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
Originally Posted by
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!
Warren Garber
Have a great day!