July 1932 Radio News Article

## July 1932 Radio News[Table of Contents]These articles are scanned and OCRed from old editions of the Radio & Television News magazine. Here is a list of the Radio & Television News articles I have already posted. All copyrights are hereby acknowledged. |

Lesson Eleven - Series and Parallel Circuits

By Alfred A. Ghirardi*

Series Circuit Figure 1. The same current flows through every part of the circuit. |

R = R

The total resistance is R = 380 + 20 + 20 + 20 = 440 ohms.

The current I flowing in the circuit is:

I = E/R = 110/440 = 0.25 amperes

Another important fact regarding the series circuit is that the current is the same through every part of the circuit, since there can be no accumulation of current at any point along the circuit. If five ammeters were connected at the points marked I in Figure 1, they would all indicate the same current I, of 0.25 amperes. Also if a series circuit is opened or broken at any point the current stops flowing.

A voltage drop occurs across each of the various resistances in a series circuit, depending on its resistance. If a voltmeter were connected across the filament of tube A, it would indicate E = I X R

Notice from Figure 1 that the voltage drop across any resistance in the circuit depends upon its resistance. Thus even though the same current flows through all parts, the voltage drop across the 380-ohm resistance is 95 volts, whereas that across each 20-ohm resistance is only 5 volts.

In radio receivers series circuits are very common in the plate circuits of vacuum tubes, as we shall see later. The adding of resistances in series is equivalent to increasing the length of the conductor, so that the total resistance is equal to the sum of the separate resistances.

Parallel CircuitFigure 2. The current divides and part flows through each branch. |

Figure 2 shows a parallel circuit consisting of the filaments of three dissimilar vacuum tubes supplied with current forced through the circuits by the e.m.f. of the storage battery, E. Only a portion of the total current circulating through the battery passes through each of the circuits, but of course the sum of the number of amperes of current flowing in the three circuits is equal to the number of amperes of current circulating through the battery, since all the currents combine again. The actual current in each wire of the circuit is indicated on the diagram. Notice how the current coming out of the positive terminal of the battery divides to go through the tube filaments and then combines again at the negative line.

Any number of electrical devices or circuits may be connected in parallel. The current returning to the negative side of the source of e.m.f. is exactly equal to the current leaving the positive side. The current is merely circulating through the circuits. The electrical devices connected in parallel may all have the same resistance or they may all have unequal resistances. If the resistances are equal, then it is evident that the total current will divide equally among the various paths, and the combined resistance of all the paths considered together is equal to one of the resistances divided by the number of resistances. Thus, if five resistances of 100 ohms each are connected in parallel, the combined resistance will be

100/5 = 20 ohms, since five paths are being presented to the flow of current instead of only one.

When the parallel resistances are not equal, the combined resistance must be found by another method, in which the conductances of the various paths are considered. When the resistances are arranged in parallel, since several paths are being offered for the passage of the current, the effect produced is the same as if we were to increase the cross-sectional area of the original conductor. The current passing through the separate resistances is proportional to the conductivity of each path.

It was earlier stated that the conductance of a circuit is equal to 1/R.

That is, the less the resistance of a wire, the greater is its conductance or ability to conduct current. Conductance is expressed in mhos. Thus if the resistance of a conductor is 5 ohms, its conductance is 1/5 = 0.2 mho.

The conductance of the entire parallel circuit is equal to the sum of the conductances of its individual branches. Thus if R stands for the combined resistance of the parallel circuit, and r

from which the combined resistance R may be calculated if the resistances of the individual branches are known. Thus in Figure 2 the combined resistance of the three filaments in parallel is:

from which R = 2.9 ohms. Ans.

Notice that the combined resistance is less than the resistance of any of the paths. This should be expected, of course, since even the path of the lowest resistance is having several additional conducting paths connected in parallel with it so that the resistance must be less. Additional paths increase the current-carrying ability of the circuit; that is, they decrease the resistance.

We see that two or more equal resistances in parallel is merely a special case of parallel circuits. Equation (2) can be used for any condition of equal or unequal resistances.

In a parallel circuit the voltage across each branch is the same as that across every other branch and is equal to that supplied by the source of e.m.f. The current which flows through each branch is simply equal to this voltage divided by the resistance of the branch. Thus in Figure 2, if the battery supplies an e.m.f. of 6 volts, the currents in the various branches are:

Therefore 1=0.3 + 0.6 + 1.2 = 2.1 amps. (This is the total current supplied by the battery.) As a check on this calculation we may calculate the total current directly from the value of the combined resistance of 2.9 ohms obtained above for the circuit. Thus

amps (which checks with the value just calculated)

In a parallel circuit, if anyone of the branches is opened, current will continue to flow through the others. The conditions existing in parallel circuits are as follows:

1. The voltage is equal across all branches.

2.The combined resistance is less than the resistance of any branch of the circuit.

3. The total current is equal to the sum of the currents through all the branches.

Parallel circuits are very common in radio receivers. In battery-operated receivers the filaments of the various tubes are usually connected in parallel across the source of e.m.f. (battery). In a.c. electric receivers the filaments of the tubes are connected in parallel across the filament winding of the transformer. The plate circuits are connected in parallel across the B supply unit.

Physics and science instructors will find these review questions and the "quiz" questions below useful as reading assignments for their classes. For other readers the questions provide an interesting pastime and permit a check on the reader's grasp of the material presented in the various articles in this issue.

The "Review Questions" cover material in this month's installment of the Radio Physics Course. The "General Quiz" questions are based on other articles in this issue as follows: Eliminating Fringe Howl in Regenerative Detectors; Design and Operation of an Interference Meter; Operating and Servicing the Stenode Quartz-Crystal Receiver; Radio Guards the Baby; What Television Needs; Class "B" Tubes-Their Significance in Future Audio Amplifier Design.

1. Four vacuum-tube filaments having the following resistances are all connected in series: 20, 4, 5, 10.

a. Draw the circuit diagram showing the connection.

b. What is the total resistance of the combination?

c. How much current will flow if the entire group is connected to a source of e.m.i. of 50 volts?

2. The resistances in question (1) are all connected in parallel.

a. Draw the circuit diagram for this connection.

b. What is the joint resistance of the combination?

c. What current will flow through each filament if the source of e.m.i, is 6 volts?

d. What is the total current taken from the battery?

3. The filaments of two 201A vacuum tubes having a resistance of 20 ohms each are connected in parallel. In series with this group is another filament having a resistance of 10 ohms. The entire group is supplied with current from a 6-volt storage battery. What is the combined resistance of all the tube filaments, and the total current flowing?

1. Why do most regenerative detectors howl when feed-back is adjusted close to the point of oscillation?

2. How may this condition be corrected?

3. What arrangement is used in one interference locator to visibly measure interference intensity?

4. How can the Stenode be used for short-wave reception?

5. How is the "wall of light" used as protection against burglary? What is one main factor which is retarding the popularization of television?

6. What is the principle of true "Class A" amplification?

Posted September 9, 2013