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 (if any) are hereby
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.
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Quiz on AC Circuit Theory
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, (VTotal
,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 Answers1. (10)2 = (8)2
+ (VR)2; VR = 6V
AC Circuit Theory Quiz
= (7+VC)2 + (12)2; VC =
3. (24)2 = (VL-6)2;
VL = 30V
4. (15)2 = (350-350)2
+ (VR)2; VR = 15V
= (15-3)2 + (16)2; VT =
6. (50)2 = (VL)2
+ (12)2 = (VC)2 + (12)2;
VL = VC
(VT)2 = (VL-VC)2
+ (12)2; VT = 12V
= (6)2 + (8)2; IT = 10 mA.
8. (20)2 = (16-IC)2
+ (16)2; IC = 4 mA.
= (17-13)2 + (3)2; IT =
10. (9)2 = (40-4-IC)2;
IC = 7 mA.