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).
Answer:
14.2
Explanation:
find horizontal force of the weight = 2.5kg x 9.8 Sin30 = 12.3 N
to prevent the sliding she needs to pull horizontally
Fh = 12.3 /Cos 30 =14.2N
Answer:
This represents radiation in ultra-violet region .
Explanation:
Energy of the orbit where n = 3 is given as follows
= -1.511 eV
Energy of the orbit where n = 1 is given as follows
= 13.6 eV
Difference of [tex]E_3 and [tex]E_1 = - 1.511+ 13.6
= 12.089 eV.
The wavelength of light having this energy in nm is given by the expression as follows
Wavelength in nm = 1244 / energy in eV
= 1244 / 12.089
= 102.90 nm
This represents radiation in ultra-violet region .
A :-) for this question , we should apply
F = GMm by d^2
( For making the calculation easy , first remove the decimals )
Given : G = 6.7 x 10^-11 Nm^2 / kg^2
= 67 x 10^-12 Nm^2 / Kg^2
M = 65 kg
m = 40 kg
d = 0.5 m
Solution -
F = GMm by d^2
F = 67 x 10^-12 x 65 x 40 by 0.5 x 0.5
F = 4355 x 40 x 10^-12 by 0.25
F = 174200 x 10^-12 by 0.25
F = 696800 x 10^-12
.:. The Gravitational force between mark and Katie is 696800 x 10^-12
The answer is increased. A power factor of one or "unity power factor" is the aim of any electric utility business from the time when if the power factor is less than one, they have to give more current to the user for an assumed amount of power use. In so doing, they gain more line harms.
