Figure 12.11.3 A \"black box\" connected to an AC voltage source. The elements i

Question

Figure 12.11.3 A "black box" connected to an AC voltage source. The elements in the circuit and their arrangement, however, are unknown. Measurements outside the black box provide the following information: V(t) = (80 V)sin omega I(t) = (1.6 A)sin(omega t + 45 degree)Does the current lead or lag the voltage Is the circuit in the black box largely capacitive or inductive Is the circuit in the black box at resonance What is the power factor Does the box contain a resistor A capacitor An inductor Compute the average power delivered to the black box by the AC source.

Explanation / Answer

given V(t) = 80 sin wt

I(t) = 1.6 sin (wt + 45)

Current leads by 45 deg

-----------------------------------------

b . since Current leads in capacitive circuit

thius must be capacitive circuit

--------------------------------------------------

c. yes it is resosnace

-------------------------------

d. Power factor cos theta = cos 45 = 0.707

------------------------------------------

e. it contains = capacitor

-----------------------------------

f. power delivered P = VI cos theta

P = 80 * 1.6 * cos 45

P = 90.5 Watts

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

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