## RLC AC circuit

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

V = I*Z; V(resistor) = I*R
Phasor angle(coil) = PHI = cos^-1(0.8); wherein ^-1 indicates inverse.

3. 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:

http://www.physicsforums.com/attachm...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:

x-comp y-comp

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 x-comp 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 y-comp 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!
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
Attached Thumbnails

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 I am uncertain about the components in your circuit! If the supply voltage is 200V and the power factor is 0.8 then the voltage across the resistance should be VsCos∅ = 200 x 0.8 = 160V But the current is given as 8A and the resistance as 10Ω which means a voltage of 80V across the resistance. Does the coil have resistance itself?
 Also.... your phasor diagram should have only 3 arrows A vertical (+y direction) to represent Vl A vertical (-y direction) to represent Vc and A horizontal (+ x direction) to represent Vr (this would be total R which makes me think your coil has a resistance other than the 10Ω stated in the question. Hope this helps you

Recognitions:
Homework Help

## RLC AC circuit

 Quote by technician Also.... your phasor diagram should have only 3 arrows A vertical (+y direction) to represent Vl A vertical (-y direction) to represent Vc and A horizontal (+ x direction) to represent Vr (this would be total R which makes me think your coil has a resistance other than the 10Ω stated in the question. Hope this helps you
The problem statement is a bit sneaky. It states that there is a coil rather than an inductor, and then says that there is an ideal capacitor. From that I think we should gather that the while the capacitor is ideal the coil is not; it will have some resistance. This is backed up by the statement assigning a power factor to the coil. The coil inductance and resistance can be calculated from the given information.

 Recognitions: Science Advisor The coil does have some resistance and so it is useful to do that part of the calculation first and then put the values obtained into a main calculation. This diagram might explain the process:
 that makes more sense !! I got sensible answers. Does anyone know the correct answers?

Recognitions: