The following image shows the details of a circuit required to answer the questi

Question

The following image shows the details of a circuit required to answer the questions:

Here are the questions:

Rate high, and fast and 5stars

The current in a half wave single phase diode rectifier shown below is given by: where V is the peak supply voltage Using the table below, calculate the circuit current i(omega t) and the voltage drop across the resistor, vr(omega t) for various values of t. Determine the point in tune at which the current drops to zero (i.e. when the diode ceases to conduct), and hence determine the inductance voltage VL(omega t) and the diode voltage vd(omega t) for the same values of t (remember that the diode voltage is zero while it is conducting).

Explanation / Answer

0

These are of simulation results and thus calculated results may vary slightly.

t(ms) i(t) Vs(t) Vr(t) VL(t) Vd(t)

0

0 0 0 0 0 2 166mA 40.355V 3.12V 36.4295V 797mV 4 537.1mA 66.0859V 10.09V 55.104V 882.9mV 6 954mA 65.4891V 17.9435V 46.61V 930mV 8 1.2138A 40.5245V 22.81V 16.75V 954mV 10 1.1962A -49.4mV 22.488V -23.49V 953mV 12 858mV -40.35V -57.407V 16.13V 920mV 14 330.6mA -66.0859V -73.1379V 6.215V 836mV 16 3.49mA -66.3437V 47.057V 65.7mV -113.467V 18 23.25mA -40.9606V -25.49V 437.16mV -15.8987V 20 -6.6uA 1.9068nV 5.2371mV -124.64uV -5.1124mV

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.