a 455n gymnast jumps upward a distance of 1.5 meters to reach the uneven parallel bars. how much work did she do before she even
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Let R₁ and R₂ be the two Resistance
R₁ - Resistance of 1st resistor
R₂ - Resistance of 2nd resistor
V = Voltage = 12V
I = O.31A ( in series)
I = 1.6 A ( in parallel)
when two resistance connected in series
Rs = R₁ + R₂
V = IRs = 12 /0.31 V
R₁+R₂ = 38.700Ω (equation1.)
When two resistance connected in parallel
V = I Rp
V= I (R₂ R₁ /R₁ +R₂)
R₁ R₂ = 290.25 Ω
As we know
(R₁ - R₂ ) ² = (R₁ + R₂ ) ² - 4R1R2
Using above values we get,
(R₁ - R₂ ) ² = (38.70) ² - 4x 290.25
(R₁ - R₂ ) ² = 336.69 Ω
R₁ - R₂ = 18.34 (equation 2 )
Using equation 1 and 2 we get
R₁+ R₂ = 38.70 52
R₁ - R₂ = 18.34
R₁ = 28.535 Ω
Using the value of R₁ in equation 1 we get
R₂ = 38·700 - 28.535
RR₂ = 10.125 Ω
R₁ = 28. 535 v and R₂ 10.125 Ω
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