|Electronics Demystified, 2nd edition|
|Explanations for Quiz Answers in Chapter 1|
|1. First, let's determine the net resistance R2 of the parallel
combination at the upper right. We can use the formula for resistances in parallel to
R2 = 1 / (1/12 + 1/18)
This resistance appears in series with the 4.8-ohm resistance. The total resistance in the circuit is therefore 4.8 + 7.2 ohms, or 12 ohms. The battery supplies 12 V so Ohm's law tells us that the current I drawn from the battery equals the battery voltage divided by the net resistance. Therefore
I = 12/12
This current passes entirely through the 4.8-ohm resistor, so the correct choice is B.
|2. The current through the parallel combination of resistors (12 ohms and 18 ohms) is the same as the current through the 4.8-ohm resistor, and the same as the total current drawn from the battery: 1.0 A. The correct choice is B.|
|3. The voltage across the 4.8-ohm resistor (let's call it E1) equals
the product of the current through it and its value in ohms. That's
= 1.0 x 4.8
The correct choice is C.
|4. The voltage across the 12-ohm resistor equals the voltage across the parallel combination (12 ohms and 18 ohms), because those two resistors are connected in parallel. We can use Kirchhoff's Voltage Law to deduce that potential difference by subtracting E1 from the battery voltage. That calculation gives us 12 - 4.8, or 7.2 V. The correct choice is A.|
|5. The voltage across the 18-ohm resistor equals the voltage across the 12-ohm resistor, because those two resistors are connected in parallel. We've found that voltage as 7.2 V. We can therefore use Ohm's Law to calculate the current through the 18-ohm resistor as 7.2 / 18, or 0.40 A. The correct choice is D.|
|6. When we solved Question 1, we found that the 4.8-ohm resistor carries 1.0 A of
current. We can therefore calculate the power P, in watts, as the product of the
voltage and the current, getting
P = 4.8 x 1.0
We can also calculate the power as the square of the current times the resistance, or
P = 1.02 x 4.8
There's a third way, too. When we solved Question 3, we found that 4.8 V appears across the 4.8-ohm resistor. We can square that voltage, and then divide by the resistance to get
P = 4.82 / 4.8
The correct answer is D.
|7. When we solved Question 5, we determined that the 18-ohm resistor carries 0.40 A of
current. We can therefore calculate the power P, in watts, as
= 0.402 x 18
The correct choice is A.
|8. Assuming that the wire loop in the system of Fig. 1-12 conducts current perfectly
(that is, it has no resistance), we can calculate the current I through the loop
by dividing the battery voltage (18 V) by the resistor value (2.4 ohms) to get
I = 18 / 2.4
The loop has one turn, so it produces 7.5 ampere-turns (7.5 At) of magnetomotive force, but our question contains choices that all appear in gilberts (Gb). To convert ampere-turns to gilberts, we multiply by 1.257. This arithmetic gives us 7.5 x 1.257, or 9.4275 Gb, which rounds to 9.428 Gb. The correct choice is D.
|9. If we double the loop diameter but don't add any turns to it, we won't change the magnetomotive force at all. Remember, the magnetomotive force produced by a loop or coil depends only on the current through it and the number of turns it contains. The loop or coil diameter makes no difference! The correct answer is D.|
|10. If we rearrange the single-turn loop into a four-turn coil but we don't change the current through the conductor, we get four times as much magnetomotive force as we had before. Magnetomotive force is directly proportional to the current through a loop or coil, and is also directly proportional to the number of turns in the loop or coil. The correct choice is C.|