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LECT10_183184

# LECT10_183184 - Resistors in Parallel Req What happens at a...

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Resistors in Parallel R eq

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What happens at a junction? Initial current I tot splits up: I 1 through R 1 and I 2 through R 2 Charge is conserved: I tot = I 1 + I 2 More charge will be able to travel through the path of least resistance If R 1 > R 2 , then I 2 > I 1 R 1 R 2 I tot I 1 I 2
Resistors in Parallel Note: Δ V across each resistor is the same I = I 1 + I 2 = Δ V / R 1 + Δ V / R 2 = Δ V(1/R 1 + 1/R 2 ) Δ V = I / (1/R 1 + 1/R 2 ) Δ V = I R eq R eq = 1 / (1/R 1 + 1/R 2 ) 1/R eq = (1/R 1 + 1/R 2 ) For N resistors in parallel: 1/R eq = 1/R 1 + 1/R 2 + … + 1/R N

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Understanding the parallel law A 1 A 2 R is prop.to 1/A A tot = A 1 + A 2 A tot prop.to 1/R 1 + 1/R 2 R tot prop.to 1/A tot 1/R tot prop.to 1/R 1 + 1/R 2
Example: Find the current in each resistor. I 1 = Δ V/R 1 = 18V/3 Ω = 6A I 2 = Δ V/R 2 = 18V/6 Ω = 3A I 3 = Δ V/R 3 = 18V/9 Ω = 2A (Total I = 11A) Find the power dissipated in each resistor: P 1 = I 1 Δ V = 6A × 18V = 108 W P 2 = I 2 Δ V = 3A × 18V = 54 W Py = I 3 Δ V = 2A × 18V = 36 W Total P = 198 W

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Example: Find R eq : 1/R eq = 1/R 1 + 1/R 2 + 1/R 3 = 1/(3 Ω ) + 1/(6 Ω ) + 1/(9 Ω ) = 11/(18 Ω ) R eq = (18/11) Ω = 1.64 Ω Find the power dissipated in the equivalent resistor: P = ( Δ V) 2 /R eq = (18V) 2 /1.64 Ω = 198 W Also, P = I Δ V = 11A × 18V = 198W
Comparing resistors and capacitors

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