Answers and Explanations

Electronics Information Practice Set 1

A

To solve this problem, the first step is to convert 60 mV to V. You can quickly make this conversion by moving the decimal point three places to the left, which yields 0.060 V. Next, convert 15 KΩ to Ω by moving the decimal point three places to the right, which yields 15,000. Since current is the unknown, the formula needed to finish the calculation is I = E ÷ R. Dividing 0.060 by 15,000 gives 0.000004 amperes. This can be converted to 0.004 mA.
D

Like the last problem, this one requires you to first convert the given measurements. Take the resistance and move the decimal. 60 KΩ becomes 60,000 Ω. Using our formula for calculating voltage, E = I × R, we get 30 A × 60,000 = 1,800,000 V, or 1.8 MV.
B

The resistors in a series circuit add up, so one 40-ohm resistor in series with a 20-ohm resistor is equal to a 60-ohm resistor. To find current flow through a circuit, divide the total circuit resistance (60 ohms) by the voltage across the circuit (60 volts) to get 1 amp.
D

The valence shell is the outer shell of an atom, which in a copper atom is the fourth shell. A copper atom has 29 electrons, so it would fill the first three shells completely and have one electron left to begin a fourth shell.
B

In the symbol, the number 2 indicates the collector.
C

The ampere is the basic unit of current, so current can be expressed in milliamperes.
B

A closed circuit has continuity and will allow current to flow in it.
C

The choices represent four different capacitor symbols: a fixed capacitor (A), variable capacitor (B), polarized capacitor (C), and trimmer capacitor (D).
B

When a P-type material is joined with an N-type material, a diode is created.
C

Direct current (DC) means that current only flows one way in a conductor. This is the type of current in circuits powered by batteries.
D

Examine the diagram carefully. Notice that switches S1, S3, and S4 all can direct current flow to either of two alternative paths. Switch S2 will bridge the current from one of the top parallel paths to the other. The circuit pictured is currently open. The S1 and S3 two-switches connect to the same parallel paths, but presently, each is connected to a different branch. If either of these switches were to flip, a complete circuit would be made, passing through one or the other of these paths. S2 is a simple on/off switch, and if closed, it too would close the circuit. However, S4, a two-switch, cannot affect the current break in the circuit.
A

This question has no numbers, which can make it tougher to grasp. It can be reasoned out if your understanding of parallel and series resistors is very strong. Otherwise, your best bet in a situation like this might be to just choose some simple numbers and do the math. The question has three identical resistors in a series-parallel circuit. Choose an easy number—three 1 Ω resistors would work just fine, and you can determine the correct answer choice by seeing if the effective resistance is greater than, less than, or equal to 1 Ω. The effective resistance of two 1 Ω resistors in parallel is Ω. Add to this an additional 1 Ω resistor in series, and the final effective resistance is Ω, an increase. Time permitting, the calculation can be repeated with a different resistance value for the three resistors, just to ensure that the resistance still increases.
C

Each answer choice would increase the rate of current flow, except one. Answer choices (A) and (D) would both decrease the overall resistance of the circuit, which, according to Ohm’s law, would increase current. Answer choice (B) would also increase current flow if resistance was not changed. However, for a direct current, an inductor in the circuit has little effect. It will resist changes in current flow but not tend to either decrease or increase it.
B

The effective resistance of several resistors wired in parallel can be determined by taking the reciprocal of the sum of the reciprocals of each individual resistance value. Since each resistor has a resistance of 2 Ω, you get + + = . Since the reciprocal of this sum is needed in parallel resistor calculations, it’s best to stick with fractions at least until the end. The reciprocal of is simply the fraction created by switching the numerator and denominator values, that is, Ω.
D

Copper isn’t the best conductor. Silver has a lower resistance and is thus better at conducting electricity than copper is. But though silver is a somewhat better conductor, it is far, far more expensive. Copper is a very good (but not the best) conductor, while also being very affordable. Aluminum is inexpensive but is a worse conductor than copper. Copper is also safer than aluminum, which, due to some of its heating and expanding properties, is more likely to cause electrical fires in household wiring setups.
B

Rearranging the power formula, I = P ÷ V = 10 ÷ 6 = 1.67 A (rounded to the nearest hundredth).

ELECTRONICS INFORMATION PRACTICE SET 2

D

The relevant formula here is P = I × V, and both the current and voltage are given, so P = 30 × 2.5 = 75 W.
A

The relevant formula here, V = I × R, needs to be rearranged in order to calculate resistance. R = V ÷ I = 120 ÷ 15 = 8 Ω.
C

These two resistors are in parallel and can be treated as a unit. Since we’re given the voltage drop across them both (the voltage drop either in total or across either resistor is the same when they’re wired in parallel), as well as the total current through them both, we can calculate the total effective resistance of the two resistors. R = V ÷ I = 18 ÷ 6 = 3 Ω. The question asks for the resistance of either resistor, though, so there’s one more step. The formula for the total effective resistance of these two resistors in parallel is . However, since we know both resistors are the same, and we know the total resistance, the formula can be written as . R is the resistance of a single resistor and is exactly what we’re trying to figure out. Multiply both sides of the equation by R to get , or . Then multiply both sides by 3 to get R = 6 Ω.

Alternatively, you might recall that in the special case of multiple identical resistors in parallel, you can use the shortcut of dividing the individual resistance by the total number of resistors. In order to have a total effective resistance of 3 Ω then, there must be two 6 Ω resistors (6 ÷ 2 = 3).
D

Household wiring and long-distance high-voltage power lines are connected via step-down transformers in the local substations. The other statements are not true. The current is AC in both power lines and household wiring, and power lines do not always use aluminum wiring.
B

The first and second color determine digits, and the last one determines how many zeroes to add. Black is zero, then orange (in either of the first two positions) is three, and then orange (in the final position) is three zeroes. A zero at the beginning of the number has no effect, as 03000 is just written as 3,000. Answer choice (C) would have been correct if the pattern were orange, black, and then orange.
A

Inductors resist changes in current and for this reason act as resistors in alternating current circuits. However, this property is also important in a direct current circuit as seen here. An open circuit has no current, but the moment the switch is closed, current will begin to flow. The inductor, resisting the sudden change in voltage, will resist current flow at first. The lamp, since it is parallel to the resisting inductor, will light up immediately without difficulty. However, the inductor will soon “get used” to the current flow, and the resistance of its path will drop significantly. As more current flows through the easy inductor path, the total current (and therefore energy use) of the circuit will increase significantly as the overall resistance drops. (D) is exactly the opposite of what will happen. Something similar to (B) could happen only if the voltage source were limited in the amount of current it could provide and the small fraction of current passing through the lamp were not enough to light the lamp once the inductor dropped its resistance, or if the total current draw caused a fuse or breaker to cut off the current. However, in neither of these cases would the lamp itself “burn out.”
A

The hair dryer has the same resistance no matter what circuit it’s attached to. You can calculate the resistance from the specifications given and then use that same resistance to determine current under a lower voltage. R = V ÷ I = 220 ÷ 10 = 22 ohms. Plugging this 22-ohm hair dryer into an American socket gives you the current as I = V ÷ R = 110 ÷ 22 = 5 amps, which is answer choice (A).

Another way to answer this question is by reasoning it out from the proportionality given by Ohm’s law: V = I × R. Given a constant resistance, voltage and current are constant, so if the voltage is halved from 220 to 110 volts, the current must also be halved from 10 to 5 amps.
C

Before making any calculations, think about whether the resistor should be wired in series or parallel if the total current draw is to drop. Adding a resistor in series increases the total resistance of the circuit; increasing the resistance makes it harder for current to flow and causes the number of amps to drop. Adding a resistor in parallel, however, always lowers resistance, because it provides an additional path for current to take. So, we’re looking for a series add, and the two options are 0.5 ohms or 1 ohm. Recalculate the current given 10 volts and a higher resistance of 4 + 0.5 = 4.5 ohms. I = V ÷ R = 10 ÷ 4.5 = 2.2 amps, which doesn’t fit with the information in the question. With 4 + 1 = 5 ohms, I = V ÷ R = 10 ÷ 5 = 2 amps, which matches. So the additional resistor is 1 ohm, and it must be placed in series.
B

This question about power takes you back to the more general definition of power, as the rate of energy flow over time. You do not need the electrical power formula, P = I × V. Instead, divide the total energy, in joules, by the total time, in seconds, to determine the total power, in watts. The information in the question is already given in the correct units, so calculate 800,000 ÷ 400 = 2,000 W.
D

You can dismiss (A) and (B), since for loads in parallel, the voltage drop is the same across all paths; the brightness of the light would not change if the lamp were wired in parallel with either a capacitor or an inductor. Loads in series, on the other hand, affect the overall resistance of the circuit and therefore the total power draw on the battery. Recall that capacitors and inductors can either let current flow freely or act as resistors, depending on the type of current. In an AC circuit, a capacitor lets current flow freely while an inductor resists current flow. An inductor placed in series with the lamp will therefore act as a resistor over which a voltage drop occurs, and the lamp will be correspondingly dimmer than if the voltage drop of the circuit occurred entirely over the lamp itself.
D

Before the switch is closed, there is only one path through the circuit, through L₂, so it is lit while the other lamp is dark. Closing the switch will provide a second possible path, and as a result, both lamps will now be lit. However, it’s also necessary to determine what will happen to the brightness of L₂ itself. The general rule is that lights in parallel will not change their brightness as more paths are added, because the voltage drop across each path remains the same and therefore the amount of current passing through each lamp also stays the same. However, the fact that there are now two lamps brightly lit instead of one does mean that the battery will drain twice as fast because the total power draw of the circuit will be doubled when the switch closes.
B

I = V ÷ R, and since the voltage across parallel loads is always equal, any difference in current will be a result of the different resistances. You can see that current and resistance are inversely proportional, meaning any change to one will have the opposite effect on the other. But this question may be a little easier to figure out by just making up a value for voltage, say 5 V. Given this value, then, the currents through each load will be I = 5 ÷ 5 = 1 A, and I = 5 ÷ 2.5 = 2 A. The 2.5 Ω resistor has twice as much current as does the 5.0 Ω resistor.
C

Since the device’s resistance is constant, the formula V = I × R shows that if voltage increases, current increases proportionally. To put it another way, the greater the voltage, the greater the “pressure” pushing those electrons forward. This has the effect of producing a higher current, assuming the resistance is the same. Only answer choice (C) mentions the higher current that must result from this situation. And yes, you should definitely only plug your device into a socket that fits so you don’t burn it out or start a fire.
B

This is the general symbol for a capacitor, which acts as a resistor to direct current but does little to disrupt alternating current. It’s the opposite of an inductor, pictured in answer choice (A), which acts as a resistor to alternating current. Choice (C) is the diode symbol, which allows current flow in only one direction. Since alternating current constantly changes direction, half of the current flow would be disrupted by the process known as rectification. Finally, choice (D) displays an open switch, which will not allow either AC or DC power to flow.
D

The total current in the circuit can be determined from the voltage source and total resistance of the circuit. This is a tricky series-parallel circuit that has to be taken in two steps. The total resistance of the parallel loads L₁ and L₂ can be calculated without the full parallel-resistors formula by using the rule that equal resistors in parallel have an effective resistance equal to their individual resistance divided by the number of parallel loads. The effective resistance of L₁₂ then is 40 ÷ 2 = 20 Ω. This is in series with R, so the total resistance of the whole circuit is 20 + 10 = 30 Ω. Now, calculate the current draw of the circuit as I = V ÷ R = 30 ÷ 30 = 1 A. Once again, the fact that L₁ and L₂ have an equal amount of resistance makes things easier. Since the two paths are equivalent, each will take half the current, meaning 0.5 A passes through either one. This is greater than the maximum current allowed by the fuse, so it will burn out, cutting the current flow through L₁ and leaving only L₂ lit.
A

Current will separate when reaching parallel paths. Since the two resistors are different, they’ll draw different amounts of current, so the statement in (B) is false. However, regardless of whether they’re the same or different, it’s the sum of the two currents that will give you the total current. This technician would have been better off placing her ammeter just after the battery or perhaps near the switch, however, as doing so would have allowed her to get the total current with a single measurement.

Review and Reflect

Look back over your work on these practice sets. If you struggled with some of the questions, ask yourself why.

Did you misunderstand a formula or simply make errors in calculations? If it’s the latter, be sure to slow down and double-check your math when you do EIectronics Information questions. If you need to review the formulas discussed in this chapter, consider making flash cards to help you remember them.

Did you have trouble remembering some of the technical terms introduced in this chapter? If so, consider making flash cards to help you memorize the meaning of those terms.

Which areas do you most need to review?