**CrktCT -0b**
Two glowing light bulbs are in a battery-operated circuit. Light bulb A has greater resistance than light bulb B. Which light bulb is brighter?
A B C) Depends on the circuit.
Answer: depends on the circuit. If the resistors are in series, then they both carry the same current, and the higher resistance will be brighter according to P = I^{2} R. If the resistors are in parallel, then they both have the same voltage drop, and the lower resistance will be brighter according to P = V^{2} / R.
**CrktCT -5**
The circuit below consists of a battery attached to two resistors in series. Resistor R_{1} is **variable**.
When R_{1} is decreased, the voltage V_{2} across R_{2}
A) increases B) decreases C) stays the same.
Answer: increases. This is easiest to see if you let R_{1} go all the way to zero! Kirchhoff's Voltage Law says that the battery voltage V=V_{1}+V_{2}. The voltage drop is split between resistors 1 and 2. So V_{2} is smaller than the battery voltage V. But if R_{1}=0, then the full battery voltage V is across R_{2}. V_{2} increased as R_{1} goes down.
**CrktCT -6**
Two resistors R_{1} = R and R_{2} = 2R are hooked to a battery in parallel. R_{2} is twice as large as R_{1}. How does the current I_{Bat} from the battery compare to the current I_{2} though R_{2}?
A) I_{B} = I_{2} B) I_{B}=2I_{2} C) I_{B}=3I_{2}
D) I_{B}=4I_{2} E) None of these.
Answer: I_{B}=3I_{2}^{ } Using Ohm’s Law for each resistor separately gives V = I_{1 }R, and V = I_{2} (2R), I_{1} = V/R, I_{2} = (1/2) (V/R). Or I_{2} = 2 I_{1}. Total current from the batter is I_{Bat} = I_{1} + I_{2} = 3I_{1}.
**CrktCT-7** A flashlight requires 2 AA (1.5V) batteries, and is arranged as shown. Notice that the switch is open. Which statement is true.?
A) The bulb has 1.5 V across it, and glows
B) The bulb has 3 V across it, and glows
C) The bulb has 3 V across it, and is dark
D) The bulb has 0 V across it, and is dark
E) The bulb has 0 V across it, and glows
Which statement is true if the switched is closed?
Which statement is true if the switched is closed and *one* of the two batteries is reversed?
Answers:
If the switch is open, the bulb has 0 V across it, and is dark.
If the switch is closed, the bulb has 3 V across it, and glows.
If one of the batteries is reversed, the total voltage difference between the ends of the two batteries is +1.5V – 1.5V = 0 V, and so the bulb has 0 V across it, and is dark.
**CrktCT-8**
In the circuit below, what happens to the brightness of bulb 1, when bulb 2 burns out? (When a bulb burns out, its resistance becomes infinite.)
A) Bulb 1 gets brighter
B) Bulb 1 gets dimmer.
C) It's brightness remains the same.
(Hint: What happens to the current from the battery when bulb 2 burns out.)
A harder question: What happens to the brightness of bulb 3, when bulb 2 burns out? [Think about the voltages across 1 and 3.]
A) Bulb 3 gets brighter B) Bulb 3 gets dimmer.
C) It's brightness remains the same.
Answers: Bulb 1 gets dimmer! When bulb 2 burns outs, the filament inside breaks and R_{2} becomes infinitely large. The total equivalent resistance which the battery sees increases (since bulb 2 is gone, there are fewer paths for the current flow, so less flow, more total resistance.) Since the battery sees a larger R_{tot}, the current from the battery I_{tot} = V/R_{tot} is reduced. Less current from the battery means less current through bulb 1, less light.
Bulb 3 gets brighter. The voltage V_{1} across Bulb 1 and the voltage V_{2} across the R_{3}/R_{2} combination must add up to V=12V, by Kirchhoff's voltage law: V = V_{1}+V_{2}. When bulb 2 burns out, bulb 1 gets dimmer (see above) so the V_{1} gets smaller. Smaller V_{1} means larger V_{2}, since V_{1} and V_{2} must add up to 12 V. Larger V_{2} means bulb 2 gets brighter.
Another way to see this: R_{2} becomes infinite and the R_{2}/R_{3 } parallel combination become larger. Bigger R_{2}/R_{3 }resistance means larger voltage drop V_{2} and smaller voltage drop V_{1}. Larger V_{2} means more current thru R_{3}, brighter bulb 3.
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