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Unformatted text preview: C HAPTER 17 Quick Quizzes 1. d, b = c, a. The current in (d) is equivalent to two positive charges moving to the left. Parts (b) and (c) each represent four charges moving in the same direction because negative charges moving to the left are equivalent to positive charges moving to the right. The current in (a) is equivalent to five positive charges moving to the right. 2. (c), (d). Neither circuit (a) nor circuit (b) applies a difference in potential across the bulb. Circuit (a) has both lead wires connected to the same battery terminal. Circuit (b) has a low resistance path (a “short”) between the two battery terminals as well as between the bulb terminals. 3. (b). The slope of the line tangent to the curve at a point is the reciprocal of the resistance at that point. Note that as V ∆ increases, the slope (and hence 1 R ) increases. Thus, the resistance decreases. 4. (b), (d). The length of the line cord will double in this event. This would tend to increase the resistance of the line cord. But the doubling of the radius of the line cord results in the increase of the cross-sectional area by a factor of 4. This would reduce the resistance more than the doubling of length increases it. The net result is a decrease in resistance. The same effect would occur for the lightbulb filament. The lowered resistance would result in a larger current in the filament, causing it to glow more brightly. 5. (a). The resistance of the shorter wire is half that of the longer wire. The power dissipated, ( 29 1 R R T T α = +- , (and hence the rate of heating) will be greater for the shorter wire. Consideration of the expression 2 I R = might initially lead one to think that the reverse would be true. However, one must realize that the currents will not be the same in the two wires. 6. I a = I b > I c = I d > I e = I f . Charges constituting the current I a leave the positive terminal of the battery and then split to flow through the two bulbs; thus, I a = I c + I e . Because the potential difference ∆ V is the same across the two bulbs and because the power delivered to a device is ( 29 I V = ∆ , the 60– W bulb with the higher power rating must carry the greater current. Because charge does not accumulate in the bulbs, all the charge flowing into a bulb from the left has to flow out on the right; consequently I c = I d and I e = I f . The two currents leaving the bulbs recombine to form the current back into the battery, I f + I d = I b ....
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