What is wrong with my KVL approach on a really simple circuit?
$begingroup$
Hello, so this is a really simple circuit question that has been bothering me. I know that by simply looking at the circuit and using logic it makes sense that the voltage drop across the resistor(as pointed in the picture) should be 0.6V. But when I actually write the Kirchoff Voltage Law(KVL) around the loop as shown in the picture, I somehow get -1.8V. I am probably messing up my convention. Would someone please explain how I would get the same 0.6V using KVL as well?
Edit: found the answer from my class notes for your refernce. Is the solution assuming opposite convention for how the current flows in the circuit so as opposed to positive to minus terminal is it assuming minus to positive?
circuit-analysis
asked 2 hours ago
CuriousJCuriousJ
112
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CuriousJ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Hello, so this is a really simple circuit question that has been bothering me. I know that by simply looking at the circuit and using logic it makes sense that the voltage drop across the resistor(as pointed in the picture) should be 0.6V. But when I actually write the Kirchoff Voltage Law(KVL) around the loop as shown in the picture, I somehow get -1.8V. I am probably messing up my convention. Would someone please explain how I would get the same 0.6V using KVL as well?
Edit: found the answer from my class notes for your refernce. Is the solution assuming opposite convention for how the current flows in the circuit so as opposed to positive to minus terminal is it assuming minus to positive?
circuit-analysis
asked 2 hours ago
CuriousJCuriousJ
112
New contributor
CuriousJ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
3
$begingroup$
You shouldn't have flipped the battery polarity. (or you mis-flipped it in your calculations.)
$endgroup$
– Peter Bennett
2 hours ago
$begingroup$
Small detail but the question lists R as 200ohms and the current as 3mA. You have it as R being 3ohms and the current as 200mA. Though when multiplied they'd give the same thing, sometimes in an exam you reference values from the circuit you redrew and if you copy the wrong values well...
$endgroup$
– Simeon R
2 hours ago
$begingroup$
Don't use the same letter and case for variables and units. Units shouldn't be inside equations in any case.
$endgroup$
– Chu
1 hour ago
add a comment |
$begingroup$
Hello, so this is a really simple circuit question that has been bothering me. I know that by simply looking at the circuit and using logic it makes sense that the voltage drop across the resistor(as pointed in the picture) should be 0.6V. But when I actually write the Kirchoff Voltage Law(KVL) around the loop as shown in the picture, I somehow get -1.8V. I am probably messing up my convention. Would someone please explain how I would get the same 0.6V using KVL as well?
Edit: found the answer from my class notes for your refernce. Is the solution assuming opposite convention for how the current flows in the circuit so as opposed to positive to minus terminal is it assuming minus to positive?
circuit-analysis
asked 2 hours ago
CuriousJCuriousJ
112
New contributor
CuriousJ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Hello, so this is a really simple circuit question that has been bothering me. I know that by simply looking at the circuit and using logic it makes sense that the voltage drop across the resistor(as pointed in the picture) should be 0.6V. But when I actually write the Kirchoff Voltage Law(KVL) around the loop as shown in the picture, I somehow get -1.8V. I am probably messing up my convention. Would someone please explain how I would get the same 0.6V using KVL as well?
Edit: found the answer from my class notes for your refernce. Is the solution assuming opposite convention for how the current flows in the circuit so as opposed to positive to minus terminal is it assuming minus to positive?
circuit-analysis
circuit-analysis
asked 2 hours ago
CuriousJCuriousJ
112
New contributor
CuriousJ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
CuriousJCuriousJ
112
New contributor
CuriousJ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 hours ago
CuriousJ
asked 2 hours ago
CuriousJCuriousJ
112
New contributor
CuriousJ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
CuriousJCuriousJ
112
asked 2 hours ago
CuriousJCuriousJ
112
112
New contributor
CuriousJ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
CuriousJ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
CuriousJ is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
3
$begingroup$
You shouldn't have flipped the battery polarity. (or you mis-flipped it in your calculations.)
$endgroup$
– Peter Bennett
2 hours ago
$begingroup$
Small detail but the question lists R as 200ohms and the current as 3mA. You have it as R being 3ohms and the current as 200mA. Though when multiplied they'd give the same thing, sometimes in an exam you reference values from the circuit you redrew and if you copy the wrong values well...
$endgroup$
– Simeon R
2 hours ago
$begingroup$
Don't use the same letter and case for variables and units. Units shouldn't be inside equations in any case.
$endgroup$
– Chu
1 hour ago
add a comment |
3
$begingroup$
You shouldn't have flipped the battery polarity. (or you mis-flipped it in your calculations.)
$endgroup$
– Peter Bennett
2 hours ago
$begingroup$
Small detail but the question lists R as 200ohms and the current as 3mA. You have it as R being 3ohms and the current as 200mA. Though when multiplied they'd give the same thing, sometimes in an exam you reference values from the circuit you redrew and if you copy the wrong values well...
$endgroup$
– Simeon R
2 hours ago
$begingroup$
Don't use the same letter and case for variables and units. Units shouldn't be inside equations in any case.
$endgroup$
– Chu
1 hour ago
3
3
$begingroup$
You shouldn't have flipped the battery polarity. (or you mis-flipped it in your calculations.)
$endgroup$
– Peter Bennett
2 hours ago
$begingroup$
You shouldn't have flipped the battery polarity. (or you mis-flipped it in your calculations.)
$endgroup$
– Peter Bennett
2 hours ago
$begingroup$
Small detail but the question lists R as 200ohms and the current as 3mA. You have it as R being 3ohms and the current as 200mA. Though when multiplied they'd give the same thing, sometimes in an exam you reference values from the circuit you redrew and if you copy the wrong values well...
$endgroup$
– Simeon R
2 hours ago
$begingroup$
Small detail but the question lists R as 200ohms and the current as 3mA. You have it as R being 3ohms and the current as 200mA. Though when multiplied they'd give the same thing, sometimes in an exam you reference values from the circuit you redrew and if you copy the wrong values well...
$endgroup$
– Simeon R
2 hours ago
$begingroup$
Don't use the same letter and case for variables and units. Units shouldn't be inside equations in any case.
$endgroup$
– Chu
1 hour ago
$begingroup$
Don't use the same letter and case for variables and units. Units shouldn't be inside equations in any case.
$endgroup$
– Chu
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I think the direction of current given in the problem is backwards, or the polarity of the voltage source is upside down. That would make 0.6V the correct answer.
I say this because it is impossible to have 1.8V across a resistor in a circuit if the only source is 1.2V (regardless of polarity).
answered 2 hours ago
Elliot AldersonElliot Alderson
6,25611021
$endgroup$
$begingroup$
Definitely the more practical answer. +1.
$endgroup$
– Metric
2 hours ago
add a comment |
$begingroup$
Your process is correct, and so is your answer (-1.8 V).
The answer provided is incorrect (0.6 V)
If the voltage source had its polarity switched in the initial circuit, then the voltage drop should be 0.6 V, but since it was not, we have
$$1.2 + V + .6 = 0 implies V = -1.8text{ V}$$
Of course, depending on how your instructor defines the polarity on the desired voltage drop, it could also be 1.8 V.
answered 2 hours ago
MetricMetric
2067
$endgroup$
$begingroup$
This could also depend on how your instructor interprets the voltage drop across a resistor with a certain current direction. Using the typical convention, your answer is right, and NOT the posted solution.
$endgroup$
– Metric
2 hours ago
$begingroup$
The typical convention, by the way, is that the voltage drop (from + to -) across a resistor with a current I going from + to - is V = IR.
$endgroup$
– Metric
2 hours ago
$begingroup$
my goodness I thought I was going insane. Thank you so much for your input.
$endgroup$
– CuriousJ
2 hours ago
$begingroup$
Glad I could reduce the insanity :P.
$endgroup$
– Metric
2 hours ago
1
$begingroup$
How can you get 1.8V across a resistor, regardless of polarity, if the only source has a voltage of 1.2V? The voltages across the resistors in the loop must add to the source voltage, and you can't have one negative and the other positive unless the current changes direction between the resistors.
$endgroup$
– Elliot Alderson
2 hours ago
|
show 4 more comments
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I think the direction of current given in the problem is backwards, or the polarity of the voltage source is upside down. That would make 0.6V the correct answer.
I say this because it is impossible to have 1.8V across a resistor in a circuit if the only source is 1.2V (regardless of polarity).
answered 2 hours ago
Elliot AldersonElliot Alderson
6,25611021
$endgroup$
$begingroup$
Definitely the more practical answer. +1.
$endgroup$
– Metric
2 hours ago
add a comment |
$begingroup$
I think the direction of current given in the problem is backwards, or the polarity of the voltage source is upside down. That would make 0.6V the correct answer.
I say this because it is impossible to have 1.8V across a resistor in a circuit if the only source is 1.2V (regardless of polarity).
answered 2 hours ago
Elliot AldersonElliot Alderson
6,25611021
$endgroup$
$begingroup$
Definitely the more practical answer. +1.
$endgroup$
– Metric
2 hours ago
add a comment |
$begingroup$
I think the direction of current given in the problem is backwards, or the polarity of the voltage source is upside down. That would make 0.6V the correct answer.
I say this because it is impossible to have 1.8V across a resistor in a circuit if the only source is 1.2V (regardless of polarity).
answered 2 hours ago
Elliot AldersonElliot Alderson
6,25611021
$endgroup$
I think the direction of current given in the problem is backwards, or the polarity of the voltage source is upside down. That would make 0.6V the correct answer.
I say this because it is impossible to have 1.8V across a resistor in a circuit if the only source is 1.2V (regardless of polarity).
answered 2 hours ago
Elliot AldersonElliot Alderson
6,25611021
answered 2 hours ago
Elliot AldersonElliot Alderson
6,25611021
answered 2 hours ago
Elliot AldersonElliot Alderson
6,25611021
answered 2 hours ago
Elliot AldersonElliot Alderson
6,25611021
6,25611021
$begingroup$
Definitely the more practical answer. +1.
$endgroup$
– Metric
2 hours ago
add a comment |
$begingroup$
Definitely the more practical answer. +1.
$endgroup$
– Metric
2 hours ago
$begingroup$
Definitely the more practical answer. +1.
$endgroup$
– Metric
2 hours ago
$begingroup$
Definitely the more practical answer. +1.
$endgroup$
– Metric
2 hours ago
add a comment |
$begingroup$
Your process is correct, and so is your answer (-1.8 V).
The answer provided is incorrect (0.6 V)
If the voltage source had its polarity switched in the initial circuit, then the voltage drop should be 0.6 V, but since it was not, we have
$$1.2 + V + .6 = 0 implies V = -1.8text{ V}$$
Of course, depending on how your instructor defines the polarity on the desired voltage drop, it could also be 1.8 V.
answered 2 hours ago
MetricMetric
2067
$endgroup$
$begingroup$
This could also depend on how your instructor interprets the voltage drop across a resistor with a certain current direction. Using the typical convention, your answer is right, and NOT the posted solution.
$endgroup$
– Metric
2 hours ago
$begingroup$
The typical convention, by the way, is that the voltage drop (from + to -) across a resistor with a current I going from + to - is V = IR.
$endgroup$
– Metric
2 hours ago
$begingroup$
my goodness I thought I was going insane. Thank you so much for your input.
$endgroup$
– CuriousJ
2 hours ago
$begingroup$
Glad I could reduce the insanity :P.
$endgroup$
– Metric
2 hours ago
1
$begingroup$
How can you get 1.8V across a resistor, regardless of polarity, if the only source has a voltage of 1.2V? The voltages across the resistors in the loop must add to the source voltage, and you can't have one negative and the other positive unless the current changes direction between the resistors.
$endgroup$
– Elliot Alderson
2 hours ago
|
show 4 more comments
$begingroup$
Your process is correct, and so is your answer (-1.8 V).
The answer provided is incorrect (0.6 V)
If the voltage source had its polarity switched in the initial circuit, then the voltage drop should be 0.6 V, but since it was not, we have
$$1.2 + V + .6 = 0 implies V = -1.8text{ V}$$
Of course, depending on how your instructor defines the polarity on the desired voltage drop, it could also be 1.8 V.
answered 2 hours ago
MetricMetric
2067
$endgroup$
$begingroup$
This could also depend on how your instructor interprets the voltage drop across a resistor with a certain current direction. Using the typical convention, your answer is right, and NOT the posted solution.
$endgroup$
– Metric
2 hours ago
$begingroup$
The typical convention, by the way, is that the voltage drop (from + to -) across a resistor with a current I going from + to - is V = IR.
$endgroup$
– Metric
2 hours ago
$begingroup$
my goodness I thought I was going insane. Thank you so much for your input.
$endgroup$
– CuriousJ
2 hours ago
$begingroup$
Glad I could reduce the insanity :P.
$endgroup$
– Metric
2 hours ago
1
$begingroup$
How can you get 1.8V across a resistor, regardless of polarity, if the only source has a voltage of 1.2V? The voltages across the resistors in the loop must add to the source voltage, and you can't have one negative and the other positive unless the current changes direction between the resistors.
$endgroup$
– Elliot Alderson
2 hours ago
|
show 4 more comments
$begingroup$
Your process is correct, and so is your answer (-1.8 V).
The answer provided is incorrect (0.6 V)
If the voltage source had its polarity switched in the initial circuit, then the voltage drop should be 0.6 V, but since it was not, we have
$$1.2 + V + .6 = 0 implies V = -1.8text{ V}$$
Of course, depending on how your instructor defines the polarity on the desired voltage drop, it could also be 1.8 V.
answered 2 hours ago
MetricMetric
2067
$endgroup$
Your process is correct, and so is your answer (-1.8 V).
The answer provided is incorrect (0.6 V)
If the voltage source had its polarity switched in the initial circuit, then the voltage drop should be 0.6 V, but since it was not, we have
$$1.2 + V + .6 = 0 implies V = -1.8text{ V}$$
Of course, depending on how your instructor defines the polarity on the desired voltage drop, it could also be 1.8 V.
answered 2 hours ago
MetricMetric
2067
answered 2 hours ago
MetricMetric
2067
answered 2 hours ago
MetricMetric
2067
answered 2 hours ago
MetricMetric
2067
2067
$begingroup$
This could also depend on how your instructor interprets the voltage drop across a resistor with a certain current direction. Using the typical convention, your answer is right, and NOT the posted solution.
$endgroup$
– Metric
2 hours ago
$begingroup$
The typical convention, by the way, is that the voltage drop (from + to -) across a resistor with a current I going from + to - is V = IR.
$endgroup$
– Metric
2 hours ago
$begingroup$
my goodness I thought I was going insane. Thank you so much for your input.
$endgroup$
– CuriousJ
2 hours ago
$begingroup$
Glad I could reduce the insanity :P.
$endgroup$
– Metric
2 hours ago
1
$begingroup$
How can you get 1.8V across a resistor, regardless of polarity, if the only source has a voltage of 1.2V? The voltages across the resistors in the loop must add to the source voltage, and you can't have one negative and the other positive unless the current changes direction between the resistors.
$endgroup$
– Elliot Alderson
2 hours ago
|
show 4 more comments
$begingroup$
This could also depend on how your instructor interprets the voltage drop across a resistor with a certain current direction. Using the typical convention, your answer is right, and NOT the posted solution.
$endgroup$
– Metric
2 hours ago
$begingroup$
The typical convention, by the way, is that the voltage drop (from + to -) across a resistor with a current I going from + to - is V = IR.
$endgroup$
– Metric
2 hours ago
$begingroup$
my goodness I thought I was going insane. Thank you so much for your input.
$endgroup$
– CuriousJ
2 hours ago
$begingroup$
Glad I could reduce the insanity :P.
$endgroup$
– Metric
2 hours ago
1
$begingroup$
How can you get 1.8V across a resistor, regardless of polarity, if the only source has a voltage of 1.2V? The voltages across the resistors in the loop must add to the source voltage, and you can't have one negative and the other positive unless the current changes direction between the resistors.
$endgroup$
– Elliot Alderson
2 hours ago
$begingroup$
This could also depend on how your instructor interprets the voltage drop across a resistor with a certain current direction. Using the typical convention, your answer is right, and NOT the posted solution.
$endgroup$
– Metric
2 hours ago
$begingroup$
This could also depend on how your instructor interprets the voltage drop across a resistor with a certain current direction. Using the typical convention, your answer is right, and NOT the posted solution.
$endgroup$
– Metric
2 hours ago
$begingroup$
The typical convention, by the way, is that the voltage drop (from + to -) across a resistor with a current I going from + to - is V = IR.
$endgroup$
– Metric
2 hours ago
$begingroup$
The typical convention, by the way, is that the voltage drop (from + to -) across a resistor with a current I going from + to - is V = IR.
$endgroup$
– Metric
2 hours ago
$begingroup$
my goodness I thought I was going insane. Thank you so much for your input.
$endgroup$
– CuriousJ
2 hours ago
$begingroup$
my goodness I thought I was going insane. Thank you so much for your input.
$endgroup$
– CuriousJ
2 hours ago
$begingroup$
Glad I could reduce the insanity :P.
$endgroup$
– Metric
2 hours ago
$begingroup$
Glad I could reduce the insanity :P.
$endgroup$
– Metric
2 hours ago
1
1
$begingroup$
How can you get 1.8V across a resistor, regardless of polarity, if the only source has a voltage of 1.2V? The voltages across the resistors in the loop must add to the source voltage, and you can't have one negative and the other positive unless the current changes direction between the resistors.
$endgroup$
– Elliot Alderson
2 hours ago
$begingroup$
How can you get 1.8V across a resistor, regardless of polarity, if the only source has a voltage of 1.2V? The voltages across the resistors in the loop must add to the source voltage, and you can't have one negative and the other positive unless the current changes direction between the resistors.
$endgroup$
– Elliot Alderson
2 hours ago
|
show 4 more comments
CuriousJ is a new contributor. Be nice, and check out our Code of Conduct.
CuriousJ is a new contributor. Be nice, and check out our Code of Conduct.
CuriousJ is a new contributor. Be nice, and check out our Code of Conduct.
CuriousJ is a new contributor. Be nice, and check out our Code of Conduct.
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3
$begingroup$
You shouldn't have flipped the battery polarity. (or you mis-flipped it in your calculations.)
$endgroup$
– Peter Bennett
2 hours ago
$begingroup$
Small detail but the question lists R as 200ohms and the current as 3mA. You have it as R being 3ohms and the current as 200mA. Though when multiplied they'd give the same thing, sometimes in an exam you reference values from the circuit you redrew and if you copy the wrong values well...
$endgroup$
– Simeon R
2 hours ago
$begingroup$
Don't use the same letter and case for variables and units. Units shouldn't be inside equations in any case.
$endgroup$
– Chu
1 hour ago