Solve KVL for R and e: 3.2 Ohms, 0 Volts

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In summary, KVL, or Kirchhoff's Voltage Law, is a fundamental principle in circuit analysis that states the sum of all voltage drops in a closed loop is equal to the sum of all voltage rises. It can be used to solve for unknown variables, such as resistance and voltage, by writing out the equation and rearranging it. KVL can be applied to any type of circuit, but it has limitations, such as assuming ideal components and not considering non-linear effects or changing magnetic fields.
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Homework Statement


The question wants to find R and e (epsilon), I found R to be 3.2 ohms but epsilon is 0 volts, which is very weird. original question is attached.[/B]
http://[ATTACH=full]200051[/ATTACH]

[h2]Homework Equations[/h2][h2]The Attempt at a Solution[/h2]
I also assumed infinite resistance for voltmeter and zero resistance for ammeter[/B]
 

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Please show the work you did to arrive at your result. You must show your attempt before help can be offered.
 

Related to Solve KVL for R and e: 3.2 Ohms, 0 Volts

1. What is KVL and how does it relate to this problem?

KVL stands for Kirchhoff's Voltage Law, which is a fundamental principle in circuit analysis. It states that in a closed loop, the sum of all voltage drops is equal to the sum of all voltage rises. This applies to the circuit in question because it is a closed loop with a known voltage and resistance.

2. How do I solve for R and e using KVL?

To solve for R and e using KVL, you will need to write out the equation using the known values and then solve for the unknown variables. In this case, you would write out the equation 0 = R*I - e, where I is the current flowing through the circuit. Then, you can rearrange the equation to solve for R and e.

3. Can you provide an example of solving KVL for R and e?

Sure, let's say we have a circuit with a 12-volt battery and a 5-ohm resistor. The current flowing through the circuit is 2 amps. We can use KVL to solve for R and e by writing out the equation: 12 = 5*2 - e. Then, we can rearrange to solve for R and e: R = (12 + e)/2 and e = 12 - 10 = 2.

4. Can KVL be used for any type of circuit?

Yes, KVL can be used for any type of circuit as long as it is a closed loop. It is a fundamental principle in circuit analysis and can be applied to both simple and complex circuits.

5. Are there any limitations to using KVL?

While KVL is a useful tool for solving circuit problems, it does have some limitations. It assumes that all components in the circuit are ideal and does not take into account any non-linear effects. It also does not consider the effects of changing magnetic fields, which can be important in certain circuits.

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