Power, Voltage, Current, Resistance Problem help! Please, Thank you.

[cnowak]

If anyone has any advise on starting this question the help would be greatly appreciated. I cannot figure out how to begin. Every time I think I am going in the right direction other parts of the question don't add up.

[rofo42]

If anyone has any advise on starting this question the help would be greatly appreciated. I cannot figure out how to begin. Every time I think I am going in the right direction other parts of the question don't add up.

This should help you out.

[cnowak]

This should help you out.

I appreciate this video but it unfortunately doesn't help because I am starting with the Power on each component and need to figure out everything else. I really just need a good starting point to get me going. I'm just not sure where to begin.

[CQCmaster]

I appreciate this video but it unfortunately doesn't help because I am starting with the Power on each component and need to figure out everything else. I really just need a good starting point to get me going. I'm just not sure where to begin.

I'll help you start but not solve the entire thing...

P = 200W
Q_L = 50 VARS (Inductive)
Q_C = 100 VARS (Capacitive)
Therefore Q(Net) = 100 - 50 = 50 VARS (Reactive Power)

S = sqrt[(true power)^2 + (reactive power)^2]

Phase Angle = sin(Q/S) OR cos(P/S)

S=IE

You know E = 120V

The rest you can calculate yourself.

De nada amigo

[jrm5116]

If anyone has any advise on starting this question the help would be greatly appreciated. I cannot figure out how to begin. Every time I think I am going in the right direction other parts of the question don't add up.

This is an older post, so hopefully you figured it out since then, but I figured I'd reply in case it could help someone else out.

So how I'd start out is to determine the overall complex power as you are given all power values.

Since you are given this, it is just the sum of the powers provided in this circuit with the exception that they did not provide a sign convention on the capacitor or inductor power. So S=200W+j((50VARs)+(-100VARS)) = 200-j50. In phasor form S = 206.16<-14.04 degrees. This is found by taking the Sqrt of (200²+50²) for the magnitude then taking Tan^-1(-50/200).

At this point you have all of your power values (P_r, P_L, P_C were given) and you just found the total power and phase angle. (As a side note if P is only referring to real power and not complex power, then P_r and P_T=200, but P_L and P_C would equal 0 and those answers would be in W only, but I do not think that was what you were looking for here, rather I made an assumption those answers were for S_X and so on.)

With your total complex power found, you can determine your total current for the circuit and since this is a series circuit the current through the entire circuit will be the same.

So S_T = 206.16<-14.04 degrees. And S=VI*.

So 206.16<-14.04 degrees = 120xI
So I = 1.667+j.4167 or 1.718<14.04
& I*=1.718<-14.04
So I_T=I_R=I_L=I_C = 1.718<14.04 degrees

With the current for the whole circuit now known, you can find the individual voltages across the elements by using the formula S=VI* on individual elements rather than the whole circuit.

For the Resistor
V=200/1.718<-14.04 = 112.94+j28.23 or 116.395<14
For the Inductor
V=(50<90)/(1.718<-14.04) = -7.059+j28.23 or 29.1<104
For the Capacitor
V=(100<-90)/(1.718<-14.04) = 14.118-j56.47 or 58.2<-75.9

Notice that when you add these values together you get 120V with no imaginary components, which is what your given voltage was.

With these voltages you can then find the Impedances, by utilizing the formula R=V/I

For the Resistor
R=116.395<14/1.718<14 = 67.76<0
For the Inductor
X_L=29.1<104/1.718<14 = 16.94<90
For the Cap
X_C=58.2<-76/1.718<14 = 33.88<-90

These added together gives you
Z_T=67.76-j16.94 = 69.845<-14.04
You can then check this by using V=IR and multiplying this impedance by your total current to confirm you get 120<0.

V=1.718<14.04*69.845<-14.04 = 120<0

There's a lot of room for error, but hopefully that makes some sense. I had to come back to edit because I had messed up some of my signs. Its important to double check on these types of problems.