Let's solve each part step by step: --- ## **Given Data** - \( R = 10\,k\Omega = 10,000\,\Omega \) - \( C = 83.3\,\mu F = 83.3 \times 10^{-6}\,F \) - \( V_s(t) = 50 \cos(2\pi 50 t) \) - Frequency, \( f = 50\,Hz \) - Peak input voltage, \( V_m = 50\,V \) --- ## **a) Peak-Peak Ripple Voltage (\( V_r \))** For a **half-wave rectifier**: \[ V_r = \frac{V_m}{f R C} \] But for peak-to-peak ripple: \[ V_{r(pp)} = \frac{V_m}{f R C} \] Plug the values: \[ V_{r(pp)} = \frac{50}{50 \times 10,000 \times 83.3 \times 10^{-6}} \] Calculate the denominator: \[ 50 \times 10,000 \times 83.3 \times 10^{-6} = 50 \times 10,000 \times .0000833 = 50 \times .833 = 41.65 \] \[ V_{r(pp)} = \frac{50}{41.65} \approx 1.2\,V \] --- ## **b) Average and Maximum Values of the Diode Current (\( i_D \))** ### **Maximum Diode Current (\( i_{D(max)} \)):** - Occurs at the start of the conduction (when the capacitor is discharged). - \( i_{D(max)} = \frac{V_m}{R} \) \[ i_{D(max)} = \frac{50}{10,000} = .005\,A = 5\,mA \] ### **Average Diode Current (\( i_{D(avg)} \)):** - The average current is the charge delivered to the capacitor per cycle divided by the period. - Charge per cycle: \( Q = C \cdot V_{r(pp)} \) - Period: \( T = 1/f \) \[ i_{D(avg)} = \frac{Q}{T} = C \cdot V_{r(pp)} \cdot f \] Plug the values: \[ i_{D(avg)} = 83.3 \times 10^{-6} \times 1.2 \times 50 \] \[ = 83.3 \times 10^{-6} \times 60 \] \[ = 4.998 \times 10^{-3} = 5.\,mA \] --- ## **c) Ripple Voltage if a Full-Wave Rectifier is Used** For a **full-wave rectifier**: \[ V_{r(pp)} = \frac{V_m}{2 f R C} \] Plug in the values: \[ V_{r(pp)} = \frac{50}{2 \times 50 \times 10,000 \times 83.3 \times 10^{-6}} \] Calculate the denominator: \[ 2 \times 50 \times 10,000 \times 83.3 \times 10^{-6} = 100 \times 10,000 \times .0000833 = 100 \times .833 = 83.3 \] \[ V_{r(pp)} = \frac{50}{83.3} \approx .6\,V \] --- ## **Summary Table** | Part | Formula | Value | |------|---------|-------| | a) | \( V_{r(pp)} = \frac{V_m}{fRC} \) | **1.2 V** | | b) | \( i_{D(max)} = \frac{V_m}{R} \) | **5 mA** | | | \( i_{D(avg)} = C V_{r(pp)} f \) | **5 mA** | | c) | \( V_{r(pp)} = \frac{V_m}{2fRC} \) | **.6 V** | --- ### **Let me know if you need the derivation for any step!**