# 5 Silly Mistakes Beginners in Electricity and Magnetism Make

*Charles is an electrical engineer who always aims to stay up to date with the latest technology trends.*

You have spent a week studying hard for this particular paper. You go into the examination room very confident and write the paper to the best of your ability. You are very hopeful of scoring nothing less than an "A". The exam result finally arrives and you have a "C". You are furious and probably think your professor marked you down because you missed three of his classes during the term. You approach your professor and ask to see your exam sheet only to realize you made silly mistakes. These mistakes cost you a lot of marks and hindered your chance to get the "A" you worked all week for.

This is a very common occurrence among students which I believe can be easily avoided. Teachers should make students aware of the possible areas where they are likely to make these errors, so they don't repeat them during examinations. Below are some of the most common mistakes students make in their electricity and magnetism tests.

## 1. Adding Resistors in Parallel

If you ask a number of students to add resistors with given values in parallel, it's likely you would get different answers from the students. It is one of the most common mistakes made in the field of electricity and is due to a simple oversight. So let's break it down.

Suppose you have two resistors of values 6Ω and 3Ω connected in parallel. You are then asked to calculate the total resistance. Most students would solve the question the right way but would only miss the answer at the last step. Let's solve the question together.

1/R_{T} = 1/R_{1} + 1/R_{2} where R_{T} = total resistance, R_{1} = 6Ω and R_{2} = 3Ω

1/R_{T} = 1/6 + 1/3 = 9/18 = 1/2Ω

Some students would leave their answer as 1/2Ω or 0.5Ω which is wrong. You were asked to find the value of the total resistance and not the reciprocal value of the total resistance. The right approach should be to find the reciprocal of 1/R_{T} (1/2Ω ) which is R_{T }(2Ω).

Hence the right value of R_{T} = 2Ω.** **

**Always remember to find the reciprocal of 1/R _{T }to get R_{T.}**

## 2. Mixing Up the Addition of Capacitors With the Addition of Resistors

This is one of the concepts that takes a while to sink in for every beginner studying about electricity. Please take note of the following equations

Adding capacitors in parallel: C_{T =} C_{1} + C_{2} + C_{3} +........

Adding capacitors in series: 1/C_{T} = 1/C_{1} + 1/C_{2} + 1/C_{3} +............

Adding resistors in series: R_{T} = R_{1} + R_{2} + R_{3} +........

Adding resistors in parallel: 1/R_{T} = 1/R_{1} + 1/R_{2} + 1/R_{3} +.......

Hence the procedure for the addition of capacitors in parallel is the same as the procedure for the addition of resistors in series. Also, the procedure for the addition of capacitors in series is the same as the procedure for the addition of resistors in parallel. This can be really confusing at first but with time you would get used to it. So let's look at the common mistake students make with the addition of capacitors by analyzing this question.

Suppose we have two capacitors of capacitance 3F and 6F connected in parallel and we are asked to find the total capacitance. Some students would not take the time to analyze the question and would assume they are dealing with resistors. Here is how such students would solve this question:

1/C_{T} = 1/C_{1} + 1/C_{2} where C_{T} = total capacitance, C_{1 =} 3F and C_{2} = 6F

1/C_{T} = 1/3 + 1/6 = 1/2 which implies that **C**_{T}** = 2F; this is absolutely wrong**

The right procedure is simply C_{T} = 3F + 6F = 9F and hence **9F is the correct answer**

Care should also be taken when given a question which has capacitors connected in series. Suppose we have two capacitors of values 20F and 30F connected in series. Please do not make this mistake:

C_{T} = 20F + 30F = 50F, **this is wrong**

The right procedure is:

1/C_{T} = 1/20 + 1/30 = 1/12; C_{T} = 12F, **this is the right answer.**

## 3. Adding Equal Voltage Sources Connected in Parallel

First and foremost, you can only place voltage sources in parallel if they have the same voltage. The primary reason or advantage for combining voltage sources in parallel is to increase the current output above that of any single source. When in parallel, the total current produced by the combined source is equal to the sum of currents of each individual source, all while maintaining the original voltage.

Some students make the mistake of adding equal voltage sources connected in parallel as if they were connected in series. It is important to note that if we had a million voltage sources, all of equal voltages and were all connected in parallel; the total voltage would be equal to the voltage of just one voltage source. Let us look at an example.

Suppose we have three equal voltage sources, V_{1} =12V, V_{2}=12V, V_{3}=12V which are all connected in parallel and we are asked to determine the total voltage. Some students would go about solving this question like this:

V_{T} = V_{1 }+ V_{2} + V_{3} where V_{T} is the total voltage

V_{T} = 12V + 12V + 12V = 36V; **V**_{T}** = 36V, which is totally wrong**

Keep in mind that the above solution would have been correct if the voltage sources were connected in series.

The right way to solving this question is to realize the fact that since they are equal voltages which are all connected in parallel, the total voltage would be equal to the voltage of only one of the voltage sources. Hence the solution is **V**_{T}** = V**_{1}** = V**_{2}** = V**_{3}** = 12V.**

## 4. Thinking Inductance Is the Same as Inductive Reactance and That Capacitance Is the Same as Capacitive Reactance

Students usually interchange these terms a lot in calculations. First, let us consider the difference between inductance and inductive reactance. Inductance is a quantity that describes a property of a circuit element. It is the property of an electrical conductor by which a change in current flowing through it induces an electromotive force in both the conductor itself and in any nearby conductors by mutual inductance. Inductive reactance, on the other hand, is the effect of that inductance at a given frequency. It is an opposition to a change in current.

The higher the inductive reactance, the larger the resistance to a change in current. A very obvious difference between these two terms can also be seen in their units. The unit of inductance is Henry(H) whereas that of inductive reactance is Ohm(Ω). Now that we have a clear understanding of the difference between these two terms, let us take a look at an example.

Suppose we have an AC-circuit that has a voltage source of voltage 10V and frequency 60Hz which is connected in series with an inductor of inductance 1H. We are then asked to determine the current through this circuit. Some students would make the mistake of taking inductance to be inductive reactance and solve the question like this:

According to Ohm's law V = IR where V = voltage, I = current and R = resistance

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

We first need to convert inductance (H) to inductive reactance (Ω) and then solve for the current. The right solution is:

X_{L} = 2πfL where X_{L} = inductive reactance f= frequency, L = inductance

X_{L} = 2×3.142×60×1 = 377Ω; I = V/X_{L}; I = 10/377; **I = 0.027A, which is correct.**

The same precaution should also be taken when dealing with capacitance and capacitive reactance. Capacitance is the property of the capacitor in a given AC-circuit whereas capacitive reactance is the opposition to the change of voltage across an element and is inversely proportional to the capacitance and frequency. The unit of capacitance is the farad (F) and that of capacitive reactance is Ohm (Ω).

When you are asked to calculate the current through an AC-circuit which consists of a voltage source connected in series with a capacitor, do not use the capacitance of the capacitor as the resistance. Rather, first convert the capacitance of the capacitor to capacitive reactance and then use it solve for the current.

## 5. Interchanging the Turns Ratio of a Transformer

A transformer is a device that is used to step-up or step-down voltages and it does this by the principle of electromagnetic induction. The turns ratio of a transformer is defined as the number of turns on its secondary divided by the number of turns on its primary. The voltage ratio of an ideal transformer is directly related to the turns ratio: V_{S}/V_{P} = N_{S}/N_{P}.

The current ratio of an ideal transformer is inversely related to the turns ratio: I_{P}/I_{S} = N_{S}/N_{P}. Where V_{S} = secondary voltage, I_{S} = secondary current, V_{P} = primary voltage, I_{P} = primary current, N_{S} = number of turns in the secondary winding and N_{P} = number of turns in the primary winding. Students can sometimes get confused and interchange the turns ratio. Let us look at an example to illustrate this.

Suppose we have a transformer with the number of turns in the primary winding being 200 and the number of turns in the secondary winding being 50. It has a primary voltage of 120V and we are asked to calculate the secondary voltage. It is very common for students to mix up the turns ratio and solve the question like this:

V_{S}/V_{P = }N_{P}/N_{S}; V_{S}/120 = 200/50; V_{S} = (200/50)×120;** V**_{S}** = 480V, which is incorrect.**

Always keep in mind that the voltage ratio of an ideal transformer is directly related to it turns ratio. The right way to solve the question would therefore 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 right answer.**

Also, the current ratio of an ideal transformer is inversely related to its turns ratio and it is very important you take note of this when solving questions. It is very common for students to use this equation:** I**_{P}**/I**_{S}** = N**_{P}**/N**_{S}**. This equation should be totally avoided.**

**© 2016 Charles Nuamah**