Parallel Resistance
Storyboard
When the resistors are connected in parallel, they are all exposed to the same potential difference which, by Ohm's law, generates different currents. The total current is the sum of the partial currents, so the total resistance is the inverse of the sum of the inverse of the individual resistances.
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Parallel resistors (Diagram)
Image
The diagram representing resistors connected in parallel has the following form:
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Resistance in parallel
Equation
Al conectarse resistencias
$I=\displaystyle\sum_iI_i$
\\n\\nComo en cada resistencia se cumple la ley de Ohm\\n\\n
$\Delta\varphi=R_iI_i$
\\n\\nla suma de corrientes se puede escribir como\\n\\n
$I=\displaystyle\sum_i\displaystyle\frac{\Delta\varphi}{R_i}$
Por ello se puede definir una resistencia total para el caso de suma paralela es con de la forma
$\displaystyle\frac{1}{ R_p }=\displaystyle\sum_i\displaystyle\frac{1}{ R_i }$ |
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Sum of resistors in parallel (2)
Equation
Since the sum of resistors in parallel is
$\displaystyle\frac{1}{ R_p }=\displaystyle\sum_i\displaystyle\frac{1}{ R_i }$ |
You have to in the case of two resistors:
$\displaystyle\frac{1}{ R_p }=\displaystyle\frac{1}{ R_1 }+\displaystyle\frac{1}{ R_2 }$ |
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Sum of resistors in parallel (3)
Equation
Since the sum of resistors in parallel is
$\displaystyle\frac{1}{ R_p }=\displaystyle\sum_i\displaystyle\frac{1}{ R_i }$ |
You have to in the case of three resistors:
$\displaystyle\frac{1}{ R_{p3} }=\displaystyle\frac{1}{ R_{1p3} }+\displaystyle\frac{1}{ R_{2p3} }+\displaystyle\frac{1}{ R_{3p3} }$ |
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Sum of resistors in parallel (4)
Equation
Since the sum of resistors in parallel is
$\displaystyle\frac{1}{ R_p }=\displaystyle\sum_i\displaystyle\frac{1}{ R_i }$ |
it is necessary that in the case of four resistances:
$\displaystyle\frac{1}{ R_p }=\displaystyle\frac{1}{ R_1 }+\displaystyle\frac{1}{ R_2 }+\displaystyle\frac{1}{ R_3 }+\displaystyle\frac{1}{ R_4 }$ |
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Sum of resistors in parallel (5)
Equation
Since the sum of resistors in parallel is
$\displaystyle\frac{1}{ R_p }=\displaystyle\sum_i\displaystyle\frac{1}{ R_i }$ |
You have to in the case of five resistors:
$\displaystyle\frac{1}{ R_p }=\displaystyle\frac{1}{ R_1 }+\displaystyle\frac{1}{ R_2 }+\displaystyle\frac{1}{ R_3 }+\displaystyle\frac{1}{ R_4 }+\displaystyle\frac{1}{ R_5 }$ |
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