Mixing
What direction does electric current flow in when the voltage drop is negative?
Clash Royale CLAN TAG#URR8PPP
up vote
4
down vote
favorite
We were given the below diagram, and we needed to determine whether the unknown component is supplying energy into the system.
$hspace200px$
I thought the unknown component was supplying energy since charge is flowing to a higher potential, but my teacher said it was the other way around.
What does the $`` -2 , mathrmV "$ label in the diagram mean? Is it:
the difference from $left(+right)$ to $left(-right);$ or
the difference from $left(-right)$ to $left(+right) ?$
I'm sure there is a convention to this sort of stuff but I'm not aware of it yet.
electricity conventions
edited Aug 28 at 6:28
Nat
3,37931731
asked Aug 27 at 21:12
Nick Yarn
436
add a comment |Â
up vote
4
down vote
favorite
We were given the below diagram, and we needed to determine whether the unknown component is supplying energy into the system.
$hspace200px$
I thought the unknown component was supplying energy since charge is flowing to a higher potential, but my teacher said it was the other way around.
What does the $`` -2 , mathrmV "$ label in the diagram mean? Is it:
the difference from $left(+right)$ to $left(-right);$ or
the difference from $left(-right)$ to $left(+right) ?$
I'm sure there is a convention to this sort of stuff but I'm not aware of it yet.
electricity conventions
edited Aug 28 at 6:28
Nat
3,37931731
asked Aug 27 at 21:12
Nick Yarn
436
I have updated my answer.
– Farcher
Aug 28 at 11:47
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
We were given the below diagram, and we needed to determine whether the unknown component is supplying energy into the system.
$hspace200px$
I thought the unknown component was supplying energy since charge is flowing to a higher potential, but my teacher said it was the other way around.
What does the $`` -2 , mathrmV "$ label in the diagram mean? Is it:
the difference from $left(+right)$ to $left(-right);$ or
the difference from $left(-right)$ to $left(+right) ?$
I'm sure there is a convention to this sort of stuff but I'm not aware of it yet.
electricity conventions
edited Aug 28 at 6:28
Nat
3,37931731
asked Aug 27 at 21:12
Nick Yarn
436
We were given the below diagram, and we needed to determine whether the unknown component is supplying energy into the system.
$hspace200px$
I thought the unknown component was supplying energy since charge is flowing to a higher potential, but my teacher said it was the other way around.
What does the $`` -2 , mathrmV "$ label in the diagram mean? Is it:
the difference from $left(+right)$ to $left(-right);$ or
the difference from $left(-right)$ to $left(+right) ?$
I'm sure there is a convention to this sort of stuff but I'm not aware of it yet.
electricity conventions
edited Aug 28 at 6:28
Nat
3,37931731
asked Aug 27 at 21:12
Nick Yarn
436
edited Aug 28 at 6:28
Nat
3,37931731
edited Aug 28 at 6:28
Nat
3,37931731
edited Aug 28 at 6:28
Nat
3,37931731
3,37931731
asked Aug 27 at 21:12
Nick Yarn
436
asked Aug 27 at 21:12
Nick Yarn
436
asked Aug 27 at 21:12
Nick Yarn
436
436
I have updated my answer.
– Farcher
Aug 28 at 11:47
add a comment |Â
I have updated my answer.
– Farcher
Aug 28 at 11:47
I have updated my answer.
– Farcher
Aug 28 at 11:47
I have updated my answer.
– Farcher
Aug 28 at 11:47
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
4
down vote
accepted
It looks like the passive sign convention is being used.
One terminal of a circuit component is labelled $(+)$ and the other terminal is labelled $(-)$.
The current direction is chosen such that it enters the $(+)$ terminal as shown in the diagram below.
The potential of the terminal labelled $(+)$ relative to the potential of the terminal labelled $(-)$ is equal to the stated voltage which is $v$ in my diagram and $-2,rm V$ in your diagram.
In your diagram the terminal labelled $(+)$ has a potential of $-2,rm V$ relative to the terminal labelled $(-)$ ie the terminal labelled $(-)$ is at a higher potential than the terminal labelled $(+)$ .
In your diagram with the current shown to be $-3,rm A$ that means that the current is flowing out of the terminal labelled $(+)$.
Having labelled the terminals and the current direction then the power is $vi$.
If the product $vi$ is positive then electrical power is being absorbed by the component ie the component is a sink.
If the product $vi$ is negative then electrical power is being generated by the circuit component, ie the component is a source.
In your example $vi=(-2)times (-3) =+6,rm W$ ie the circuit component is absorbing $6,rm W$ of electrical power eg the circuit component could be a resistance of resistance $frac 23,Omega$ which is converting electrical energy to heat at a rate of $6,rm W$.
Update
To show what happens in a complete circuit I have added a cell of emf $2,rm V$ in series with the component and included some labels related to the cell.
I have also labelled two nodes $A$ and $B$.
The $(+)$, $(-)$ and $2,rm V$ labels relating to the cell state that the potential of node $B$ is greater than the potential of node $A$ by $2,rm V$.
The $(+)$, $(-)$ and $-2,rm V$ labels relating to the component state that the potential of node $A$ is less than the potential of node $B$ by $2,rm V$ ie the potential of node $B$ is greater than the potential of node $A$ by $2,rm V$, which is the same as for the cell labels.
We can apply KIrchhoff's voltage law for the loop starting at node $B$ and going clockwise. $(+2) + (-2)=0$
The current direction and size for the cell are a consequence of applying Kirchhoff's current law.
For the cell the power is $(+2)times (-3) = -6 ,rm W$.
The cell is a source of electrical energy $-6,rm W$ at the same rate as the component is a sink of electric energy $+6,rm W$.
One way to learn more about this is to enroll on the edX Circuits and Electronics 1: Basic Circuit Analysis course and there is bonus of an excellent textbook which is associated with the course.
answered Aug 27 at 22:44
Farcher
44.2k33388
add a comment |Â
up vote
1
down vote
in these kind of problems first always try to find which component is acting as source (energy supplier) and which is acting as sink (energy consumer)and then,the rest follows
absorb negative signs of voltage by interchanging polarities of upper and lower terminals respectively and that of the current by changing it's direction
now draw modified diagram such that current $+3A$ will leave $+ve$ terminal of battery making it act like source i.e, it supplies/delivers energy/ to the unknown component which makes unknown component as sink i.e, it dissipates/absorbs energy.
answered Aug 27 at 22:31
veeresh pandey
2418
add a comment |Â
up vote
0
down vote
I just wanted to know what the -2v actually means.
Operationally, it means that if you place the red (positive) lead of your voltmeter on the $+$ labelled terminal and the black (negative or reference lead) of your voltmeter on the other, the voltmeter will read $-2,mathrmV$.
This tells you that the $-$ labelled terminal is more positive than the $+$ labelled terminal.
In anticipation of a follow-up question, the $-3,mathrmA$ means that if you place an ammeter in series with the $+$ labelled terminal of the circuit element such that the red lead is 'to the left', the ammeter will read $-3,mathrmA$.
This tells you that the current is directed directed out of the $+$ labelled terminal (in the opposite direction of the arrow).
As others have pointed out, with the chosen reference polarity and direction for the voltage and current respectively (passive sign convention), the power associated with the circuit element is
$$P = -2,mathrmV cdot -3,mathrmA = 6,mathrmW$$
Since the power is positive, power is delivered to the circuit element from the external circuit.
answered Aug 28 at 0:01
Alfred Centauri
45.1k343134
add a comment |Â
up vote
0
down vote
I would have thought:
$$
beginalign
P &= V cdot I \[5px]
P &= -fracmathrmdUmathrmdt
endalign \ , \
Delta U = -intPcdot mathrmdt = - intVcdot Icdot mathrmdt
$$
$ Vcdot I > 0$ since the current is entering in the device through the positive reference of the voltage and I supposed a passive sign convention, so$$
-intVcdot Icdot mathrmdt < 0
qquad textand qquad
Delta U < 0
, ,
$$so: it absorbs energy.
I have also done it by studying the system by simplifying it in this way, adjusting the signs:
About Convention
The number which represents the voltage is the number attached to the "+" symbol on the drawing. It means that the voltage written is referred to the positive pin of the bipole. The difference between the pin + and the pin - is the written one near the symbol as: $ V(+) - V(-) = $ written number, in this case $V(+) - V(-) = -2 , mathrmV.$ The current is inverted in verse if changed in sign.
The passive sign convention is used, but someone can use a different one.
(Please feel free to comment if there are mistakes)
edited Aug 28 at 6:21
Nat
3,37931731
answered Aug 27 at 21:40
Costantino
11312
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
It looks like the passive sign convention is being used.
One terminal of a circuit component is labelled $(+)$ and the other terminal is labelled $(-)$.
The current direction is chosen such that it enters the $(+)$ terminal as shown in the diagram below.
The potential of the terminal labelled $(+)$ relative to the potential of the terminal labelled $(-)$ is equal to the stated voltage which is $v$ in my diagram and $-2,rm V$ in your diagram.
In your diagram the terminal labelled $(+)$ has a potential of $-2,rm V$ relative to the terminal labelled $(-)$ ie the terminal labelled $(-)$ is at a higher potential than the terminal labelled $(+)$ .
In your diagram with the current shown to be $-3,rm A$ that means that the current is flowing out of the terminal labelled $(+)$.
Having labelled the terminals and the current direction then the power is $vi$.
If the product $vi$ is positive then electrical power is being absorbed by the component ie the component is a sink.
If the product $vi$ is negative then electrical power is being generated by the circuit component, ie the component is a source.
In your example $vi=(-2)times (-3) =+6,rm W$ ie the circuit component is absorbing $6,rm W$ of electrical power eg the circuit component could be a resistance of resistance $frac 23,Omega$ which is converting electrical energy to heat at a rate of $6,rm W$.
Update
To show what happens in a complete circuit I have added a cell of emf $2,rm V$ in series with the component and included some labels related to the cell.
I have also labelled two nodes $A$ and $B$.
The $(+)$, $(-)$ and $2,rm V$ labels relating to the cell state that the potential of node $B$ is greater than the potential of node $A$ by $2,rm V$.
The $(+)$, $(-)$ and $-2,rm V$ labels relating to the component state that the potential of node $A$ is less than the potential of node $B$ by $2,rm V$ ie the potential of node $B$ is greater than the potential of node $A$ by $2,rm V$, which is the same as for the cell labels.
We can apply KIrchhoff's voltage law for the loop starting at node $B$ and going clockwise. $(+2) + (-2)=0$
The current direction and size for the cell are a consequence of applying Kirchhoff's current law.
For the cell the power is $(+2)times (-3) = -6 ,rm W$.
The cell is a source of electrical energy $-6,rm W$ at the same rate as the component is a sink of electric energy $+6,rm W$.
One way to learn more about this is to enroll on the edX Circuits and Electronics 1: Basic Circuit Analysis course and there is bonus of an excellent textbook which is associated with the course.
answered Aug 27 at 22:44
Farcher
44.2k33388
add a comment |Â
up vote
4
down vote
accepted
It looks like the passive sign convention is being used.
One terminal of a circuit component is labelled $(+)$ and the other terminal is labelled $(-)$.
The current direction is chosen such that it enters the $(+)$ terminal as shown in the diagram below.
The potential of the terminal labelled $(+)$ relative to the potential of the terminal labelled $(-)$ is equal to the stated voltage which is $v$ in my diagram and $-2,rm V$ in your diagram.
In your diagram the terminal labelled $(+)$ has a potential of $-2,rm V$ relative to the terminal labelled $(-)$ ie the terminal labelled $(-)$ is at a higher potential than the terminal labelled $(+)$ .
In your diagram with the current shown to be $-3,rm A$ that means that the current is flowing out of the terminal labelled $(+)$.
Having labelled the terminals and the current direction then the power is $vi$.
If the product $vi$ is positive then electrical power is being absorbed by the component ie the component is a sink.
If the product $vi$ is negative then electrical power is being generated by the circuit component, ie the component is a source.
In your example $vi=(-2)times (-3) =+6,rm W$ ie the circuit component is absorbing $6,rm W$ of electrical power eg the circuit component could be a resistance of resistance $frac 23,Omega$ which is converting electrical energy to heat at a rate of $6,rm W$.
Update
To show what happens in a complete circuit I have added a cell of emf $2,rm V$ in series with the component and included some labels related to the cell.
I have also labelled two nodes $A$ and $B$.
The $(+)$, $(-)$ and $2,rm V$ labels relating to the cell state that the potential of node $B$ is greater than the potential of node $A$ by $2,rm V$.
The $(+)$, $(-)$ and $-2,rm V$ labels relating to the component state that the potential of node $A$ is less than the potential of node $B$ by $2,rm V$ ie the potential of node $B$ is greater than the potential of node $A$ by $2,rm V$, which is the same as for the cell labels.
We can apply KIrchhoff's voltage law for the loop starting at node $B$ and going clockwise. $(+2) + (-2)=0$
The current direction and size for the cell are a consequence of applying Kirchhoff's current law.
For the cell the power is $(+2)times (-3) = -6 ,rm W$.
The cell is a source of electrical energy $-6,rm W$ at the same rate as the component is a sink of electric energy $+6,rm W$.
One way to learn more about this is to enroll on the edX Circuits and Electronics 1: Basic Circuit Analysis course and there is bonus of an excellent textbook which is associated with the course.
answered Aug 27 at 22:44
Farcher
44.2k33388
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
It looks like the passive sign convention is being used.
One terminal of a circuit component is labelled $(+)$ and the other terminal is labelled $(-)$.
The current direction is chosen such that it enters the $(+)$ terminal as shown in the diagram below.
The potential of the terminal labelled $(+)$ relative to the potential of the terminal labelled $(-)$ is equal to the stated voltage which is $v$ in my diagram and $-2,rm V$ in your diagram.
In your diagram the terminal labelled $(+)$ has a potential of $-2,rm V$ relative to the terminal labelled $(-)$ ie the terminal labelled $(-)$ is at a higher potential than the terminal labelled $(+)$ .
In your diagram with the current shown to be $-3,rm A$ that means that the current is flowing out of the terminal labelled $(+)$.
Having labelled the terminals and the current direction then the power is $vi$.
If the product $vi$ is positive then electrical power is being absorbed by the component ie the component is a sink.
If the product $vi$ is negative then electrical power is being generated by the circuit component, ie the component is a source.
In your example $vi=(-2)times (-3) =+6,rm W$ ie the circuit component is absorbing $6,rm W$ of electrical power eg the circuit component could be a resistance of resistance $frac 23,Omega$ which is converting electrical energy to heat at a rate of $6,rm W$.
Update
To show what happens in a complete circuit I have added a cell of emf $2,rm V$ in series with the component and included some labels related to the cell.
I have also labelled two nodes $A$ and $B$.
The $(+)$, $(-)$ and $2,rm V$ labels relating to the cell state that the potential of node $B$ is greater than the potential of node $A$ by $2,rm V$.
The $(+)$, $(-)$ and $-2,rm V$ labels relating to the component state that the potential of node $A$ is less than the potential of node $B$ by $2,rm V$ ie the potential of node $B$ is greater than the potential of node $A$ by $2,rm V$, which is the same as for the cell labels.
We can apply KIrchhoff's voltage law for the loop starting at node $B$ and going clockwise. $(+2) + (-2)=0$
The current direction and size for the cell are a consequence of applying Kirchhoff's current law.
For the cell the power is $(+2)times (-3) = -6 ,rm W$.
The cell is a source of electrical energy $-6,rm W$ at the same rate as the component is a sink of electric energy $+6,rm W$.
One way to learn more about this is to enroll on the edX Circuits and Electronics 1: Basic Circuit Analysis course and there is bonus of an excellent textbook which is associated with the course.
answered Aug 27 at 22:44
Farcher
44.2k33388
It looks like the passive sign convention is being used.
One terminal of a circuit component is labelled $(+)$ and the other terminal is labelled $(-)$.
The current direction is chosen such that it enters the $(+)$ terminal as shown in the diagram below.
The potential of the terminal labelled $(+)$ relative to the potential of the terminal labelled $(-)$ is equal to the stated voltage which is $v$ in my diagram and $-2,rm V$ in your diagram.
In your diagram the terminal labelled $(+)$ has a potential of $-2,rm V$ relative to the terminal labelled $(-)$ ie the terminal labelled $(-)$ is at a higher potential than the terminal labelled $(+)$ .
In your diagram with the current shown to be $-3,rm A$ that means that the current is flowing out of the terminal labelled $(+)$.
Having labelled the terminals and the current direction then the power is $vi$.
If the product $vi$ is positive then electrical power is being absorbed by the component ie the component is a sink.
If the product $vi$ is negative then electrical power is being generated by the circuit component, ie the component is a source.
In your example $vi=(-2)times (-3) =+6,rm W$ ie the circuit component is absorbing $6,rm W$ of electrical power eg the circuit component could be a resistance of resistance $frac 23,Omega$ which is converting electrical energy to heat at a rate of $6,rm W$.
Update
To show what happens in a complete circuit I have added a cell of emf $2,rm V$ in series with the component and included some labels related to the cell.
I have also labelled two nodes $A$ and $B$.
The $(+)$, $(-)$ and $2,rm V$ labels relating to the cell state that the potential of node $B$ is greater than the potential of node $A$ by $2,rm V$.
The $(+)$, $(-)$ and $-2,rm V$ labels relating to the component state that the potential of node $A$ is less than the potential of node $B$ by $2,rm V$ ie the potential of node $B$ is greater than the potential of node $A$ by $2,rm V$, which is the same as for the cell labels.
We can apply KIrchhoff's voltage law for the loop starting at node $B$ and going clockwise. $(+2) + (-2)=0$
The current direction and size for the cell are a consequence of applying Kirchhoff's current law.
For the cell the power is $(+2)times (-3) = -6 ,rm W$.
The cell is a source of electrical energy $-6,rm W$ at the same rate as the component is a sink of electric energy $+6,rm W$.
One way to learn more about this is to enroll on the edX Circuits and Electronics 1: Basic Circuit Analysis course and there is bonus of an excellent textbook which is associated with the course.
answered Aug 27 at 22:44
Farcher
44.2k33388
edited Aug 28 at 12:15
answered Aug 27 at 22:44
Farcher
44.2k33388
answered Aug 27 at 22:44
Farcher
44.2k33388
answered Aug 27 at 22:44
Farcher
44.2k33388
44.2k33388
add a comment |Â
add a comment |Â
up vote
1
down vote
in these kind of problems first always try to find which component is acting as source (energy supplier) and which is acting as sink (energy consumer)and then,the rest follows
absorb negative signs of voltage by interchanging polarities of upper and lower terminals respectively and that of the current by changing it's direction
now draw modified diagram such that current $+3A$ will leave $+ve$ terminal of battery making it act like source i.e, it supplies/delivers energy/ to the unknown component which makes unknown component as sink i.e, it dissipates/absorbs energy.
answered Aug 27 at 22:31
veeresh pandey
2418
add a comment |Â
up vote
1
down vote
in these kind of problems first always try to find which component is acting as source (energy supplier) and which is acting as sink (energy consumer)and then,the rest follows
absorb negative signs of voltage by interchanging polarities of upper and lower terminals respectively and that of the current by changing it's direction
now draw modified diagram such that current $+3A$ will leave $+ve$ terminal of battery making it act like source i.e, it supplies/delivers energy/ to the unknown component which makes unknown component as sink i.e, it dissipates/absorbs energy.
answered Aug 27 at 22:31
veeresh pandey
2418
add a comment |Â
up vote
1
down vote
up vote
1
down vote
in these kind of problems first always try to find which component is acting as source (energy supplier) and which is acting as sink (energy consumer)and then,the rest follows
absorb negative signs of voltage by interchanging polarities of upper and lower terminals respectively and that of the current by changing it's direction
now draw modified diagram such that current $+3A$ will leave $+ve$ terminal of battery making it act like source i.e, it supplies/delivers energy/ to the unknown component which makes unknown component as sink i.e, it dissipates/absorbs energy.
answered Aug 27 at 22:31
veeresh pandey
2418
in these kind of problems first always try to find which component is acting as source (energy supplier) and which is acting as sink (energy consumer)and then,the rest follows
absorb negative signs of voltage by interchanging polarities of upper and lower terminals respectively and that of the current by changing it's direction
now draw modified diagram such that current $+3A$ will leave $+ve$ terminal of battery making it act like source i.e, it supplies/delivers energy/ to the unknown component which makes unknown component as sink i.e, it dissipates/absorbs energy.
answered Aug 27 at 22:31
veeresh pandey
2418
edited Aug 27 at 22:37
answered Aug 27 at 22:31
veeresh pandey
2418
answered Aug 27 at 22:31
veeresh pandey
2418
answered Aug 27 at 22:31
veeresh pandey
2418
2418
add a comment |Â
add a comment |Â
up vote
0
down vote
I just wanted to know what the -2v actually means.
Operationally, it means that if you place the red (positive) lead of your voltmeter on the $+$ labelled terminal and the black (negative or reference lead) of your voltmeter on the other, the voltmeter will read $-2,mathrmV$.
This tells you that the $-$ labelled terminal is more positive than the $+$ labelled terminal.
In anticipation of a follow-up question, the $-3,mathrmA$ means that if you place an ammeter in series with the $+$ labelled terminal of the circuit element such that the red lead is 'to the left', the ammeter will read $-3,mathrmA$.
This tells you that the current is directed directed out of the $+$ labelled terminal (in the opposite direction of the arrow).
As others have pointed out, with the chosen reference polarity and direction for the voltage and current respectively (passive sign convention), the power associated with the circuit element is
$$P = -2,mathrmV cdot -3,mathrmA = 6,mathrmW$$
Since the power is positive, power is delivered to the circuit element from the external circuit.
answered Aug 28 at 0:01
Alfred Centauri
45.1k343134
add a comment |Â
up vote
0
down vote
I just wanted to know what the -2v actually means.
Operationally, it means that if you place the red (positive) lead of your voltmeter on the $+$ labelled terminal and the black (negative or reference lead) of your voltmeter on the other, the voltmeter will read $-2,mathrmV$.
This tells you that the $-$ labelled terminal is more positive than the $+$ labelled terminal.
In anticipation of a follow-up question, the $-3,mathrmA$ means that if you place an ammeter in series with the $+$ labelled terminal of the circuit element such that the red lead is 'to the left', the ammeter will read $-3,mathrmA$.
This tells you that the current is directed directed out of the $+$ labelled terminal (in the opposite direction of the arrow).
As others have pointed out, with the chosen reference polarity and direction for the voltage and current respectively (passive sign convention), the power associated with the circuit element is
$$P = -2,mathrmV cdot -3,mathrmA = 6,mathrmW$$
Since the power is positive, power is delivered to the circuit element from the external circuit.
answered Aug 28 at 0:01
Alfred Centauri
45.1k343134
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I just wanted to know what the -2v actually means.
Operationally, it means that if you place the red (positive) lead of your voltmeter on the $+$ labelled terminal and the black (negative or reference lead) of your voltmeter on the other, the voltmeter will read $-2,mathrmV$.
This tells you that the $-$ labelled terminal is more positive than the $+$ labelled terminal.
In anticipation of a follow-up question, the $-3,mathrmA$ means that if you place an ammeter in series with the $+$ labelled terminal of the circuit element such that the red lead is 'to the left', the ammeter will read $-3,mathrmA$.
This tells you that the current is directed directed out of the $+$ labelled terminal (in the opposite direction of the arrow).
As others have pointed out, with the chosen reference polarity and direction for the voltage and current respectively (passive sign convention), the power associated with the circuit element is
$$P = -2,mathrmV cdot -3,mathrmA = 6,mathrmW$$
Since the power is positive, power is delivered to the circuit element from the external circuit.
answered Aug 28 at 0:01
Alfred Centauri
45.1k343134
I just wanted to know what the -2v actually means.
Operationally, it means that if you place the red (positive) lead of your voltmeter on the $+$ labelled terminal and the black (negative or reference lead) of your voltmeter on the other, the voltmeter will read $-2,mathrmV$.
This tells you that the $-$ labelled terminal is more positive than the $+$ labelled terminal.
In anticipation of a follow-up question, the $-3,mathrmA$ means that if you place an ammeter in series with the $+$ labelled terminal of the circuit element such that the red lead is 'to the left', the ammeter will read $-3,mathrmA$.
This tells you that the current is directed directed out of the $+$ labelled terminal (in the opposite direction of the arrow).
As others have pointed out, with the chosen reference polarity and direction for the voltage and current respectively (passive sign convention), the power associated with the circuit element is
$$P = -2,mathrmV cdot -3,mathrmA = 6,mathrmW$$
Since the power is positive, power is delivered to the circuit element from the external circuit.
answered Aug 28 at 0:01
Alfred Centauri
45.1k343134
edited Aug 28 at 0:06
answered Aug 28 at 0:01
Alfred Centauri
45.1k343134
answered Aug 28 at 0:01
Alfred Centauri
45.1k343134
answered Aug 28 at 0:01
Alfred Centauri
45.1k343134
45.1k343134
add a comment |Â
add a comment |Â
up vote
0
down vote
I would have thought:
$$
beginalign
P &= V cdot I \[5px]
P &= -fracmathrmdUmathrmdt
endalign \ , \
Delta U = -intPcdot mathrmdt = - intVcdot Icdot mathrmdt
$$
$ Vcdot I > 0$ since the current is entering in the device through the positive reference of the voltage and I supposed a passive sign convention, so$$
-intVcdot Icdot mathrmdt < 0
qquad textand qquad
Delta U < 0
, ,
$$so: it absorbs energy.
I have also done it by studying the system by simplifying it in this way, adjusting the signs:
About Convention
The number which represents the voltage is the number attached to the "+" symbol on the drawing. It means that the voltage written is referred to the positive pin of the bipole. The difference between the pin + and the pin - is the written one near the symbol as: $ V(+) - V(-) = $ written number, in this case $V(+) - V(-) = -2 , mathrmV.$ The current is inverted in verse if changed in sign.
The passive sign convention is used, but someone can use a different one.
(Please feel free to comment if there are mistakes)
edited Aug 28 at 6:21
Nat
3,37931731
answered Aug 27 at 21:40
Costantino
11312
add a comment |Â
up vote
0
down vote
I would have thought:
$$
beginalign
P &= V cdot I \[5px]
P &= -fracmathrmdUmathrmdt
endalign \ , \
Delta U = -intPcdot mathrmdt = - intVcdot Icdot mathrmdt
$$
$ Vcdot I > 0$ since the current is entering in the device through the positive reference of the voltage and I supposed a passive sign convention, so$$
-intVcdot Icdot mathrmdt < 0
qquad textand qquad
Delta U < 0
, ,
$$so: it absorbs energy.
I have also done it by studying the system by simplifying it in this way, adjusting the signs:
About Convention
The number which represents the voltage is the number attached to the "+" symbol on the drawing. It means that the voltage written is referred to the positive pin of the bipole. The difference between the pin + and the pin - is the written one near the symbol as: $ V(+) - V(-) = $ written number, in this case $V(+) - V(-) = -2 , mathrmV.$ The current is inverted in verse if changed in sign.
The passive sign convention is used, but someone can use a different one.
(Please feel free to comment if there are mistakes)
edited Aug 28 at 6:21
Nat
3,37931731
answered Aug 27 at 21:40
Costantino
11312
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I would have thought:
$$
beginalign
P &= V cdot I \[5px]
P &= -fracmathrmdUmathrmdt
endalign \ , \
Delta U = -intPcdot mathrmdt = - intVcdot Icdot mathrmdt
$$
$ Vcdot I > 0$ since the current is entering in the device through the positive reference of the voltage and I supposed a passive sign convention, so$$
-intVcdot Icdot mathrmdt < 0
qquad textand qquad
Delta U < 0
, ,
$$so: it absorbs energy.
I have also done it by studying the system by simplifying it in this way, adjusting the signs:
About Convention
The number which represents the voltage is the number attached to the "+" symbol on the drawing. It means that the voltage written is referred to the positive pin of the bipole. The difference between the pin + and the pin - is the written one near the symbol as: $ V(+) - V(-) = $ written number, in this case $V(+) - V(-) = -2 , mathrmV.$ The current is inverted in verse if changed in sign.
The passive sign convention is used, but someone can use a different one.
(Please feel free to comment if there are mistakes)
edited Aug 28 at 6:21
Nat
3,37931731
answered Aug 27 at 21:40
Costantino
11312
I would have thought:
$$
beginalign
P &= V cdot I \[5px]
P &= -fracmathrmdUmathrmdt
endalign \ , \
Delta U = -intPcdot mathrmdt = - intVcdot Icdot mathrmdt
$$
$ Vcdot I > 0$ since the current is entering in the device through the positive reference of the voltage and I supposed a passive sign convention, so$$
-intVcdot Icdot mathrmdt < 0
qquad textand qquad
Delta U < 0
, ,
$$so: it absorbs energy.
I have also done it by studying the system by simplifying it in this way, adjusting the signs:
About Convention
The number which represents the voltage is the number attached to the "+" symbol on the drawing. It means that the voltage written is referred to the positive pin of the bipole. The difference between the pin + and the pin - is the written one near the symbol as: $ V(+) - V(-) = $ written number, in this case $V(+) - V(-) = -2 , mathrmV.$ The current is inverted in verse if changed in sign.
The passive sign convention is used, but someone can use a different one.
(Please feel free to comment if there are mistakes)
edited Aug 28 at 6:21
Nat
3,37931731
answered Aug 27 at 21:40
Costantino
11312
edited Aug 28 at 6:21
Nat
3,37931731
edited Aug 28 at 6:21
Nat
3,37931731
edited Aug 28 at 6:21
Nat
3,37931731
3,37931731
answered Aug 27 at 21:40
Costantino
11312
answered Aug 27 at 21:40
Costantino
11312
answered Aug 27 at 21:40
Costantino
11312
11312
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f425194%2fwhat-direction-does-electric-current-flow-in-when-the-voltage-drop-is-negative%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
I have updated my answer.
– Farcher
Aug 28 at 11:47