Students are experimenting with circuits in their physics class and they build the two working circuits pictured below. The batt
ery in each circuit is identical and each bulb (resistor) they add provides the same amount of resistance.
They notice that every time they added bulb 3 (R3) to Circuit A, the brightness of all the bulbs dimmed but when they added bulb 3 (R3), nothing changed in the brightness of the bulb. Which explanation of this observation is most complete?
Question options:
The brightness of the bulbs is all about the total amount of voltage at each resistor: less voltage = dimmer bulbs. Circuit A is a series circuit, so the voltage of the battery is divided across all of the resistors. Circuit B is a parallel circuit so the voltage across each resistor in parallel is the same as the battery.
The brightness of the bulbs is all about the total amount of current in the circuit: less current = dimmer bulbs. Circuit A is a series circuit, the more resistors you add, the less total current you get. Circuit B is a parallel circuit so the more resistors you add, the more total current you get.
The brightness of the bulbs is all about the total amount of resistance in the circuit: more resistance = dimmer bulbs. Circuit A is a series circuit, so every time you add a resistor the total resistance increases. Circuit B is a parallel circuit, so every time you add a resistor the total resistance decreases.
The brightness of the bulbs has to do with power which considers both the voltage and the current: less voltage x less current = dimmer bulbs. In circuit A, the voltage is divided across the resistors and the current decreases as resistance increases. In circuit B, the voltage is the same in each parallel section of the circuit and the current through that section of the circuit only depends on the resistor in that section.
They notice that every time they added bulb 3 (R3) to Circuit A, the brightness of all the bulbs dimmed but when they added bulb 3 (R3), nothing changed in the brightness of the bulb. Which explanation of this observation is most complete?
Question options:
The brightness of the bulbs is all about the total amount of voltage at each resistor: less voltage = dimmer bulbs. Circuit A is a series circuit, so the voltage of the battery is divided across all of the resistors. Circuit B is a parallel circuit so the voltage across each resistor in parallel is the same as the battery.
The brightness of the bulbs is all about the total amount of current in the circuit: less current = dimmer bulbs. Circuit A is a series circuit, the more resistors you add, the less total current you get. Circuit B is a parallel circuit so the more resistors you add, the more total current you get.
The brightness of the bulbs is all about the total amount of resistance in the circuit: more resistance = dimmer bulbs. Circuit A is a series circuit, so every time you add a resistor the total resistance increases. Circuit B is a parallel circuit, so every time you add a resistor the total resistance decreases.
The brightness of the bulbs has to do with power which considers both the voltage and the current: less voltage x less current = dimmer bulbs. In circuit A, the voltage is divided across the resistors and the current decreases as resistance increases. In circuit B, the voltage is the same in each parallel section of the circuit and the current through that section of the circuit only depends on the resistor in that section.
1 answer:
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Answer:
a) h=3.16 m, b) v_{cm }^ = 6.43 m / s
Explanation:
a) For this exercise we can use the conservation of mechanical energy
Starting point. Highest on the hill
Em₀ = U = mg h
final point. Lowest point
= K
Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere
K = ½ m + ½
w²
angular and linear speed are related
v = w r
w = v / r
K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²
Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)
as there are no friction losses, mechanical energy is conserved
Em₀ = Em_{f}
mg h = ½ v_{cm }^{2} (m + I_{cm} / r²) (1)
h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)
for the moment of inertia of a basketball we can approximate it to a spherical shell
I_{cm} = ⅔ m r²
we substitute
h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)
h = ½ v_{cm }^{2}/g 5/3
h = 5/6 v_{cm }^{2} / g
let's calculate
h = 5/6 6.1 2 / 9.8
h = 3.16 m
b) this part of the exercise we solve the speed of equation 1
v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)
in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia
I_{cm} = ½ m r²
we substitute
v_{cm } = √ [2gh / (1 + ½)]
v_{cm } = √(4/3 gh)
let's calculate
v_{cm } = √ (4/3 9.8 3.16)
v_{cm }^ = 6.43 m / s