A few different ways to do this:
Way #1:
The current in the series loop is (12 V) / (total resistance) .
(Turns out to be 2 Amperes, but the question isn't asking for that.)
In a series loop, the current is the same at every point, so it's
the same current through each resistor.
The power dissipated by a resistor is (current)² · (resistance),
and the current is the same everywhere in the circuit, so the
smallest resistance will dissipate the least power. That's R1 .
And by the way, it's not "drawing" the most power. It's dissipating it.
Way #2:
Another expression for the power dissipated by a resistance is
(voltage across the resistance)² / (resistance) .
In a series loop, the voltage across each resistor is
[ (individual resistance) / (total resistance ] x battery voltage.
So the power dissipated by each resistor is
(individual resistance)² x [(battery voltage) / (total resistance)²]
This expression is smallest for the smallest individual resistance.
(The other two quantities are the same for each individual resistor.)
So again, the least power is dissipated by the smallest individual resistance.
That's R1 .
Way #3: (Einstein's way)
If we sat back and relaxed for a minute, stared at the ceiling, let our minds
wander, puffed gently on our pipe, and just daydreamed about this question
for a minute or two, we might have easily guessed at the answer.
===> When you wire up a battery and a light bulb in series, the part
that dissipates power, and gets so hot that it radiates heat and light, is
the light bulb (some resistance), not the wire (very small resistance).
Hello there! :)
Room temperature is approximately 20°C.
We can automatically eliminate choices B and D since they are not equal to 20°C.
Since some choices use the Kelvin scale, we can convert from Celsius to Kelvin using a simple formula:
K = C° + 273
Find room temperature in degrees <u>Kelvin</u>:
K = 20° + 273
K = 293°
Thus, the correct choice would be <u>C. 293K.</u>
Answer:
This question is incomplete
Explanation:
This question is incomplete because the telescope's focal length was not provided. The formula to be used here is
Magnification = telescope's focal length/eyepiece's focal length
The eyepiece's focal length was provided in the question as 0.38 m.
NOTE: Magnification can be described as the power of an instrument (in this case telescope) to enlarge an object. It has no unit and thus the two focal lengths mentioned in the formula above must be in the same unit (preferably meters since one of them is in meters already).
Out of the choices given, the statement about how light travels is "<span>Light can travel in a vacuum, and it travels faster if the light source is moving."</span>
the answer in my opinion would be A
