Determine total reactive resistance of this circuit

[whiskers]

This question appeared on the NETA III exam.

Find the total reactance in this circuit. I distinctly remember it because it was the very first question to hit me and I said "Gee thanks alot".

I wasn't quite ready for it.

[rofo42]

This question appeared on the NETA III exam.

Find the total reactance in this circuit. I distinctly remember it because it was the very first question to hit me and I said "Gee thanks alot".

I wasn't quite ready for it.

I'll take a stab at this...

Capacitors in series are found using the reciprocal or product over sum method. This gives a total of 10 uF. (20*20 = 400. 400/40=10)
Capacitors in parallel are added together. 10 uF + 25 uF = 35 uF

Capacitive Reactance (X_c) = 1 / 2*pie*F*C ( 2 * 3.14 * 60 * .000035 = .013188) 1/.013188 = 75.82
X_c = 76

Inductors in parallel are found using the reciprocal or product over sum method. 1/250 + 1/250 = .008 (1/.008 = 125 mH
Inductors in series are added together. 500 mH + 125 mH = 625 mH

Inductive Reactance = 2*pie*F*H (2*3.14*60*.625 = 235.5)
X_l = 236

X_c + X_l (76 + 236) = 312 ohms

I would be surprised if I did that correctly. Someone correct me if needed!

[Schracky]

X_c + X_l (76 + 236) = 312 ohms

I would be surprised if I did that correctly. Someone correct me if needed!

I think you're correct up to this point.

Looking at the equation sheet, it should be X_L - X_C

So doing that I get 160 ohms.

I'd be willing to bet that both 160 and 312 are possible answers on the test.

[Warrengarber]

Your math all works out to me. A time saver when performing this and from my experience with similar questions on the test is that they will use two identical, in this case capacitors in series and reactors in parallel, values. When this is the case the calculated value is half of one of the identical capacitors.

Capacitor Example: Look at Rofo42's capacitors 20ufx20uf=400uf over the sum of 20uf+20uf= 40uf, 400uf/40uf=10uf or half of one of the 20uf capacitors. (Capacitors in series)

Inductor Example: 1/250mH+1/250mH=0.004mH+0.004mH=0.008mH then 1/0.008 with a result of 125mH or half of one of the two parallel 250mH reactors (Reactors in Parallel)

It is all about the time it takes to answer the questions on the test. Remembering these shortcuts during the stress of taking the test and knowing that that one wrong answer may be the make or break is not easy. A question like this could take 5 minutes to answer when doing all of the math but when you only have an average of 72 seconds to answer a question shortcuts count. I hope that helps and great job Rofo42

I'll take a stab at this...

Capacitors in series are found using the reciprocal or product over sum method. This gives a total of 10 uF. (20*20 = 400. 400/40=10)
Capacitors in parallel are added together. 10 uF + 25 uF = 35 uF

Capacitive Reactance (X_c) = 1 / 2*pie*F*C ( 2 * 3.14 * 60 * .000035 = .013188) 1/.013188 = 75.82
X_c = 76

Inductors in parallel are found using the reciprocal or product over sum method. 1/250 + 1/250 = .008 (1/.008 = 125 mH
Inductors in series are added together. 500 mH + 125 mH = 625 mH

Inductive Reactance = 2*pie*F*H (2*3.14*60*.625 = 235.5)
X_l = 236

X_c + X_l (76 + 236) = 312 ohms

I would be surprised if I did that correctly. Someone correct me if needed!

[whiskers]

Your math all works out to me. A time saver when performing this and from my experience with similar questions on the test is that they will use two identical, in this case capacitors in series and reactors in parallel, values. When this is the case the calculated value is half of one of the identical capacitors.

Capacitor Example: Look at Rofo42's capacitors 20ufx20uf=400uf over the sum of 20uf+20uf= 40uf, 400uf/40uf=10uf or half of one of the 20uf capacitors. (Capacitors in series)

Inductor Example: 1/250mH+1/250mH=0.004mH+0.004mH=0.008mH then 1/0.008 with a result of 125mH or half of one of the two parallel 250mH reactors (Reactors in Parallel)

It is all about the time it takes to answer the questions on the test. Remembering these shortcuts during the stress of taking the test and knowing that that one wrong answer may be the make or break is not easy. A question like this could take 5 minutes to answer when doing all of the math but when you only have an average of 72 seconds to answer a question shortcuts count. I hope that helps and great job Rofo42

Yes the math does work out.

However as stated, two inductors in parallel of equal value = 1/2 the inductance of both and 2 capacitors in series of equal value, likewise = 1/2 the capacitance of both. No need for the math.

Whiskers

[Kalbi_Rob]

I'll take a stab at this...

Capacitors in series are found using the reciprocal or product over sum method. This gives a total of 10 uF. (20*20 = 400. 400/40=10)
Capacitors in parallel are added together. 10 uF + 25 uF = 35 uF

Capacitive Reactance (X_c) = 1 / 2*pie*F*C ( 2 * 3.14 * 60 * .000035 = .013188) 1/.013188 = 75.82
X_c = 76

Inductors in parallel are found using the reciprocal or product over sum method. 1/250 + 1/250 = .008 (1/.008 = 125 mH
Inductors in series are added together. 500 mH + 125 mH = 625 mH

Inductive Reactance = 2*pie*F*H (2*3.14*60*.625 = 235.5)
X_l = 236

X_c + X_l (76 + 236) = 312 ohms

I would be surprised if I did that correctly. Someone correct me if needed!

I disagree with the above post, the math is almost good. Only problem I see is you cannot just add Capacitive Impedance and Inductive Impedance together, remember the phase angles are up to 180 degrees apart.

Z_T²=X_R²+(|X_L-X_C|)²

So, since there is no X_R

Z_T=(236-76)=160 Ohms

[Warrengarber]

Thank you for the catch. I see on the formula sheet that they have the formula for Z and you are very correct.

Z= the square root of R squared plus (XL-Xc) squared

I disagree with the above post, the math is almost good. Only problem I see is you cannot just add Capacitive Impedance and Inductive Impedance together, remember the phase angles are up to 180 degrees apart.

Z_T²=X_R²+(|X_L-X_C|)²

So, since there is no X_R

Z_T=(236-76)=160 Ohms

[rofo42]

I disagree with the above post, the math is almost good. Only problem I see is you cannot just add Capacitive Impedance and Inductive Impedance together, remember the phase angles are up to 180 degrees apart.

Z_T²=X_R²+(|X_L-X_C|)²

So, since there is no X_R

Z_T=(236-76)=160 Ohms

I did not know this. Thank you for the information!

[whiskers]

I did not know this. Thank you for the information!

Well, that throws a wrench in the works but, the question was not looking for the impedance of the circuit but simply the reactive component of the circuit at 60 hz.

While the math is correct about the phase angles, I'm not sure what the exact wording of the question was or what 4 choices were available, wish I knew.

[Kalbi_Rob]

Well, that throws a wrench in the works but, the question was not looking for the impedance of the circuit but simply the reactive component of the circuit at 60 hz.

While the math is correct about the phase angles, I'm not sure what the exact wording of the question was or what 4 choices were available, wish I knew.

Impedance must be calculated for each component (i.e. the capacitive impedance, inductive impedance and resistive impedance) to calculate the total reactance of the circuit. Without the frequency, you cannot determine the reactance of the circuit as inductors and capacitors react differently at different frequencies.

This page explains it alot better than I do:
https://www.wikihow.com/Calculate-Impedance