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Circuit solving
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Find the power of each source and provide your answers based on passive convention
so the problem was that the powers didnt add up to zero at the end, so I got some of the current and voltage values wrong...
Also, I dont know which power values are suppose to be negatives.
circuit-analysis
asked 3 hours ago
LKim
111
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up vote
2
down vote
favorite
Find the power of each source and provide your answers based on passive convention
so the problem was that the powers didnt add up to zero at the end, so I got some of the current and voltage values wrong...
Also, I dont know which power values are suppose to be negatives.
circuit-analysis
asked 3 hours ago
LKim
111
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Find the power of each source and provide your answers based on passive convention
so the problem was that the powers didnt add up to zero at the end, so I got some of the current and voltage values wrong...
Also, I dont know which power values are suppose to be negatives.
circuit-analysis
asked 3 hours ago
LKim
111
Find the power of each source and provide your answers based on passive convention
so the problem was that the powers didnt add up to zero at the end, so I got some of the current and voltage values wrong...
Also, I dont know which power values are suppose to be negatives.
circuit-analysis
circuit-analysis
asked 3 hours ago
LKim
111
asked 3 hours ago
LKim
111
asked 3 hours ago
LKim
111
asked 3 hours ago
LKim
111
asked 3 hours ago
LKim
111
111
add a comment |Â
add a comment |Â
3 Answers
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oldest
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up vote
1
down vote
I'm not sure where you went wrong, but the thing can be solved fairly quickly using superposition.
First set the current source to 0 and just solve with the voltage source present.
$15 rm V/3 rm kOmega=5 rm mA$. So we'll say $i_1a=-5 rm mA$ and $i_2a=5 rm mA$.
Now set the voltage source to 0, and just consider the current source. This can be solved by inspection because using the current divider rule, the 2 kohm resistor will get 1/3 of the current and the 1 kohm resistor wil get 2/3 of the current. So $i_1b = 2 rm mA$ and $i_2b = 4 rm mA$.
Combining the two solutions, $i_1 = -3 rm mA$ and $i_2 = 9 rm mA$.
From there you can easily calculate the current through the voltage source ($i_1$) and the voltage across the current source ($1 rm kOmegatimes 7 rm mA$), and then you can calculate the power in or out of each component.
answered 2 hours ago
The Photon
79.5k394189
@KingDuken, thanks. Tried to do the math in my head while scrolling back and forth between the answer editor and the question post.
– The Photon
2 hours ago
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up vote
1
down vote
While superposition is, I believe, the best way to approach this problem (as The Photon shows), since there is just one unknown node voltage, node voltage analysis is almost as good.
If you find the node voltage $V_A$ at the junction of the current source and two resistors, you have all you need to find the powers.
The node voltage equation is:
$$frac15,mathrmV - V_A2,mathrmkOmega + 6,mathrmmA = fracV_A1,mathrmkOmega$$
Solving for $V_A$ is just algebra:
$$V_A = left(frac15,mathrmV2,mathrmkOmega + 6,mathrmmAright)/left(frac12,mathrmkOmega + frac11,mathrmkOmega right)= 9,mathrmV$$
Can you take it from here?
answered 1 hour ago
Alfred Centauri
23.8k11554
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up vote
1
down vote
1) The first problem I can see is when using Cramer's rule. The correct expression should be (using your equations):
$$i_1=dfracmatrix0.006&1\-15&-1000biggbigg=-3textmA $$
You can then find $i_2=9textmA$.
2) Also you are forgetting to square the current to find the power in the resistors.
With that:
$$(2textkOmega)(-3textmA)^2+(1textkOmega)(9textmA)^2+(15textV)(-3textmA)+(-6textmA)(9textV) =0$$
Notice that I had to place a minus sign for the last term $-(6textmA)(9textV)$ and that is because the current is "comes out" of the terminal labeled as + in your schematic, which means the current source is supplying power. The voltage is actually supplying power too but you have the reference direction for the current going into the + term, that is why we didn't have to place a minus sign in front of that term, but anyway the current came back with a negative sign so everything still holds.
answered 1 hour ago
sixcab
2,6831515
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
I'm not sure where you went wrong, but the thing can be solved fairly quickly using superposition.
First set the current source to 0 and just solve with the voltage source present.
$15 rm V/3 rm kOmega=5 rm mA$. So we'll say $i_1a=-5 rm mA$ and $i_2a=5 rm mA$.
Now set the voltage source to 0, and just consider the current source. This can be solved by inspection because using the current divider rule, the 2 kohm resistor will get 1/3 of the current and the 1 kohm resistor wil get 2/3 of the current. So $i_1b = 2 rm mA$ and $i_2b = 4 rm mA$.
Combining the two solutions, $i_1 = -3 rm mA$ and $i_2 = 9 rm mA$.
From there you can easily calculate the current through the voltage source ($i_1$) and the voltage across the current source ($1 rm kOmegatimes 7 rm mA$), and then you can calculate the power in or out of each component.
answered 2 hours ago
The Photon
79.5k394189
@KingDuken, thanks. Tried to do the math in my head while scrolling back and forth between the answer editor and the question post.
– The Photon
2 hours ago
add a comment |Â
up vote
1
down vote
I'm not sure where you went wrong, but the thing can be solved fairly quickly using superposition.
First set the current source to 0 and just solve with the voltage source present.
$15 rm V/3 rm kOmega=5 rm mA$. So we'll say $i_1a=-5 rm mA$ and $i_2a=5 rm mA$.
Now set the voltage source to 0, and just consider the current source. This can be solved by inspection because using the current divider rule, the 2 kohm resistor will get 1/3 of the current and the 1 kohm resistor wil get 2/3 of the current. So $i_1b = 2 rm mA$ and $i_2b = 4 rm mA$.
Combining the two solutions, $i_1 = -3 rm mA$ and $i_2 = 9 rm mA$.
From there you can easily calculate the current through the voltage source ($i_1$) and the voltage across the current source ($1 rm kOmegatimes 7 rm mA$), and then you can calculate the power in or out of each component.
answered 2 hours ago
The Photon
79.5k394189
@KingDuken, thanks. Tried to do the math in my head while scrolling back and forth between the answer editor and the question post.
– The Photon
2 hours ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I'm not sure where you went wrong, but the thing can be solved fairly quickly using superposition.
First set the current source to 0 and just solve with the voltage source present.
$15 rm V/3 rm kOmega=5 rm mA$. So we'll say $i_1a=-5 rm mA$ and $i_2a=5 rm mA$.
Now set the voltage source to 0, and just consider the current source. This can be solved by inspection because using the current divider rule, the 2 kohm resistor will get 1/3 of the current and the 1 kohm resistor wil get 2/3 of the current. So $i_1b = 2 rm mA$ and $i_2b = 4 rm mA$.
Combining the two solutions, $i_1 = -3 rm mA$ and $i_2 = 9 rm mA$.
From there you can easily calculate the current through the voltage source ($i_1$) and the voltage across the current source ($1 rm kOmegatimes 7 rm mA$), and then you can calculate the power in or out of each component.
answered 2 hours ago
The Photon
79.5k394189
I'm not sure where you went wrong, but the thing can be solved fairly quickly using superposition.
First set the current source to 0 and just solve with the voltage source present.
$15 rm V/3 rm kOmega=5 rm mA$. So we'll say $i_1a=-5 rm mA$ and $i_2a=5 rm mA$.
Now set the voltage source to 0, and just consider the current source. This can be solved by inspection because using the current divider rule, the 2 kohm resistor will get 1/3 of the current and the 1 kohm resistor wil get 2/3 of the current. So $i_1b = 2 rm mA$ and $i_2b = 4 rm mA$.
Combining the two solutions, $i_1 = -3 rm mA$ and $i_2 = 9 rm mA$.
From there you can easily calculate the current through the voltage source ($i_1$) and the voltage across the current source ($1 rm kOmegatimes 7 rm mA$), and then you can calculate the power in or out of each component.
answered 2 hours ago
The Photon
79.5k394189
edited 2 hours ago
answered 2 hours ago
The Photon
79.5k394189
answered 2 hours ago
The Photon
79.5k394189
answered 2 hours ago
The Photon
79.5k394189
79.5k394189
@KingDuken, thanks. Tried to do the math in my head while scrolling back and forth between the answer editor and the question post.
– The Photon
2 hours ago
add a comment |Â
@KingDuken, thanks. Tried to do the math in my head while scrolling back and forth between the answer editor and the question post.
– The Photon
2 hours ago
@KingDuken, thanks. Tried to do the math in my head while scrolling back and forth between the answer editor and the question post.
– The Photon
2 hours ago
@KingDuken, thanks. Tried to do the math in my head while scrolling back and forth between the answer editor and the question post.
– The Photon
2 hours ago
add a comment |Â
up vote
1
down vote
While superposition is, I believe, the best way to approach this problem (as The Photon shows), since there is just one unknown node voltage, node voltage analysis is almost as good.
If you find the node voltage $V_A$ at the junction of the current source and two resistors, you have all you need to find the powers.
The node voltage equation is:
$$frac15,mathrmV - V_A2,mathrmkOmega + 6,mathrmmA = fracV_A1,mathrmkOmega$$
Solving for $V_A$ is just algebra:
$$V_A = left(frac15,mathrmV2,mathrmkOmega + 6,mathrmmAright)/left(frac12,mathrmkOmega + frac11,mathrmkOmega right)= 9,mathrmV$$
Can you take it from here?
answered 1 hour ago
Alfred Centauri
23.8k11554
add a comment |Â
up vote
1
down vote
While superposition is, I believe, the best way to approach this problem (as The Photon shows), since there is just one unknown node voltage, node voltage analysis is almost as good.
If you find the node voltage $V_A$ at the junction of the current source and two resistors, you have all you need to find the powers.
The node voltage equation is:
$$frac15,mathrmV - V_A2,mathrmkOmega + 6,mathrmmA = fracV_A1,mathrmkOmega$$
Solving for $V_A$ is just algebra:
$$V_A = left(frac15,mathrmV2,mathrmkOmega + 6,mathrmmAright)/left(frac12,mathrmkOmega + frac11,mathrmkOmega right)= 9,mathrmV$$
Can you take it from here?
answered 1 hour ago
Alfred Centauri
23.8k11554
add a comment |Â
up vote
1
down vote
up vote
1
down vote
While superposition is, I believe, the best way to approach this problem (as The Photon shows), since there is just one unknown node voltage, node voltage analysis is almost as good.
If you find the node voltage $V_A$ at the junction of the current source and two resistors, you have all you need to find the powers.
The node voltage equation is:
$$frac15,mathrmV - V_A2,mathrmkOmega + 6,mathrmmA = fracV_A1,mathrmkOmega$$
Solving for $V_A$ is just algebra:
$$V_A = left(frac15,mathrmV2,mathrmkOmega + 6,mathrmmAright)/left(frac12,mathrmkOmega + frac11,mathrmkOmega right)= 9,mathrmV$$
Can you take it from here?
answered 1 hour ago
Alfred Centauri
23.8k11554
While superposition is, I believe, the best way to approach this problem (as The Photon shows), since there is just one unknown node voltage, node voltage analysis is almost as good.
If you find the node voltage $V_A$ at the junction of the current source and two resistors, you have all you need to find the powers.
The node voltage equation is:
$$frac15,mathrmV - V_A2,mathrmkOmega + 6,mathrmmA = fracV_A1,mathrmkOmega$$
Solving for $V_A$ is just algebra:
$$V_A = left(frac15,mathrmV2,mathrmkOmega + 6,mathrmmAright)/left(frac12,mathrmkOmega + frac11,mathrmkOmega right)= 9,mathrmV$$
Can you take it from here?
answered 1 hour ago
Alfred Centauri
23.8k11554
edited 1 hour ago
answered 1 hour ago
Alfred Centauri
23.8k11554
answered 1 hour ago
Alfred Centauri
23.8k11554
answered 1 hour ago
Alfred Centauri
23.8k11554
23.8k11554
add a comment |Â
add a comment |Â
up vote
1
down vote
1) The first problem I can see is when using Cramer's rule. The correct expression should be (using your equations):
$$i_1=dfracmatrix0.006&1\-15&-1000biggbigg=-3textmA $$
You can then find $i_2=9textmA$.
2) Also you are forgetting to square the current to find the power in the resistors.
With that:
$$(2textkOmega)(-3textmA)^2+(1textkOmega)(9textmA)^2+(15textV)(-3textmA)+(-6textmA)(9textV) =0$$
Notice that I had to place a minus sign for the last term $-(6textmA)(9textV)$ and that is because the current is "comes out" of the terminal labeled as + in your schematic, which means the current source is supplying power. The voltage is actually supplying power too but you have the reference direction for the current going into the + term, that is why we didn't have to place a minus sign in front of that term, but anyway the current came back with a negative sign so everything still holds.
answered 1 hour ago
sixcab
2,6831515
add a comment |Â
up vote
1
down vote
1) The first problem I can see is when using Cramer's rule. The correct expression should be (using your equations):
$$i_1=dfracmatrix0.006&1\-15&-1000biggbigg=-3textmA $$
You can then find $i_2=9textmA$.
2) Also you are forgetting to square the current to find the power in the resistors.
With that:
$$(2textkOmega)(-3textmA)^2+(1textkOmega)(9textmA)^2+(15textV)(-3textmA)+(-6textmA)(9textV) =0$$
Notice that I had to place a minus sign for the last term $-(6textmA)(9textV)$ and that is because the current is "comes out" of the terminal labeled as + in your schematic, which means the current source is supplying power. The voltage is actually supplying power too but you have the reference direction for the current going into the + term, that is why we didn't have to place a minus sign in front of that term, but anyway the current came back with a negative sign so everything still holds.
answered 1 hour ago
sixcab
2,6831515
add a comment |Â
up vote
1
down vote
up vote
1
down vote
1) The first problem I can see is when using Cramer's rule. The correct expression should be (using your equations):
$$i_1=dfracmatrix0.006&1\-15&-1000biggbigg=-3textmA $$
You can then find $i_2=9textmA$.
2) Also you are forgetting to square the current to find the power in the resistors.
With that:
$$(2textkOmega)(-3textmA)^2+(1textkOmega)(9textmA)^2+(15textV)(-3textmA)+(-6textmA)(9textV) =0$$
Notice that I had to place a minus sign for the last term $-(6textmA)(9textV)$ and that is because the current is "comes out" of the terminal labeled as + in your schematic, which means the current source is supplying power. The voltage is actually supplying power too but you have the reference direction for the current going into the + term, that is why we didn't have to place a minus sign in front of that term, but anyway the current came back with a negative sign so everything still holds.
answered 1 hour ago
sixcab
2,6831515
1) The first problem I can see is when using Cramer's rule. The correct expression should be (using your equations):
$$i_1=dfracmatrix0.006&1\-15&-1000biggbigg=-3textmA $$
You can then find $i_2=9textmA$.
2) Also you are forgetting to square the current to find the power in the resistors.
With that:
$$(2textkOmega)(-3textmA)^2+(1textkOmega)(9textmA)^2+(15textV)(-3textmA)+(-6textmA)(9textV) =0$$
Notice that I had to place a minus sign for the last term $-(6textmA)(9textV)$ and that is because the current is "comes out" of the terminal labeled as + in your schematic, which means the current source is supplying power. The voltage is actually supplying power too but you have the reference direction for the current going into the + term, that is why we didn't have to place a minus sign in front of that term, but anyway the current came back with a negative sign so everything still holds.
answered 1 hour ago
sixcab
2,6831515
edited 1 hour ago
answered 1 hour ago
sixcab
2,6831515
answered 1 hour ago
sixcab
2,6831515
answered 1 hour ago
sixcab
2,6831515
2,6831515
add a comment |Â
add a comment |Â
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