User:


Resistors in series (2)
Storyboard

When several resistors are connected in series, the current is the same in all resistors due to the conservation of loads. Therefore, in each resistance a potential drop equal to the electrical resistance multiplied by the current is experienced and whose sum must be the total potential difference. Therefore, the total resistance of a series of resistors is equal to the sum of these.

>Model

ID:(1396, 0)


Mechanisms
Iframe

>Top



Knowledge

Code
Concept

Mechanisms

ID:(16030, 0)


Model
Top

>Top



Parameters

Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$R_1$
R_1
Resistance 1
Ohm
$R_2$
R_2
Resistance 2
Ohm
$R_s$
R_s
Resistance in Series
Ohm

Variables

Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$I$
I
Current
A
$\Delta\varphi_1$
Dphi_1
Difference of potential 1
V
$\Delta\varphi_2$
Dphi_2
Difference of potential 2
V
$\Delta\varphi$
Dphi
Potential difference
V

Calculations


First, select the equation: to , then, select the variable: to

Calculations

Symbol
Equation
Solved
Translated

Calculations

Symbol
Equation
Solved
Translated

Variable Given Calculate Target : Equation To be used


Equations

#
Equation
1

$ \Delta\varphi = \Delta\varphi_1 + \Delta\varphi_2 $

Dphi = Dphi_1 + Dphi_2

2

$ \Delta\varphi = R_s I $

Dphi = R * I

3

$ \Delta\varphi_1 = R_1 I $

Dphi = R * I

4

$ \Delta\varphi_2 = R_2 I $

Dphi = R * I

5

$ R_s = R_1 + R_2 $

R_s = R_1 + R_2

ID:(16019, 0)


Series resistance (2)
Equation

>Top, >Model


In the case of two resistors connected in series, the resistance in Series ($R_s$) is equal to the sum of the resistance 1 ($R_1$) and the resistance 2 ($R_2$). This relationship is expressed as:

$ R_s = R_1 + R_2 $

$R_1$
Resistance 1
$Ohm$
5500
$R_2$
Resistance 2
$Ohm$
5501
$R_s$
Resistance in Series
$Ohm$
5498

None

ID:(16004, 0)


Sum of potential difference (2)
Equation

>Top, >Model


By the principle of energy conservation, the potential difference ($\Delta\varphi$) is equal to the sum of the difference of potential 1 ($\Delta\varphi_1$) and the difference of potential 2 ($\Delta\varphi_2$). This can be expressed through the following relationship:

$ \Delta\varphi = \Delta\varphi_1 + \Delta\varphi_2 $

$\Delta\varphi_1$
Difference of potential 1
$V$
5538
$\Delta\varphi_2$
Difference of potential 2
$V$
5539
$\Delta\varphi$
Potential difference
$V$
5477

ID:(16012, 0)


Ohm's law (1)
Equation

>Top, >Model


Traditional Ohm's law establishes a relationship between the potential difference ($\Delta\varphi$) and the current ($I$) through the resistance ($R$), using the following expression:

$ \Delta\varphi = R_s I $

$ \Delta\varphi = R I $

$I$
Current
$A$
5483
$\Delta\varphi$
Potential difference
$V$
5477
$R$
$R_s$
Resistance in Series
$Ohm$
5498

None

ID:(3214, 1)


Ohm's law (2)
Equation

>Top, >Model


Traditional Ohm's law establishes a relationship between the potential difference ($\Delta\varphi$) and the current ($I$) through the resistance ($R$), using the following expression:

$ \Delta\varphi_1 = R_1 I $

$ \Delta\varphi = R I $

$I$
Current
$A$
5483
$\Delta\varphi$
$\Delta\varphi_1$
Difference of potential 1
$V$
5538
$R$
$R_1$
Resistance 1
$Ohm$
5500

None

ID:(3214, 2)


Ohm's law (3)
Equation

>Top, >Model


Traditional Ohm's law establishes a relationship between the potential difference ($\Delta\varphi$) and the current ($I$) through the resistance ($R$), using the following expression:

$ \Delta\varphi_2 = R_2 I $

$ \Delta\varphi = R I $

$I$
Current
$A$
5483
$\Delta\varphi$
$\Delta\varphi_2$
Difference of potential 2
$V$
5539
$R$
$R_2$
Resistance 2
$Ohm$
5501

None

ID:(3214, 3)