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Homework Statement
A series circuit consisting of a 10 ohm resistor, a coil and an ideal capacitor is connected across a 200 V, 60 Hz power supply. The current drawn is 8 A. The voltage across the coil is 120 V and its power factor is 0.8 lagging. Determine (a) the voltage across the capacitor, and (b) the resultant voltage across the resistor and the coil.
Homework Equations
V = I*Z; V(resistor) = I*R
Phasor angle(coil) = PHI = cos^1(0.8); wherein ^1 indicates inverse.
The Attempt at a Solution
Solving first V(resistor):
V(resistor) = I*R
" = 8 A(10 ohms)
V(resistor) = 80 V
Phasor angle(coil) = PHI = cos^1(0.8)
Phasor angle(coil) = PHI = 36.87 degrees
Here's the phasor diagram:
https://www.physicsforums.com/attachment.php?attachmentid=42443&stc=1&d=1325682519
By using cosine law to solve for V(RL):
V(RL)^2 = V(resistor)^2 + V(coil)^2  2V(resistor)V(coil)cos(180  PHI)
V(RL)^2 = 80^2 + 120^2  2(80)(120)cos(180  36.87)
V(RL) = 190.16 V => (b)
Now solving for THETA(the phasor angle of V(RL)) by using sine law:
sin(THETA)/V(coil) = sin(180  PHI)/V(RL)
Solving for THETA:
THETA = 22.25 degrees
Then by using vector component method in order to solve for V(capacitor) and the ideal capacitor's capacitance:
xcomp ycomp
V(RL) 190.16cos(22.25) 190.16sin(22.25)
V(C) V(C)*cos(90) V(C)*sin(90)
V(source) 200cos(BETA) 200sin(BETA)
Using summation of xcomp to solve for angle BETA:
190.16cos(22.25) + V(C)*cos(90) = 200cos(BETA)
BETA = 28.36 degrees (must be negative since BETA's value must be always between the value of THETA and 90 degrees)
Using summation of ycomp to now solve for V(C):
190.16sin(22.25) + V(C)*sin(90) = 200sin(28.36)
V(C) = 167.01 V => (a)
So am I doing the right thing? If not, please show your solution. Thanks!
Homework Statement
Homework Equations
The Attempt at a Solution
Attachments

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