Answer:
Explanation:
Given that,
Two resistor has resistance in the ratio 2:3
Then,
R1 : R2 = 2:3
R1 / R2 =⅔
3 •R1 = 2• R2
Let R2 = R
Then,
R1 = ⅔R2 = 2/3 R
So, if the resistor are connected in series
Let know the current that will flow in the circuit
Series connection will have a equivalent resistance of
Req = R1 + R2
Req = R + ⅔ R = 5/3 R
Req = 5R / 3
Let a voltage V be connect across then, the current that flows can be calculated using ohms law
V = iR
I = V/Req
I = V / (5R /3)
I = 3V / 5R
This the current that flows in the two resistors since the same current flows in series connection
Now, using ohms law again to calculated voltage in each resistor
V= iR
For R1 = ⅔R
V1 =i•R1
V1 = 3V / 5R × 2R / 3
V1 = 3V × 2R / 5R × 3
V1 = 2V / 5
For R2 = R
V2 = i•R2
V2 = 3V / 5R × R
V2 = 3V × R / 5R
V2 = 3V / 5
Then,
Ratio of voltage 1 to voltage 2
V1 : V2 = V1 / V2 = 2V / 5 ÷ 3V / 5
V1 : V2 = 2V / 5 × 5 / 3V.
V1 : V2 =2 / 3
V1:V2 = 2:3
The ratio of their voltages is also 2:3
Answer:
Just to help, periods on the periodic table are those running horizontally from left to right
Answer:
The power dissipated in the 3 Ω resistor is P= 5.3watts.
Explanation:
After combine the 3 and 6 Ω resistor in parallel, we have an 2 Ω and a 4 Ω resistor in series.
The resultating resistor is of Req=6Ω.
I= V/Req
I= 2A
the parallel resistors have a potential drop of Vparallel=4 volts.
I(3Ω) = Vparallel/R(3Ω)
I(3Ω)= 1.33A
P= I(3Ω)² * R(3Ω)
P= 5.3 Watts
Answer:
Orbital Eccentricity
Planet Orbital Eccentricity
(Point in Orbit Closest to Sun)
measured in AU's
Mercury 0.206
Venus 0.007
Earth 0.017
Mars 0.093
Jupiter 0.048
Saturn 0.056
Uranus 0.047
Neptune 0.009
Pluto 0.248
Explanation:
link to information:
https://www.enchantedlearning.com/subjects/astronomy/glossary/Eccentricity.shtml
