Please help me! I need FULL STEPS AND ORGANIZATED PLEASE . Thanks a lot. Problem
ID: 1660871 • Letter: P
Question
Please help me! I need FULL STEPS AND ORGANIZATED PLEASE. Thanks a lot.
Problem 3: The following graph shows the voltage and current through a single element of the circuit Voltage Current Vp Ip 0 1. What is the frequency of the AC generator? If Vp 69 V, lp 1.8 A, t1-0.07s 2. What is the average power dissipated by this element? 3. Is the element of the circuit described above a resistance, or an inductor ora capacitor? Use the graph and what you know about the circuits to determine what type of element is shown in the graph. If it is a resistance, what is the value of the resistance in Ohms?, if it is an inductor. What is the value of the inductance (in H)? If i is a capacitor, what is the value of the capacitance (in F)? 4. Is it possible to add a second element in series to the circuit, such that the circuit goes into resonance? Explain 5. If possible, what is the second element a resistance (R), coil (L), or capacitor (C)?Explanation / Answer
Q1.
from the graph, it can be seen that(1+(3/4)) of time period =t1=0.07 seconds
==>time period=0.07/(7/4)=0.04 seconds
frequency=1/0.04=25 Hz
Q2.
average power=Vrms*Irms*cos(theta)
where Vrms=rms value of voltage=Vpeak/sqrt(2)=69/sqrt(2)=48.79 volts
Irms=rms value of current=Ipeak/sqrt(2)=1.8/sqrt(2)=1.2728 A
as can be seen from the graph,
voltage=Vp*cos(2*pi*f*t)
and current=Ip*sin(2*pi*f*t)=Ip*cos(2*pi*f*t-90)
so voltage leads current by 90 degrees
so they are separated by an angle of 90 degrees .
hence theta=90 degrees
average power=Vrms*Irms*cos(90)=0
Q3.
as voltage lead current by 90 degrees, the element is an inductor.
inductive reactance=Vp/Ip=69/1.8=38.33 ohms
==>2*pi*f*inductance=38.33
==>inductance=38.33/(2*pi*25)=0.244 H=244 mH
Q4. resonance has lowest impedance.
as the impedance is inductive, by adding a capacitor in series, net impedance can be made zero. there by creating resonance condition.
Q5.
a capacitor needs to be the second element.
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