Engineering Understanding Kirchhoff's Voltage Law in Circuit Analysis

AI Thread Summary
The discussion focuses on understanding Kirchhoff's Voltage Law (KVL) in circuit analysis, particularly in relation to a specific circuit problem. Participants clarify that the current through resistors in a closed loop can vary, and they emphasize the importance of simplifying the circuit to find the equivalent resistance. The conversation includes the application of Ohm's Law (V=IR) to determine current and suggests using mesh analysis to solve for currents in different loops. It is noted that the current at point A must be less than 1 A due to the resistances present. Overall, the thread provides insights into circuit simplification and analysis techniques for beginners.
luysion
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Hello.
The question appears not to complex! however I am been on break (and still am haha) but I saw this in an old exam paper and its bugging me that I don't know how to do it!

heres the question;
http://img59.imageshack.us/img59/9868/dfdfmu.jpg

Kirchhoff voltage law comes to mind. However I am confusing myself. I thought that the current across each resistor in one of the closed loops can't be the same? or can it?
anywhos ignore my little rant. Id really appreciate if somone could explain to me the answer cheers.
 
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Is there a way to simplify the circuit to find the current from the battery? Look at how the resistors are arranged.
 
I am new to this as well in my circuits class, so anyone correct me If I am wrong.
At point A the voltage from the battery hasn't reached any electrical resistance yet, correct?
So would not the voltage be the same since as it started 2 volts because it hasn't reached resistors yet?
And solving for current would be just doing a little algebraic manipulation of one form of the equation for voltage

V=IR and resolving to find current I=\frac{V}{R}

?

I think I might be way off... :(
 
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Start from the right of the circuit. There are some resistors in series.

What would be the size of one resistor that could replace these three resistors?

[URL]http://upload.wikimedia.org/math/0/1/5/01541fd01585b4f8ba5ac819e4abc042.png[/URL]


Then, take this new resistor and put it in parallel with the 4 ohm resistor. What is the formula for two resistors in parallel?
dc55458c0154c67e7e8eed2b2e5b835a.png



So, you could put one resistor in here that is equivalent to the 4 resistors.


Now, you have 3 resistors in series. How do you get the equivalent resistance of these 3 resistors?
[URL]http://upload.wikimedia.org/math/0/1/5/01541fd01585b4f8ba5ac819e4abc042.png[/URL]

So, now there is one resistor across a battery. Ohms Law tells you what the current will be.
7564ef88e7c926febabebd837d9f744a.png
 
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hey i think u should use mesh analysis for solving this circuuit. consider separate currents I1 and I2 for the two loops.now apply kirchhoffs voltage law in both the loops sepreately.solve for I1 and I2.
the value of I1 will give u the current at point A.
 
No real need to calculate. The current has to pass the two 1 Ohm resistors on the left, and then at least one more resistor. Hence, the resistance is larger than 2 Ohms, this means the current must be smaller than 2V/2Ohm = 1 A. I only see one answer smaller than one.
 
It should be mental arithmetic anyway.

1 + 2 + 1 = 4

4 // 4 = 2

1 + 2 + 1 = 4

2 / 4 = 0.5
 
hey sorry late reply guys, thanks so much for the help :)
 

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