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# Voltage QuizDecember 1961 Popular Electronics

 December 1961 Popular Electronics Table of Contents Wax nostalgic about and learn from the history of early electronics. See articles from Popular Electronics, published October 1954 - April 1985. All copyrights are hereby acknowledged.

Popular Electronics magazine's quizmeister Robert P. Balin created this "Voltage Quiz" for the December 1961 issue. Fortunately, only fixed resistors (6 Ω each) and battery voltage source(s) are involved. The goal is to calculate the voltage at point "A" with respect to ground. Analyzing some of the circuits is made easier by rearranging the layout to remove wiring overlaps. Circuit 5 might be the simplest of all, and can be done fairly easily in your head. Others, like circuit 1, are best approached by writing two mesh equations per Kirchhoff's voltage law (the sum of all voltages in a closed circuit equals zero) and solving the resulting two equations in two unknowns (basic algebra). Spoiler: Upon redrawing a couple circuits you might discover an identical instance of a previous one, making the solution the same as well. A big list of other quizzes is given below.

## Voltage Quiz

By Robert P. Balin

Can you determine the voltage and polarity of point A with respect to ground in each of the seven circuits shown below? All resistors measure six ohms and each battery produces six volts d.c. You'll find it helpful to use pencil and paper in working out the problems. Write the answers in the spaces provided under the diagrams.

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Quizzes from vintage electronics magazines such as Popular Electronics, Electronics-World, QST, and Radio News were published over the years - some really simple and others not so simple. Robert P. Balin created most of the quizzes for Popular Electronics. This is a listing of all I have posted thus far.

In each problem, the schematic diagram shown on page 69 must be redrawn for the sake of clarity. The solution to the problem should then become apparent. Remember that each resistor is six ohms.

1) Kirchhoff's law states that the algebraic sum of the products of the current and resistance in any closed path in a network is equal to the voltages connected in the same path. Setting up two simultaneous equations, we solve for current 12, then determine the voltage drop across R4 which is voltage A:

(1) 6= 6I1+ 6(I1 - I2)

(2) 0= 6(I2 - I1) + 12I2

Now separate variables I1 and I2:

(1) 6= 12I2 - 6I2

(2) 0 = -6I1 + 18I2

Divide equation (1) by 2 and add equations (1) and (2):

3 = 15I2

I2 = 0.2 ampere

Now solve for voltage A:

Ea =R4 x -I2

Ea = 6 x -0.2 = -1.2 volts

2) The symmetry of the redrawn circuit suggests that voltage A is at ground potential (or 0 volt). Remove R5 from the circuit and note that currents flowing through resistor branches R1-R4 and R2-R3 are equal. Hence, the voltages at the midpoint of both branches are equal or point A is at ground potential. Now insert R5 (which could be any value from zero ohm to infinity). Since point A is at ground potential, no current will flow through R5. This can be proved using Kirchhoff's law and setting up three simultaneous equations.

3) The redrawn version of the schematic is identical to the one shown above in Answer 1 except that the battery is connected backwards. So, voltage A is the same as in Answer 1 change in sign, or +1.2 volts.

4) Redrawn, the circuit is simplicity itself. It is easy to determine the current flowing through the loop consisting of the battery, R1, R2 and R3 (1/3 ampere). The voltage at point B is determined by the IR drop across R3 (-2 volts) . Since no current flows through resistor R4 to point A, both point A and point B are at the same potential: -2 volts.

5) A glance at the redrawn circuit suggests that resistors R1, R2, R3 and R5 be combined mathematically into one resistor (found to be 6 ohms) and, resorting to Ohm's law, the voltage at point A is +3 volts.

6) The original schematic, impressively complicated in appearance, becomes straightforward when redrawn as above. It is simply a problem of combining resistors R2 and R3 mathematically and computing current flow. Once done, the voltage drop across R1 is computed to be +2 volts.

7) Here again it is wise to combine resistors R1, R2 and R4 mathematically into one resistor and then resort to Ohm's law. The current through R3 is six-tenths of an ampere. Thus, we find that the voltage drop across the R1, R2 and R4 resistor combination is +2.4 volts.

Posted October 24, 2023