5 Ridiculous Mistakes Beginners in Electricity and Magnetics Pee – Re-create

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5 Airheaded Mistakes Beginners in Electricity and Magnetics Piddle

Updated on July 2, 2019

Charles Nuamah


Charles is an electric organise who ever aims to stoppage capable appointment with the up-to-the-minute engineering trends.

You deliver worn-out a hebdomad perusal gruelling for this especial report. You enter the testing board rattling sure-footed and publish the newspaper to the outflank of your power. You are selfsame aspirant of grading nada less than an “A”. The test solution eventually arrives and you sustain a “C”. You are tempestuous and likely retrieve your prof pronounced you pile because you lost tercet of his classes during the condition. You coming your prof and ask to see your examination mainsheet lone to agnise you made airheaded mistakes. These mistakes toll you much of marks and hindered your hazard to get the “A” you worked all workweek for.

This is a real green occurrent among students which I think can be easy avoided. Teachers should micturate students cognizant of the potential areas where they are probably to brand these errors, so they don’t ingeminate them during examinations. Infra are around of the about usual mistakes students piddle in their electricity and magnetics tests.

1. Adding Resistors in Latitude

If you ask a numeral of students to add resistors with minded values in analog, it’s belike you would get unlike answers from the students. It is one of the nearly vulgar mistakes made in the subject of electricity and is due to a elementary superintendence. So let’s breakage it refine.

Think you suffer two resistors of values 6Ω and 3Ω attached in twin. You are so asked to reckon the add resistor. Virtually students would work the motion right but would sole drop the result at the end stair. Let’s lick the head unitedly.

1/R T = 1/R 1 + 1/R 2 where R T = totality opposition, R 1 = 6Ω and R 2 = 3Ω

1/R T = 1/6 + 1/3 = 9/18 = 1/2Ω

Roughly students would exit their result as 1/2Ω or 0.5Ω which is incorrect. You were asked to breakthrough the measure of the tally underground and not the mutual esteem of the summate underground. The compensate overture should be to discovery the mutual of 1/R T (1/2Ω ) which is R T (2Ω).

Therefore the rightfulness evaluate of R T = 2Ω.

E’er commemorate to obtain the mutual of 1/R T to get R T.

2. Intermixture Up the Summation of Capacitors With the Add-on of Resistors

This is one of the concepts that takes a spell to penetrate for every tiro perusing some electricity. Delight note of the pursuit equations

Adding capacitors in analog: C T = C 1 + C 2 + C 3 +……..

Adding capacitors serial: 1/C T = 1/C 1 + 1/C 2 + 1/C 3 +…………

Adding resistors serial: R T = R 1 + R 2 + R 3 +……..

Adding resistors in analog: 1/R T = 1/R 1 + 1/R 2 + 1/R 3 +…….

So the process for the increase of capacitors in collimate is the like as the process for the plus of resistors nonparallel. Likewise, the operation for the edubirdie reviews improver of capacitors serial is the like as the process for the increase of resistors in twin. This can be rattling puzzling at kickoff but with sentence you would get victimised thereto. So let’s take the plebeian misunderstanding students shuffle with the accession of capacitors by analyzing this enquiry.

Opine we birth two capacitors of condenser 3F and 6F affiliated in duplicate and we are asked to get the add capacitor. Approximately students would not takings the metre to canvas the doubtfulness and would take they are dealings with resistors. Hither is how such students would resolve this motion:

1/C T = 1/C 1 + 1/C 2 where C T = amount capacitor, C 1 = 3F and C 2 = 6F

1/C T = 1/3 + 1/6 = 1/2 which implies that C T = 2F; this is perfectly faulty

The rightfulness routine is plainly C T = 3F + 6F = 9F and thus 9F is the adjust resolution

Charge should too be interpreted when minded a enquiry which has capacitors attached nonparallel. Think we sustain two capacitors of values 20F and 30F machine-accessible nonparallel. Delight do not pee this fault:

C T = 20F + 30F = 50F, this is improper

The redress process is:

1/C T = 1/20 + 1/30 = 1/12; C T = 12F, this is the compensate solvent.

3. Adding Compeer Emf Sources Attached in Analog

Low and world-class, you can solitary position emf sources in collimate if they deliver the like emf. The main grounds or reward for combination emf sources in analog is to step-up the stream yield supra that of any i germ. When in analog, the add flow produced by the combined germ is capable the sum of currents of apiece single germ, all spell maintaining the archetype potential.

Approximately students shuffling the error of adding equate potential sources machine-accessible in twin as if they were attached nonparallel. It is significant to line that if we had a trillion potential sources, all of match voltages and were all affiliated in analogue; the aggregate emf would be capable the potential of equitable one emf beginning. Let us deal an exemplar.

Think we deliver ternary equate potential sources, V 1 =12V, V 2 =12V, V 3 =12V which are all machine-accessible in duplicate and we are asked to shape the totality emf. Roughly students would approach resolution this inquiry ilk this:

V T = V 1 + V 2 + V 3 where V T is the summate emf

V T = 12V + 12V + 12V = 36V; V T = 36V, which is completely damage

Hold in judgment that the supra root would sustain been right if the emf sources were attached serial.

Decently to resolution this doubt is to recognize the fact that since they are peer voltages which are all machine-accessible in twin, the summate emf would be capable the emf of lone one of the emf sources. So the root is V T = V 1 = V 2 = V 3 = 12V.

4. Reasoning Induction Is the Like as Inducive Reactance and That Capacitor Is the Like as Capacitive Reactance

Students ordinarily flip these price lots in calculations. Kickoff, let us view the deviation betwixt inductor and inducive reactance. Induction is a amount that describes a place of a tour component. It is the holding of an electric director by which a modify in flow flow done it induces an voltage in both the director itself and in any nearby conductors by common induction. Inducive reactance, on the otc paw, is the force of that induction at a presumption oftenness. It is an confrontation to a commute in flow.

The higher the inducive reactance, the bigger the impedance to a commute in flow. A rattling obvious divergence ‘tween these two footing can besides be seen in their units. The whole of induction is H(H) whereas that of inducive reactance is Ohm(Ω). Now that we bear a open intellect of the divergence ‘tween these two footing, let us issue a take an illustration.

Conjecture we let an AC-circuit that has a emf reference of emf 10V and frequence 60Hz which is affiliated nonparallel with an inductance of induction 1H. We are so asked to mold the flow done edubirdie reviews this tour. Approximately students would pee-pee the fault of winning inductor to be inducive reactance and clear the motion same this:

According to Ohm’s law V = IR where V = potential, I = flow and R = opposition

V = 10V R = 1H; I = V/R; I = 10/1; I = 10A; which is damage.

We kickoff demand to commute inductor (H) to inducive reactance (Ω) so lick for the flow. The rightfulness answer is:

X L = 2πfL where X L = inducive reactance f= oftenness, L = inductor

X L = 2×3.142×60×1 = 377Ω; I = V/X L ; I = 10/377; I = 0.027A, which is castigate.

The like safeguard should besides be interpreted when transaction with capacity and capacitive reactance. Capacitor is the belongings of the capacitance in a granted AC-circuit whereas capacitive reactance is the opponent to the modify of potential crosswise an constituent and is reciprocally relative to the capacitor and oftenness. The whole of capacity is the f (F) and that of capacitive reactance is Ohm (Ω).

When you are asked to reckon the flow done an AC-circuit which consists of a emf reference attached nonparallel with a condenser, do not use the capacity of the capacitance as the underground. Instead, outset convince the capacity of the condenser to capacitive reactance then use it resolve for the stream.

5. Interchanging the Turns Proportion of a Transformer

A transformer is a gimmick that is victimised to increase or diminution voltages and it does this by the precept of electromagnetic trigger. The turns proportion of a transformer is outlined as the bit of turns on its petty dual-lane by the figure of turns on its master. The emf proportion of an nonpareil transformer is direct related the turns proportion: V S /V P = N S /N P .

The flow proportion of an nonsuch transformer is reciprocally related the turns proportion: I P /I S = N S /N P . Where V S = junior-grade potential, I S = petty flow, V P = basal potential, I P = basal stream, N S = bit of turns in the secondary and N P = numeral of turns in the primary. Students can sometimes get garbled and alternate the turns proportion. Let us take an instance to exemplify this.

Conjecture we deliver a transformer with the bit of turns in the primary organism 200 and the issue of turns in the secondary beingness 50. It has a basal emf of 120V and we are asked to cipher the petty emf. It is really commons for students to confuse the turns proportion and clear the doubtfulness alike this:

V S /V P = N P /N S ; V S /120 = 200/50; V S = (200/50)×120; V S = 480V, which is faulty.

Constantly dungeon in brain that the emf proportion of an idealistic transformer is direct related it turns proportion. Properly to clear the inquiry would thence be:

V S /V P = N S /N P ; V S /120 = 50/200; V S = (50/200)×120; V S = 30V, which is the rectify solvent.

Likewise, the stream proportion of an nonsuch transformer is reciprocally related its turns proportion and it is identical authoritative you observe of this when resolution questions. It is identical green for students to use this par: I P /I S = N P /N S . This equality should be all avoided.

© 2016 Charles Nuamah



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