Your RF Cafe


5th MOB: Airplanes and Rockets: Equine Kingdom: 

RF Cafe SoftwareCalculator WorkbookRF Workbench Smith Chartâ„˘ for Visio Smith Chartâ„˘ for Excel RF & EE Symbols Word RF Stencils for Visio 
December 1970 Popular Electronics[Table of Contents]People old and young enjoy waxing nostalgic about and learning some of the history of early electronics. Popular Electronics was published from October 1954 through April 1985. All copyrights are hereby acknowledged. See all articles from Popular Electronics. 
Here is a fairly simple quiz on AC circuit analysis. If you are not already comfortable with adding series and parallel circuits containing resistors, capacitors, and inductors, you will appreciate the simple formula presented that will keep the sweat level down ;) OK, pick up you pencils... now.
By Robert P. Balin
Voltage measurements made in a series ac circuit seldom add up as simply as they do in a dc circuit. You may even find the voltage across a coil or capacitor to be greater than the supply voltage! Nevertheless, Ohm's Law and Kirchhoff's Law do apply, and careful measurements will show that the supply voltage and the various voltage drops around a series ac circuit are related in an unusual way: the square of the supply voltage is equal to the square of the difference between the voltage on the coil and the voltage on the capacitor, plus the square of the voltage on the resistor.
This relationship, (V_{Total})^{2} = (V_{L}  V_{C})^{2} + (V_{R})^{2},can be used to find any unknown voltage if all others are known.
In parallel ac circuits, the currents add up the same way as voltages do in a series ac circuit. Brush up on your ac theory and see if you can solve the missing voltage or current in the circuits below. Where necessary, the voltages and currents are related by the 3:4:5 ratio to provide easy, whole number answers. Only simple algebra is required. Vectors, phasors, and quadratic equations are not necessary to find the solutions.
(See answers below)
Circuit Quiz Answers
1. (10)^{2}= (8)^{2} + (VR)^{2}; VR = 6V
2. (20)^{2}= (7+VC)^{2}+ (12)^{2}; VC = 9V
3. (24)^{2}= (VL6)^{2}; VL = 30V
4. (15)^{2}= (350350)^{2}+ (VR)^{2}; VR = 15V
5. (VT)^{2}= (153)^{2}+ (16)^{2}; VT = 20V
6. (50)^{2}= (VL)^{2}+ (12)^{2}= (VC)^{2}+ (12)2; VL = VC
(VT)^{2}= (VLVC)^{2}+ (12)^{2}; VT = 12V
7. (IT)^{2}= (6)^{2}+ (8)^{2}; IT = 10 mA.
8. (20)^{2}= (16IC)^{2}+ (16)^{2}; IC = 4 mA.
9. (IT)^{2}= (1713)^{2}+ (3)^{2}; IT = 5 mA.
10. (9)^{2}= (404IC)^{2}; IC = 7 mA.
Posted June 21, 2013