#6 #18 #24 #30 #34 #38 #42 #44 #46 #48 #50 #566. When you stack three flashlight batteries in the same direction, you get a voltage of 3 x 1.5 Volts = 4.5 Volts. What voltage do you get if one of the batteries is turned end-for-end?
The net voltage would be 1.5 Volts + 1.5 Volts - 1.5 Volts = 1.5 Volts, since two of the voltages (or batteries) cancel.
18. What happens when a light bulb burns out?
The filament burns out or breaks creating a gap in the circuit. Since the circuit is incomplete, the light bulb no longer shines.
24. How does the current in a Y-shaped junction compare with the current out of the junction?
As discussed in class, since charge is conserved, current is also conserved. Thus, the current out of a Y-shaped junction is the sum of the currents in the two wires leading to the junction.
30. If the resistance connected to a battery is doubled, what happens to the current?
Ohm's law states that the voltage is equal to the current times the resistance. Since the battery voltage is constant, the current is cut in half.
34. Why would sea water be a better conductor than fresh water?
The ions from the salt (like sodium+ and chlorine-) make sea water a better conductor.
38. How would you wire two batteries and two bulbs to produce light for the longest time?
Since batteries last longer when connected in parallel, the batteries should be connected in parallel. The more resistance there is to the current, the longer the battery will last. The largest resistance is obtained when the light bulbs are connected in series. Thus, the light bulbs should be connected in series to the batteries already connected to one another in parallel.
42. Three identical light bulbs are wired in series. What happens to the brightness of the other two when one of the bulbs burns out?
When a light bulb burns out, it leaves a gap in the circuit. This causes the other two light bulbs to go out (like the old-style holiday bulbs we discussed in class).
44. Which bulbs are the brightest and dimmest in the circuit below?
Recall that power is equal to current times voltage or P = iV. Ohm's law tells us that voltage is current times resistance or V = iR. That means that power is current squared times resistance.
Since the current through bulb A is split into two parts at the junction, the greatest current flows through bulb A. Thus, bulb A will be the brightest. Since C and D are connected in series they will each be of the same brightness and dimmer than bulb B.
46. What happens to the bulbs B and C in the circuit shown below when bulb A burns out?
Since bulbs B and C are still connected to the same voltage V, their brightness is unchanged when bulb A burns out.
48. What happens to the brightness of A, B, and C in the circuit below when bulb D is removed from its socket?
When bulb D is removed, bulb C goes out because its circuit to the battery has a gap in it. In the circuit that remains, bulbs A and B have equal brightness. Bulb A is dimmer than it was before D was removed, while B is brighter.
50. A box has three identical bulbs mounted on its top with the wires hidden inside the box. Initially bulb A is the brightest and bulbs B and C are equally bright. If you unscrew A, bulbs B and C remain the same. If you unscrew B, A remains the same and C goes out. If you unscrew C, A remain the same and B goes out. If you unscrew B and C, A remains the same. How are the bulbs wired?
Since B and C remain lit when A goes out, A makes a circuit that is separate from B and C. We are told that when B goes out, so does C and vice-versa. Thus, B and C are in series with one another. The rest of the given information can be used to check the circuit below, but is not needed.
56. What happens to the power if a resistor is connected to two batteries in series rather than a single battery?
We know that power is equal to the current times the voltage, P= iV. From Ohm's law we have V = iR or i = V/R. Substituting this into the power relationship, we see that the power is equal to the voltage squared divided by the resistance. Since the two batteries in series have twice the voltage of the lone battery and the resistance does not change, the power quadruples.