question archive QUESTION 1 If the resistance of an electric bulb is 500 Ω and voltage across its ends is 250 V then power consumed by it is A
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QUESTION 1
1 points
QUESTION 2
1 points
QUESTION 3
1 points
QUESTION 4
1 points
QUESTION 5
1 points
QUESTION 6
1 points
QUESTION 7
1 points
QUESTION 8
1 points
QUESTION 9
1 points
QUESTION 10
2 points
QUESTION 11
2 points
QUESTION 12
QUESTION 15
1 points
QUESTION 16
2 points
QUESTION 17
1 points
QUESTION 18
1 points
QUESTION 19
QUESTION 1
B.125 W
QUESTION 2
B.The larger of the two reactances is dominant
QUESTION 3
D.zero
QUESTION 4
C.0.8
QUESTION 5
A.Inverse to
QUESTION 6
A.Computers
QUESTION 7
False
QUESTION 8
False
QUESTION 9
D.Impedance
QUESTION 10
PF = 0.57
QUESTION 11
C = 9.21 µF
QUESTION 12
Irms = 0.0357 A
P = 0.714 W
PF = 1
QUESTION 15
I cannot answer this because there is no data given and no Q:14
QUESTION 16
f = 2250.79 Hz
QUESTION 17
True
QUESTION 18
True
QUESTION 19
C.5 kHz
Step-by-step explanation
QUESTION 1
Power is:
P = V2/R
Substitute the values:
P = 2502/500 = 125 W
Answer:
B.125 W
QUESTION 2
Answer:
B.The larger of the two reactances is dominant
QUESTION 3
Answer:
D.zero
QUESTION 4
Power Factor is:
PF = true power/apparent power
PF = 800W/1000VAR = 0.8
Answer:
C.0.8
QUESTION 5
Answer:
A.Inverse to
QUESTION 6
Answer:
A.Computers
QUESTION 7
Answer:
False
QUESTION 8
Answer.
False
QUESTION 9
Answer:
D.Impedance
QUESTION 10
Power Factor is:
PF = XL/Z = 200/350
PF = 0.57
QUESTION 11
Reactive power is:
Q = V2/XC
50 = 1202/Xc
XC= 288 ohms
Capacitance can be solved using the capacitive reactance formula:
XC = 1/2πfC
288 = 1/2π(60)C
C = 9.21 µF
QUESTION 12
Impedance is:
Z2 = R2 + (XL -XC)2
Z2 = 5602 +(1000-1000)2
Z = 560 Ω
For the RMS current:
Irms = Vs / Z = 20/560
Irms = 0.0357 A
Active power is:
P = VI = (20)(0.0357)
P = 0.714 W
Power Factor is:
PF = R/Z = 560/560
PF = 1
QUESTION 15
I cannot answer this because there is no data given and no Q:14
QUESTION 16
Resonant frequency is:
f = 1/(2π√LC)
f = 1/[2π√((0.5x10-3)(10×10-6))]
f = 2250.79 Hz
QUESTION 17
Answer:
True
QUESTION 18
Answer:
True
QUESTION 19
Bandwidth is:
BW = fu - fL
BW = 12000 - 7000
BW = 5000 Hz or 5kHz
Answer:
C.5 kHz