Hi all, i am stuck on how to find the resistor and coil values on this circuit. i have been given the following: Voltage = 100v phase angle 0° at 20hz Current = 1.5A phase angle -30° Capacitor = 50μF any help would be much appreciated many thanks
Very easy. Phasor analysis. But what is the voltage 100V across and current 1.5A through? I assume the voltage 100V is the input voltage source since it has the frequency term. But what is 1.5A through? After you figure that out, convert everything to impedances and the voltage source to a phasor. ZL = jwL Ω and ZC = 1/jωC are the inductor and capacitor impedances respectively. where ω= 20hz *2π = 125.6 rad/s Do you know what I'm talking about? Not sure if you learned this yet. If you do phasor analysis on your circuit you can easily find the resistance and inductance values. It would also help if you gave us the schematic of your circuit.
ye i can sort of see what you mean all it says is the current flowing in the circuit is 1.5A phase angle -30 i have made a quick sketch from the data for you
i expect the coil to be of of a greater size than the capacitor due to the -30° angle of the current, am i right in thinking that?
Hi By size I think you mean impedance: yes the coil will have a greater impedance than the capacitor to give you a current lagging the voltage. It should be easy to calculate. The reactive parts of your circuit are as previous contributor said j(ωL-1/(ωC)). The phase angle is obtained from the imaginary part of the ac divided by the real which gives you the ratio -1/sqrt(3) in this case (remember triangles?). You know C so you should be able to find L and R. Hope this is a bit more of a clue.
Ok so its a series circuit. All you need to do is KVL using phasor analysis. Or even a voltage divider. I will go ahead and do it for you this time, but I want you to learn from this. Pay attention to what I do. Step by step.
OK a bit more help. I mentioned that the phase angle is tan-1(Im /Re) which we know is -30 deg. SInce the total imaginary is jwL-j/(wC) we can set [wL-1/(wC)]/R=-1/sqrt(3) We know that the magnitude of the current is 100/1.5 = 66.67 ohms but this is made up from the reactive and resistive bits. The mag of Z=sqrt(R**2 + (wL-1/wC)**2) =66.67 ohms Substitute for wL-1/(wC) from the phase .... cheers ...mine's a pint
I am only a 2 yr electronic student, 30 years ago and never studied since. We only studied RCL circuits 1 quarter. I was never to smart anyway. This discussion sounds so intresting that I just have to make a fool of myself. I though that you had two have 2 of any 3 in order to find the third in a regular current, voltage, resistance circuit. If this is true in and RCL circuit as well? Would you not have to test for an actual voltage across the resistor and or inductor to see what the volt drop is and then R=V/I? If the volt drop on R1 is 50 v then R= 33.333 ohm. God only knows what phasor analysis is? never heard of this. It do sound intresting though. mojojerry
For a series RLC circuit, we know the following: leads by 90° which, in turn leads by 90° leads by 90° which, in turn leads by 90° and are in phase with each other. So, first we have: and we know Which gives us Now, we can find and and Now, for the tricky part: Since we know the angular relationship between and , we know that is 90° ahead of which turns out to be (-120°) + 90° = (-30°) Also, we know that is 90° ahead of , which is (-30°) + 90° = 60° Now, all we have to do is find the magnitude of those values: Since and are 180° apart, the sum of their angles must be either 60° or -120°, therefore it must be 60° so that, when added to [E_R] (which has an angle of -30°), we get a vector resulting with an angle of 0°. Also, the x-components of the sum of , , and must be 100 and the sum of their y-components must be zero. This gives us: and substituting, we get or, more correctly, From here, you can divide by to get which gives you and, you can calculate by: or Divide to get This gives us R = 57.735Ω and XL = 192.488Ω Since XL=2∏fL, and f=20Hz, we can solve for L:
Yes, you're right. I inadvertently used the magnitude of (86.603) as the value of the resistance, forgetting to divide by the current first. I made the appropriate corrections to my earlier post.