 \ \text{V} \] --- ## **Summary Table** | Quantity | Formula | Value | |-------------------------|--------------------------|-----------| | \( V_{RL,avg} \) | \( \frac{6.4}{\pi} (\sqrt{3} - \frac{\pi}{3}) \) | **1.40 V** | | \( V_{RS,peak} \) | -1.6 + 3.2 | **1.6 V** | | \( V_{RL,peak} \) | -6.4 + 12.8 | **6.4 V** | --- If you need further explanation or want to see the waveform, let me know!