# RLC AC circuit

• Engineering

## 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:

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!

## The Attempt at a Solution

#### Attachments

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Related Engineering and Comp Sci Homework Help News on Phys.org
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?

Last edited:
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

gneill
Mentor
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.

vk6kro
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:

http://dl.dropbox.com/u/4222062/series%20cct%205jan.PNG [Broken]

Last edited by a moderator:
that makes more sense !! I got sensible answers.
Does anyone know the correct answers?

vk6kro
that makes more sense !! I got sensible answers.
Does anyone know the correct answers?
I have solutions, but I wonder about giving them before the OP works them out?

Could I send them to you via PM?

@vk6kro, are my answers incorrect? if so, could you please show to us the final answers? i am also in doubt of my answers.

vk6kro

Did you understand the diagrams I posted above?
First you need to work out the size of the resistance of the coil and the inductive reactance of the coil.

Then you should put those values into the main part of the problem.

The diagrams actually show you how to do it. Try to follow a logical procedure and you should get it.

As a check, you should get 132 uF for the capacitor size.