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?
ABC) 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 = I2 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 = V2 / R.
CrktCT -5
The circuit below consists of a battery attached to two resistors in series. Resistor R1 is variable.
When R1 is decreased, the voltage V2 across R2
A) increasesB) decreasesC) stays the same.
Answer: increases. This is easiest to see if you let R1 go all the way to zero! Kirchhoff's Voltage Law says that the battery voltage V=V1+V2. The voltage drop is split between resistors 1 and 2. So V2 is smaller than the battery voltage V. But if R1=0, then the full battery voltage V is across R2. V2 increased as R1 goes down.
CrktCT -6
Two resistors R1 = R and R2 = 2R are hooked to a battery in parallel. R2 is twice as large as R1. How does the current IBat from the battery compare to the current I2 though R2?
A) IB = I2B) IB=2I2C) IB=3I2
D) IB=4I2E) None of these.
Answer: IB=3I2 Using Ohm’s Law for each resistor separately gives V = I1 R, and V = I2(2R), I1 = V/R, I2 = (1/2) (V/R). Or I2 = 2 I1. Total current from the batter is IBat = I1 + I2 = 3I1.
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 brighterB) 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 R2 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 Rtot, the current from the battery Itot = V/Rtot is reduced. Less current from the battery means less current through bulb 1, less light.
Bulb 3 gets brighter. The voltage V1 across Bulb 1 and the voltage V2 across the R3/R2 combination must add up to V=12V, by Kirchhoff's voltage law: V = V1+V2. When bulb 2 burns out, bulb 1 gets dimmer (see above) so the V1 gets smaller. Smaller V1 means larger V2, since V1 and V2 must add up to 12 V. Larger V2 means bulb 2 gets brighter.
Another way to see this: R2 becomes infinite and the R2/R3 parallel combination become larger. Bigger R2/R3 resistance means larger voltage drop V2 and smaller voltage drop V1. Larger V2 means more current thru R3, brighter bulb 3.