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How to calculate the node voltages for this circuit using the voltage divider rule
Circuit Nodal VoltageFormat Approach to NODAL ANALYSISIs it safe to use a voltage divider for Arduino and DC motor controller?How to solve this resistive network using divider rules and ohms lawTransfer function in a 2L2C circuitT-Feedback Circuit Vo/Vi (Why is my approach wrong)?Analysis of a VDB circuitArduino Nano: How to decide voltage divider values that are safe for the input of the ADCRelation between Output voltage and current in a transresistance amplifierhow to calculate voltage divider
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
I am working on the following circuit:
Simplifying the circuit:
After simplifying, my answers do not match that of values I get from simulating the circuit in multisim. Here's my calculation:
$$Voltage V_B$$
$$V_B=6(frac50005000+(32.35714286times10^3))$$
$$V_B=0.8030592734V$$
$$6-0.8030592734=5.20V$$
$$Voltage Vc$$
$$V_c=5.2(frac(32.35714286times10^3)5000+(32.35714286times10^3))$$
$$V_c=4.50V$$
The simulations I did on multisim show that the voltages for Vb and Vc are:
$$V_B=5.161V$$
$$V_c=5.035V$$
I don't know if its the simplification I'm doing wrong or my approach towards the voltage divider rule. I need help.
circuit-analysis voltage-divider
asked May 1 at 15:14
AugieJavax98AugieJavax98
1315
$endgroup$
add a comment |
$begingroup$
I am working on the following circuit:
Simplifying the circuit:
After simplifying, my answers do not match that of values I get from simulating the circuit in multisim. Here's my calculation:
$$Voltage V_B$$
$$V_B=6(frac50005000+(32.35714286times10^3))$$
$$V_B=0.8030592734V$$
$$6-0.8030592734=5.20V$$
$$Voltage Vc$$
$$V_c=5.2(frac(32.35714286times10^3)5000+(32.35714286times10^3))$$
$$V_c=4.50V$$
The simulations I did on multisim show that the voltages for Vb and Vc are:
$$V_B=5.161V$$
$$V_c=5.035V$$
I don't know if its the simplification I'm doing wrong or my approach towards the voltage divider rule. I need help.
circuit-analysis voltage-divider
asked May 1 at 15:14
AugieJavax98AugieJavax98
1315
$endgroup$
2
$begingroup$
Your first circuit has a resistor between C and ground, but your 2nd does not, so they are certainly not equivalent.
$endgroup$
– The Photon
May 1 at 15:21
add a comment |
$begingroup$
I am working on the following circuit:
Simplifying the circuit:
After simplifying, my answers do not match that of values I get from simulating the circuit in multisim. Here's my calculation:
$$Voltage V_B$$
$$V_B=6(frac50005000+(32.35714286times10^3))$$
$$V_B=0.8030592734V$$
$$6-0.8030592734=5.20V$$
$$Voltage Vc$$
$$V_c=5.2(frac(32.35714286times10^3)5000+(32.35714286times10^3))$$
$$V_c=4.50V$$
The simulations I did on multisim show that the voltages for Vb and Vc are:
$$V_B=5.161V$$
$$V_c=5.035V$$
I don't know if its the simplification I'm doing wrong or my approach towards the voltage divider rule. I need help.
circuit-analysis voltage-divider
asked May 1 at 15:14
AugieJavax98AugieJavax98
1315
$endgroup$
I am working on the following circuit:
Simplifying the circuit:
After simplifying, my answers do not match that of values I get from simulating the circuit in multisim. Here's my calculation:
$$Voltage V_B$$
$$V_B=6(frac50005000+(32.35714286times10^3))$$
$$V_B=0.8030592734V$$
$$6-0.8030592734=5.20V$$
$$Voltage Vc$$
$$V_c=5.2(frac(32.35714286times10^3)5000+(32.35714286times10^3))$$
$$V_c=4.50V$$
The simulations I did on multisim show that the voltages for Vb and Vc are:
$$V_B=5.161V$$
$$V_c=5.035V$$
I don't know if its the simplification I'm doing wrong or my approach towards the voltage divider rule. I need help.
circuit-analysis voltage-divider
circuit-analysis voltage-divider
asked May 1 at 15:14
AugieJavax98AugieJavax98
1315
asked May 1 at 15:14
AugieJavax98AugieJavax98
1315
asked May 1 at 15:14
AugieJavax98AugieJavax98
1315
asked May 1 at 15:14
AugieJavax98AugieJavax98
1315
asked May 1 at 15:14
AugieJavax98AugieJavax98
1315
1315
2
$begingroup$
Your first circuit has a resistor between C and ground, but your 2nd does not, so they are certainly not equivalent.
$endgroup$
– The Photon
May 1 at 15:21
add a comment |
2
$begingroup$
Your first circuit has a resistor between C and ground, but your 2nd does not, so they are certainly not equivalent.
$endgroup$
– The Photon
May 1 at 15:21
2
2
$begingroup$
Your first circuit has a resistor between C and ground, but your 2nd does not, so they are certainly not equivalent.
$endgroup$
– The Photon
May 1 at 15:21
$begingroup$
Your first circuit has a resistor between C and ground, but your 2nd does not, so they are certainly not equivalent.
$endgroup$
– The Photon
May 1 at 15:21
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Your first circuit has a resistor between C and GND. You seem to have added it to your bottom resistor. You also seem to have your 1k||3k resistor calculation incorrect. 3k*1k/3k+1k is 750 ohms. Thus you should end up with a circuit like this:
simulate this circuit – Schematic created using CircuitLab
From here, it is simpler to do Ohms Law to find the current in the circuit, then find the voltage drops of all the resistors.
If you must use the voltage divider rule, then you need to know if you are finding the voltage drop(s) across the resistor(s), or the voltage at the points with respect to GND, because that will make a difference to how you calculate it.
Of course, you always have the option to simplify to 2 resistors with the R2 component in the voltage divider as (R2+R3) too. It depends what your task is.
Another thing I noticed is you used your answer from Vb as your input voltage for your second divider equation. You should still use 6V as the supply for both equations. If you do that, you'll end up calculating answers that agree with your simulation.
I ended up with Vb = 5.1608V and Vc = 5.035V
answered May 1 at 15:37
MCGMCG
7,09531851
$endgroup$
$begingroup$
I am asked to calculate the node voltages for B and C using the voltage divider rule
$endgroup$
– AugieJavax98
May 1 at 15:44
$begingroup$
So that's the node voltages with respect to GND I assume? In that case, it should be fairly straightforward now. You have demonstrated you know the voltage divider rule. Simplify the circuit to how I have it and apply it again, You should get answers that match your simulation, as I did
$endgroup$
– MCG
May 1 at 15:50
$begingroup$
Yes it is with respect to GND. I will try again.
$endgroup$
– AugieJavax98
May 1 at 15:55
$begingroup$
When I get the answer from the divider equation, I have to subtract it from the source voltage right?
$endgroup$
– AugieJavax98
May 1 at 16:00
$begingroup$
No, you do the normal voltage divider equation, and you get your answer. You don't have to subtract anything from the source. You should get answers that match your sims
$endgroup$
– MCG
May 1 at 16:01
|
show 5 more comments
$begingroup$
The calculation for 1k and 3k resistors looks wrong:
$R_eq=frac1frac11000+frac13000=frac3000*10001000+3000=750Omega$
answered May 1 at 15:19
laptop2dlaptop2d
29.3k123787
$endgroup$
$begingroup$
R8 is missing as well..
$endgroup$
– Eugene Sh.
May 1 at 15:20
$begingroup$
R8 got rolled into the 32k, the total for all three would be 30.75k
$endgroup$
– laptop2d
May 1 at 15:21
$begingroup$
Well, then it can't have BC voltage
$endgroup$
– Eugene Sh.
May 1 at 15:22
$begingroup$
I see your point
$endgroup$
– laptop2d
May 1 at 15:22
add a comment |
$begingroup$
First off, you incorrectly reduced the combination of 1k||3k to 2.35k. This is incorrect on it's face since whenever you have resistors in parallel, the combined resistance is less than either resistance. You should use the formula (1/R1 + 1/R2 + ... + 1/Rn)^-1 to reduce the parallel circuit. When this is applied you get a more reasonable value of 750.
Second, you mislabeled node C. On the top circuit it is the node between R6||R7 and R8. On the bottom circuit, it is tied to ground.
answered May 1 at 15:24
OscillonoscopeOscillonoscope
1244
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your first circuit has a resistor between C and GND. You seem to have added it to your bottom resistor. You also seem to have your 1k||3k resistor calculation incorrect. 3k*1k/3k+1k is 750 ohms. Thus you should end up with a circuit like this:
simulate this circuit – Schematic created using CircuitLab
From here, it is simpler to do Ohms Law to find the current in the circuit, then find the voltage drops of all the resistors.
If you must use the voltage divider rule, then you need to know if you are finding the voltage drop(s) across the resistor(s), or the voltage at the points with respect to GND, because that will make a difference to how you calculate it.
Of course, you always have the option to simplify to 2 resistors with the R2 component in the voltage divider as (R2+R3) too. It depends what your task is.
Another thing I noticed is you used your answer from Vb as your input voltage for your second divider equation. You should still use 6V as the supply for both equations. If you do that, you'll end up calculating answers that agree with your simulation.
I ended up with Vb = 5.1608V and Vc = 5.035V
answered May 1 at 15:37
MCGMCG
7,09531851
$endgroup$
$begingroup$
I am asked to calculate the node voltages for B and C using the voltage divider rule
$endgroup$
– AugieJavax98
May 1 at 15:44
$begingroup$
So that's the node voltages with respect to GND I assume? In that case, it should be fairly straightforward now. You have demonstrated you know the voltage divider rule. Simplify the circuit to how I have it and apply it again, You should get answers that match your simulation, as I did
$endgroup$
– MCG
May 1 at 15:50
$begingroup$
Yes it is with respect to GND. I will try again.
$endgroup$
– AugieJavax98
May 1 at 15:55
$begingroup$
When I get the answer from the divider equation, I have to subtract it from the source voltage right?
$endgroup$
– AugieJavax98
May 1 at 16:00
$begingroup$
No, you do the normal voltage divider equation, and you get your answer. You don't have to subtract anything from the source. You should get answers that match your sims
$endgroup$
– MCG
May 1 at 16:01
|
show 5 more comments
$begingroup$
Your first circuit has a resistor between C and GND. You seem to have added it to your bottom resistor. You also seem to have your 1k||3k resistor calculation incorrect. 3k*1k/3k+1k is 750 ohms. Thus you should end up with a circuit like this:
simulate this circuit – Schematic created using CircuitLab
From here, it is simpler to do Ohms Law to find the current in the circuit, then find the voltage drops of all the resistors.
If you must use the voltage divider rule, then you need to know if you are finding the voltage drop(s) across the resistor(s), or the voltage at the points with respect to GND, because that will make a difference to how you calculate it.
Of course, you always have the option to simplify to 2 resistors with the R2 component in the voltage divider as (R2+R3) too. It depends what your task is.
Another thing I noticed is you used your answer from Vb as your input voltage for your second divider equation. You should still use 6V as the supply for both equations. If you do that, you'll end up calculating answers that agree with your simulation.
I ended up with Vb = 5.1608V and Vc = 5.035V
answered May 1 at 15:37
MCGMCG
7,09531851
$endgroup$
$begingroup$
I am asked to calculate the node voltages for B and C using the voltage divider rule
$endgroup$
– AugieJavax98
May 1 at 15:44
$begingroup$
So that's the node voltages with respect to GND I assume? In that case, it should be fairly straightforward now. You have demonstrated you know the voltage divider rule. Simplify the circuit to how I have it and apply it again, You should get answers that match your simulation, as I did
$endgroup$
– MCG
May 1 at 15:50
$begingroup$
Yes it is with respect to GND. I will try again.
$endgroup$
– AugieJavax98
May 1 at 15:55
$begingroup$
When I get the answer from the divider equation, I have to subtract it from the source voltage right?
$endgroup$
– AugieJavax98
May 1 at 16:00
$begingroup$
No, you do the normal voltage divider equation, and you get your answer. You don't have to subtract anything from the source. You should get answers that match your sims
$endgroup$
– MCG
May 1 at 16:01
|
show 5 more comments
$begingroup$
Your first circuit has a resistor between C and GND. You seem to have added it to your bottom resistor. You also seem to have your 1k||3k resistor calculation incorrect. 3k*1k/3k+1k is 750 ohms. Thus you should end up with a circuit like this:
simulate this circuit – Schematic created using CircuitLab
From here, it is simpler to do Ohms Law to find the current in the circuit, then find the voltage drops of all the resistors.
If you must use the voltage divider rule, then you need to know if you are finding the voltage drop(s) across the resistor(s), or the voltage at the points with respect to GND, because that will make a difference to how you calculate it.
Of course, you always have the option to simplify to 2 resistors with the R2 component in the voltage divider as (R2+R3) too. It depends what your task is.
Another thing I noticed is you used your answer from Vb as your input voltage for your second divider equation. You should still use 6V as the supply for both equations. If you do that, you'll end up calculating answers that agree with your simulation.
I ended up with Vb = 5.1608V and Vc = 5.035V
answered May 1 at 15:37
MCGMCG
7,09531851
$endgroup$
Your first circuit has a resistor between C and GND. You seem to have added it to your bottom resistor. You also seem to have your 1k||3k resistor calculation incorrect. 3k*1k/3k+1k is 750 ohms. Thus you should end up with a circuit like this:
simulate this circuit – Schematic created using CircuitLab
From here, it is simpler to do Ohms Law to find the current in the circuit, then find the voltage drops of all the resistors.
If you must use the voltage divider rule, then you need to know if you are finding the voltage drop(s) across the resistor(s), or the voltage at the points with respect to GND, because that will make a difference to how you calculate it.
Of course, you always have the option to simplify to 2 resistors with the R2 component in the voltage divider as (R2+R3) too. It depends what your task is.
Another thing I noticed is you used your answer from Vb as your input voltage for your second divider equation. You should still use 6V as the supply for both equations. If you do that, you'll end up calculating answers that agree with your simulation.
I ended up with Vb = 5.1608V and Vc = 5.035V
answered May 1 at 15:37
MCGMCG
7,09531851
edited May 1 at 15:53
answered May 1 at 15:37
MCGMCG
7,09531851
answered May 1 at 15:37
MCGMCG
7,09531851
answered May 1 at 15:37
MCGMCG
7,09531851
7,09531851
$begingroup$
I am asked to calculate the node voltages for B and C using the voltage divider rule
$endgroup$
– AugieJavax98
May 1 at 15:44
$begingroup$
So that's the node voltages with respect to GND I assume? In that case, it should be fairly straightforward now. You have demonstrated you know the voltage divider rule. Simplify the circuit to how I have it and apply it again, You should get answers that match your simulation, as I did
$endgroup$
– MCG
May 1 at 15:50
$begingroup$
Yes it is with respect to GND. I will try again.
$endgroup$
– AugieJavax98
May 1 at 15:55
$begingroup$
When I get the answer from the divider equation, I have to subtract it from the source voltage right?
$endgroup$
– AugieJavax98
May 1 at 16:00
$begingroup$
No, you do the normal voltage divider equation, and you get your answer. You don't have to subtract anything from the source. You should get answers that match your sims
$endgroup$
– MCG
May 1 at 16:01
|
show 5 more comments
$begingroup$
I am asked to calculate the node voltages for B and C using the voltage divider rule
$endgroup$
– AugieJavax98
May 1 at 15:44
$begingroup$
So that's the node voltages with respect to GND I assume? In that case, it should be fairly straightforward now. You have demonstrated you know the voltage divider rule. Simplify the circuit to how I have it and apply it again, You should get answers that match your simulation, as I did
$endgroup$
– MCG
May 1 at 15:50
$begingroup$
Yes it is with respect to GND. I will try again.
$endgroup$
– AugieJavax98
May 1 at 15:55
$begingroup$
When I get the answer from the divider equation, I have to subtract it from the source voltage right?
$endgroup$
– AugieJavax98
May 1 at 16:00
$begingroup$
No, you do the normal voltage divider equation, and you get your answer. You don't have to subtract anything from the source. You should get answers that match your sims
$endgroup$
– MCG
May 1 at 16:01
$begingroup$
I am asked to calculate the node voltages for B and C using the voltage divider rule
$endgroup$
– AugieJavax98
May 1 at 15:44
$begingroup$
I am asked to calculate the node voltages for B and C using the voltage divider rule
$endgroup$
– AugieJavax98
May 1 at 15:44
$begingroup$
So that's the node voltages with respect to GND I assume? In that case, it should be fairly straightforward now. You have demonstrated you know the voltage divider rule. Simplify the circuit to how I have it and apply it again, You should get answers that match your simulation, as I did
$endgroup$
– MCG
May 1 at 15:50
$begingroup$
So that's the node voltages with respect to GND I assume? In that case, it should be fairly straightforward now. You have demonstrated you know the voltage divider rule. Simplify the circuit to how I have it and apply it again, You should get answers that match your simulation, as I did
$endgroup$
– MCG
May 1 at 15:50
$begingroup$
Yes it is with respect to GND. I will try again.
$endgroup$
– AugieJavax98
May 1 at 15:55
$begingroup$
Yes it is with respect to GND. I will try again.
$endgroup$
– AugieJavax98
May 1 at 15:55
$begingroup$
When I get the answer from the divider equation, I have to subtract it from the source voltage right?
$endgroup$
– AugieJavax98
May 1 at 16:00
$begingroup$
When I get the answer from the divider equation, I have to subtract it from the source voltage right?
$endgroup$
– AugieJavax98
May 1 at 16:00
$begingroup$
No, you do the normal voltage divider equation, and you get your answer. You don't have to subtract anything from the source. You should get answers that match your sims
$endgroup$
– MCG
May 1 at 16:01
$begingroup$
No, you do the normal voltage divider equation, and you get your answer. You don't have to subtract anything from the source. You should get answers that match your sims
$endgroup$
– MCG
May 1 at 16:01
|
show 5 more comments
$begingroup$
The calculation for 1k and 3k resistors looks wrong:
$R_eq=frac1frac11000+frac13000=frac3000*10001000+3000=750Omega$
answered May 1 at 15:19
laptop2dlaptop2d
29.3k123787
$endgroup$
$begingroup$
R8 is missing as well..
$endgroup$
– Eugene Sh.
May 1 at 15:20
$begingroup$
R8 got rolled into the 32k, the total for all three would be 30.75k
$endgroup$
– laptop2d
May 1 at 15:21
$begingroup$
Well, then it can't have BC voltage
$endgroup$
– Eugene Sh.
May 1 at 15:22
$begingroup$
I see your point
$endgroup$
– laptop2d
May 1 at 15:22
add a comment |
$begingroup$
The calculation for 1k and 3k resistors looks wrong:
$R_eq=frac1frac11000+frac13000=frac3000*10001000+3000=750Omega$
answered May 1 at 15:19
laptop2dlaptop2d
29.3k123787
$endgroup$
$begingroup$
R8 is missing as well..
$endgroup$
– Eugene Sh.
May 1 at 15:20
$begingroup$
R8 got rolled into the 32k, the total for all three would be 30.75k
$endgroup$
– laptop2d
May 1 at 15:21
$begingroup$
Well, then it can't have BC voltage
$endgroup$
– Eugene Sh.
May 1 at 15:22
$begingroup$
I see your point
$endgroup$
– laptop2d
May 1 at 15:22
add a comment |
$begingroup$
The calculation for 1k and 3k resistors looks wrong:
$R_eq=frac1frac11000+frac13000=frac3000*10001000+3000=750Omega$
answered May 1 at 15:19
laptop2dlaptop2d
29.3k123787
$endgroup$
The calculation for 1k and 3k resistors looks wrong:
$R_eq=frac1frac11000+frac13000=frac3000*10001000+3000=750Omega$
answered May 1 at 15:19
laptop2dlaptop2d
29.3k123787
edited May 1 at 15:21
answered May 1 at 15:19
laptop2dlaptop2d
29.3k123787
answered May 1 at 15:19
laptop2dlaptop2d
29.3k123787
answered May 1 at 15:19
laptop2dlaptop2d
29.3k123787
29.3k123787
$begingroup$
R8 is missing as well..
$endgroup$
– Eugene Sh.
May 1 at 15:20
$begingroup$
R8 got rolled into the 32k, the total for all three would be 30.75k
$endgroup$
– laptop2d
May 1 at 15:21
$begingroup$
Well, then it can't have BC voltage
$endgroup$
– Eugene Sh.
May 1 at 15:22
$begingroup$
I see your point
$endgroup$
– laptop2d
May 1 at 15:22
add a comment |
$begingroup$
R8 is missing as well..
$endgroup$
– Eugene Sh.
May 1 at 15:20
$begingroup$
R8 got rolled into the 32k, the total for all three would be 30.75k
$endgroup$
– laptop2d
May 1 at 15:21
$begingroup$
Well, then it can't have BC voltage
$endgroup$
– Eugene Sh.
May 1 at 15:22
$begingroup$
I see your point
$endgroup$
– laptop2d
May 1 at 15:22
$begingroup$
R8 is missing as well..
$endgroup$
– Eugene Sh.
May 1 at 15:20
$begingroup$
R8 is missing as well..
$endgroup$
– Eugene Sh.
May 1 at 15:20
$begingroup$
R8 got rolled into the 32k, the total for all three would be 30.75k
$endgroup$
– laptop2d
May 1 at 15:21
$begingroup$
R8 got rolled into the 32k, the total for all three would be 30.75k
$endgroup$
– laptop2d
May 1 at 15:21
$begingroup$
Well, then it can't have BC voltage
$endgroup$
– Eugene Sh.
May 1 at 15:22
$begingroup$
Well, then it can't have BC voltage
$endgroup$
– Eugene Sh.
May 1 at 15:22
$begingroup$
I see your point
$endgroup$
– laptop2d
May 1 at 15:22
$begingroup$
I see your point
$endgroup$
– laptop2d
May 1 at 15:22
add a comment |
$begingroup$
First off, you incorrectly reduced the combination of 1k||3k to 2.35k. This is incorrect on it's face since whenever you have resistors in parallel, the combined resistance is less than either resistance. You should use the formula (1/R1 + 1/R2 + ... + 1/Rn)^-1 to reduce the parallel circuit. When this is applied you get a more reasonable value of 750.
Second, you mislabeled node C. On the top circuit it is the node between R6||R7 and R8. On the bottom circuit, it is tied to ground.
answered May 1 at 15:24
OscillonoscopeOscillonoscope
1244
$endgroup$
add a comment |
$begingroup$
First off, you incorrectly reduced the combination of 1k||3k to 2.35k. This is incorrect on it's face since whenever you have resistors in parallel, the combined resistance is less than either resistance. You should use the formula (1/R1 + 1/R2 + ... + 1/Rn)^-1 to reduce the parallel circuit. When this is applied you get a more reasonable value of 750.
Second, you mislabeled node C. On the top circuit it is the node between R6||R7 and R8. On the bottom circuit, it is tied to ground.
answered May 1 at 15:24
OscillonoscopeOscillonoscope
1244
$endgroup$
add a comment |
$begingroup$
First off, you incorrectly reduced the combination of 1k||3k to 2.35k. This is incorrect on it's face since whenever you have resistors in parallel, the combined resistance is less than either resistance. You should use the formula (1/R1 + 1/R2 + ... + 1/Rn)^-1 to reduce the parallel circuit. When this is applied you get a more reasonable value of 750.
Second, you mislabeled node C. On the top circuit it is the node between R6||R7 and R8. On the bottom circuit, it is tied to ground.
answered May 1 at 15:24
OscillonoscopeOscillonoscope
1244
$endgroup$
First off, you incorrectly reduced the combination of 1k||3k to 2.35k. This is incorrect on it's face since whenever you have resistors in parallel, the combined resistance is less than either resistance. You should use the formula (1/R1 + 1/R2 + ... + 1/Rn)^-1 to reduce the parallel circuit. When this is applied you get a more reasonable value of 750.
Second, you mislabeled node C. On the top circuit it is the node between R6||R7 and R8. On the bottom circuit, it is tied to ground.
answered May 1 at 15:24
OscillonoscopeOscillonoscope
1244
answered May 1 at 15:24
OscillonoscopeOscillonoscope
1244
answered May 1 at 15:24
OscillonoscopeOscillonoscope
1244
answered May 1 at 15:24
OscillonoscopeOscillonoscope
1244
1244
add a comment |
add a comment |
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v QlLFN,902f7w90XLC4yzFDmThsmc,xXrEiponl
2
$begingroup$
Your first circuit has a resistor between C and ground, but your 2nd does not, so they are certainly not equivalent.
$endgroup$
– The Photon
May 1 at 15:21