Using KVL to Solve for Vo: Kirchoff's Voltage Law Homework"

In summary, the problem is to determine the voltage (Vo) in a circuit using KVL. One way to solve this is by realizing it is a potential divider with an overall potential difference of 6V. Another way is by calculating the current through the resistors and using Ohm's law to find Vo. The potential divider equation can also be used to find Vo, resulting in a value of 5.25V. However, using Ohm's law results in a value of 4.5V, which may be the correct solution.
  • #1
tanky322
43
0

Homework Statement



Determine Vo in the circuit below:

Schematic.jpg

Homework Equations



V1 +V2+V3+...Vn=0

The Attempt at a Solution



I think this should be solved using KVL, but I have no idea on where to exactly start.

By looking at the problem, I am thinking it will be 6.0V because 9.0V-3.0V=6.0V,
but don't think that's right.

Any help would be greatly appreciated
 
Last edited:
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  • #2
Realise that you've got a potential divider across the two resistors, with an overall p.d. of 6V. 6V split in 6:2... so what's the voltage in the middle?
 
  • #3
genneth said:
Realise that you've got a potential divider across the two resistors, with an overall p.d. of 6V. 6V split in 6:2... so what's the voltage in the middle?

Im kind of lost on what you just said, could you elaborate a little?

Andrew
 
  • #4
Calculate the current through the two resistors and then the voltage drop across the 6 ohm resistor.
 
  • #5
Potential divider...

Well, what you have is two resistors in series, which is a potential divider.
In this case one end of the divider is not attached to the "ground", but to some other voltage. You can apply usual potential divider equations, just remember to add your reference voltage (3.0V) to it.

The difference between voltages at ends of the divider is 6V.

Voltage in the middle, referenced to 3V (right end of the divider) is

[tex]V_{m(3V)}=6V\frac{2.0\Omega}{6.0\Omega+2.0\Omega}=1.5V[/tex]

This is referenced to the right end of the divider. If you want it referenced to "ground", i.e. the [tex]V_o[/tex] voltage, just add 3V to it.

You can also reference everything to 9V. Then the divider is powered from -6V, and

[tex]V_{m(9V)}=-6V\frac{6.0\Omega}{6.0\Omega+2.0\Omega}=-4.5V[/tex]

This is referenced to the left end of the divider. To reference it to "ground", add 9V to it.

You will obtain the same answer either way.

The resistive divider "equation" I've used above can be derived from Ohm's and Kirchhof's laws.
 
  • #6
kuba said:
Well, what you have is two resistors in series, which is a potential divider.
In this case one end of the divider is not attached to the "ground", but to some other voltage. You can apply usual potential divider equations, just remember to add your reference voltage (3.0V) to it.

The difference between voltages at ends of the divider is 6V.

Voltage in the middle, referenced to 3V (right end of the divider) is

[tex]V_{m(3V)}=6V\frac{2.0\Omega}{6.0\Omega+2.0\Omega}=1.5V[/tex]

This is referenced to the right end of the divider. If you want it referenced to "ground", i.e. the [tex]V_o[/tex] voltage, just add 3V to it.

You can also reference everything to 9V. Then the divider is powered from -6V, and

[tex]V_{m(9V)}=-6V\frac{6.0\Omega}{6.0\Omega+2.0\Omega}=-4.5V[/tex]

This is referenced to the left end of the divider. To reference it to "ground", add 9V to it.

You will obtain the same answer either way.

The resistive divider "equation" I've used above can be derived from Ohm's and Kirchhof's laws.

Thanks for the replies! My only question is that we havnet discussed potential dividers yet. Is this the only way of doing this or is it possible in some other way?
 
  • #7
tanky322 said:
Thanks for the replies! My only question is that we havnet discussed potential dividers yet. Is this the only way of doing this or is it possible in some other way?

As I said in my previous post, calculate the current through the 2 resistors:
[tex]i = \frac {9-3}{6 + 2}[/tex]
The voltage is:
[tex]V_0 = 9 - 6 i[/tex]
 
  • #8
So using the potential divider formula, this is what I came to:

Vo=(V1(R2/(R1+R2))+V2

Plugging that in it gives me 5.25V.
 
  • #9
CEL said:
As I said in my previous post, calculate the current through the 2 resistors:
[tex]i = \frac {9-3}{6 + 2}[/tex]
The voltage is:
[tex]V_0 = 9 - 6 i[/tex]


Doing it this way I get i=6/8=.75A
Vo=9-6(.75)=4.5V

Im thinking that this way is correct.

Thanks! Sorry I didnt even notice your post.
 

FAQ: Using KVL to Solve for Vo: Kirchoff's Voltage Law Homework"

What is KVL and how is it used to solve for Vo?

KVL stands for Kirchoff's Voltage Law and it is a fundamental law in circuit analysis. It states that the algebraic sum of voltages around a closed loop in a circuit must equal zero. To solve for Vo using KVL, we set up a loop in the circuit and use the equation V1 + V2 + V3 + ... + Vn = 0, where V1, V2, V3, etc. are the voltages across each component in the loop. We can rearrange this equation to solve for Vo.

Can KVL be applied to any type of circuit?

Yes, KVL can be applied to any type of circuit, including DC, AC, and mixed circuits. It is a fundamental law that applies to all circuits, regardless of their complexity.

What is the difference between KVL and KCL (Kirchoff's Current Law)?

KVL deals with voltages in a circuit, while KCL deals with currents. KVL states that the sum of voltages around a closed loop must equal zero, while KCL states that the algebraic sum of currents entering and leaving a node must equal zero. In other words, KVL deals with the voltage drops in a circuit, while KCL deals with the current flow at a specific point in the circuit.

Are there any limitations to using KVL to solve for Vo?

While KVL is a powerful tool for solving circuit problems, it does have some limitations. It can only be applied to linear circuits, meaning circuits where the components obey Ohm's Law. Additionally, it can only be applied to circuits with a single closed loop.

Can KVL be used to determine the voltage at any point in a circuit?

Yes, KVL can be used to determine the voltage at any point in a circuit, as long as the point is part of the closed loop being analyzed. By setting up a loop that includes the desired point, we can use KVL to determine the voltage at that point. However, if the point is not part of the loop, KVL cannot be used to determine the voltage at that point.

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