Electronics Information Practice Test 1

Answer: D

Correct answer: Direct current. In direct current, charge flows steadily in one direction, from the negative terminal through the circuit to the positive in terms of electron flow. Batteries, solar cells and the output of phone chargers supply DC. Its counterpart is AC, alternating current, which reverses direction many times per second, as in household outlets that alternate 60 times per second in the U.S. Recognizing the pair DC and AC as direct and alternating current covers most abbreviations on test questions.

• A: This is not the meaning of DC.
• B: Charge distribution does not expand DC.
• C: This is not the electrical current classification.
• D (correct): DC is current flowing in one direction.

Answer: C

Correct answer: Coulomb. The coulomb counts a quantity of charge, roughly the charge of about 6.24 × 10¹⁸ electrons. It is easy to confuse with the ampere, but an ampere is a rate, one coulomb per second, just as a gallon is an amount and gallons per minute is a flow. Volts describe energy per coulomb, and watts describe energy per second, so neither measures charge itself.

• A: Watts measure power.
• B: Volts measure potential difference.
• C (correct): Coulombs measure charge. Additional check: A coulomb is an amount of electric charge, while an ampere is the rate of transporting that amount.
• D: Amperes measure charge rate.

Answer: C

Correct answer: 60 Hz. The secondary voltage is induced by core flux that changes at exactly the rate set by the primary. A transformer changes voltage level, not frequency, so a 60 Hz input gives a 60 Hz output.

• A: An output of 0 Hz would be DC, and a transformer cannot turn AC into DC.
• B: Stepping the voltage down does not divide the frequency; the secondary follows the primary flux exactly.
• C (correct): The core flux rises and falls at the primary's rate, so the induced secondary voltage alternates at the same 60 Hz.
• D: A transformer does not double frequency; changing frequency needs other equipment.

Answer: D

Correct answer: Electric charge. An ampere-hour equals one ampere of current flowing for one hour, so it measures a quantity of charge. It becomes energy only after multiplying by the battery voltage.

• A: Power is a rate measured in watts, not a quantity measured in ampere-hours.
• B: Potential, or voltage, is measured in volts and is a separate value on the label.
• C: Energy needs the voltage too; volts times ampere-hours gives watt-hours of energy.
• D (correct): An ampere-hour is one ampere flowing for one hour, which is an amount of charge.

Answer: A

Correct answer: 5 V. Potential difference is energy transferred per unit charge, so

Dividing 15 J by 3 C gives 5 J/C, and one joule per coulomb is one volt. The unit labels on the other choices are a clue: watts, amperes, and coulombs are not units of potential difference. A good habit is to check that the unit of your answer matches the quantity asked for.

• A (correct): V = 15 J/3 C = 5 $\frac{J}{C}$ = 5 V.
• B: The reciprocal is not potential difference.
• C: 45 is a product with an incorrect unit.
• D: Subtraction yields no valid charge answer.

Answer: A

Correct answer: 20 mA. The Zener holds its node at 6.2 V, so the resistor has the rest of the supply across it.

With no load, all of that current flows through the Zener.

• A (correct): The resistor carries 15 V minus 6.2 V, or 8.8 V, giving 8.8 V divided by 440 Ω, which is 20 mA, all of it through the Zener.
• B: This divides the full 15 V by the resistor and ignores the voltage held by the Zener.
• C: This adds the Zener voltage to the supply instead of subtracting it.
• D: This divides the Zener voltage by the resistor instead of the voltage left across the resistor.

Answer: B

Correct answer: 120 Hz. A bridge rectifier uses both half-cycles and sends each through the load in the same direction. That gives two humps for every input cycle, so the output pulses at twice 60 Hz, or 120 Hz.

• A: This would be true of a half-wave rectifier, which gives one pulse per input cycle.
• B (correct): A bridge turns both half-cycles into same-polarity humps, giving two pulses per input cycle, or 120 Hz.
• C: This halves the frequency; a bridge produces more pulses per cycle, not fewer.
• D: A rectifier does not turn the input into a constant 0 Hz signal; its output still pulses.

Answer: B

Correct answer: 5 A. Each branch has the full 30 V. The 10 Ω branch carries 30 divided by 10, or 3 A, and the 15 Ω branch carries 30 divided by 15, or 2 A. The source supplies their sum, 5 A.

• A: Dividing 30 V by 25 Ω treats the two resistors as if they were in series.
• B (correct): The branches carry 3 A and 2 A, and those add to 5 A at the source.
• C: 2 A is only the current in the 15 Ω branch; the source must supply both branches.
• D: 3 A is only the current in the 10 Ω branch; the 15 Ω branch adds its own current.

Answer: A

Correct answer: A PNP transistor. On a bipolar transistor, the arrow sits on the emitter. If it points out of the symbol, the transistor is NPN; if it points in, it is PNP.

• A (correct): An emitter arrow pointing in identifies a PNP bipolar transistor.
• B: NPN emitter arrows point out of the symbol: not pointing in.
• C: A photodiode is a triangle and bar with arrows pointing in, and it has only two leads.
• D: Field-effect transistors are drawn with gate, drain and source rather than base, collector and emitter.

Answer: D

Correct answer: 6 A. For two branches, the current in one is the total times the other branch's resistance over the sum: 9 A times 12/18 gives 6 A. Check: the 12 Ω branch gets 3 A, and both branches have 36 V across them.

• A: Splitting the current equally ignores that the smaller resistance carries more current.
• B: 3 A is the current in the 12 Ω branch; the ratio was applied backward.
• C: The full 9 A is the total; part of it goes through the 12 Ω branch.
• D (correct): The currents split in the inverse ratio 12 to 6, or 2 to 1, so the 6 Ω branch gets two thirds of 9 A.

Answer: B

Correct answer: A short circuit: very low resistance, so current can become very large. A short circuit is an unintended path with very low resistance that bypasses the load. Because R is so small, current becomes very large until a fuse or breaker opens. Open means no path; short means too easy a path.

• A: This mislabels the fault and wrongly says the bypassed appliance gets more current.
• B (correct): An unintended low-resistance path around the load is a short circuit, and the small resistance allows a large, dangerous current.
• C: An open circuit is a break in the path; here the conductors touch, creating too easy a path instead.
• D: A short has very low resistance, not high, and the current rises rather than falls.

Answer: C

Correct answer: The second stores nine times as much. Because energy is E = ½CV², energy grows with the square of voltage. With C the same for both, only the voltage changes, so the energy ratio is the voltage ratio squared:

The tempting answer is three times, which assumes energy grows in direct proportion to voltage. A quick check: doubling the voltage would quadruple the energy. This square law is why high-voltage capacitors are especially dangerous.

• A: Equal capacitance does not imply equal energy at different voltages.
• B: Increasing voltage raises energy in this model.
• C (correct): E = 1/2 CV² and equal C mean energy ratio is the voltage ratio squared: 3² = 9.
• D: This ignores the square on voltage.

Answer: A

Correct answer: 15 V. Add the resistances, find the current, then apply Ohm's law to the 10 Ω resistor.

Check: 9 V + 15 V + 24 V = 48 V.

• A (correct): The total is 32 Ω, so I = 1.5 A, and 1.5 A × 10 Ω = 15 V.
• B: This splits 48 V evenly three ways, which only works for identical resistors.
• C: This finds current from the full 48 V on the 6 Ω resistor alone; no drop can exceed the source.
• D: This is the drop across the 16 Ω resistor, not the 10 Ω resistor.

Answer: D

Correct answer: 8 V. The change in field is

T, so

Wb per turn.

• A: This finds the rate of flux change per turn but forgets to multiply by 40 turns.
• B: This uses the final field of 0.1 T as the change instead of the 0.4 T drop.
• C: This uses the starting field of 0.5 T as the change instead of the 0.4 T drop.
• D (correct): The flux change per turn is 0.4 times 0.03, or 0.012 Wb, and 40 times 0.012 divided by 0.06 is 8 V.

Answer: C

Correct answer: More than 0.25 A, because the cooler filament has lower resistance. A tungsten filament's resistance grows as it heats. At 60 V the filament is cooler, so its resistance is below the hot value of

With less resistance, 60 V drives more than

so the lamp is nonohmic and simple proportion does not apply.

• A: Scaling with voltage holds only for constant resistance, and a filament's resistance changes with temperature.
• B: A tungsten filament's resistance falls as it cools, so the current is not reduced below the estimate.
• C (correct): At 60 V the filament runs cooler, so its resistance is below its hot value and more than half the current flows.
• D: A lamp does not force a fixed current; its current still depends on the voltage applied.