# Help with Circuit Homework: VAB=125V, R1=17.0, R2=84.0, R3=24.0

• dgoudie
In summary, the first question was to find the resistor across AB, and the computer said it was 82 ohms. The next question was about a current divider and I couldn't get it. The third question was about homework and it states that the first question was found correctly, but the computer gives a bit of error room for the answer.
dgoudie

## Homework Statement

R1=17.0, R2=84.0 , R3=24.0
http://capaserv.physics.mun.ca/capalibrary/Graphics/Gtype1021/A5Problem27.gif
The first question was find the Req across AB, I got that to be 82.0 ohms, which the computer said was right,

Next question is :
If VAB = 125.0 V, what is the current flowing between points C and D through R2?

Hint:Find the main current, and do a loop equation for loop ACDBA

V=IR

## The Attempt at a Solution

I tried just 125V/R2, which was wrong and I tried to do a loop equation(havent been taught this yet tried to do from book) and did 125/(17+84) but that was also wrong. If anyone can help me out it would be greatly appreciated (tried 1.49 and 1.24 A)

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Where must the current come from? The source right and it must all exit at the other end.

The current flowing in the loop ACDB ≡ the current flowing through the initial R1 and the bottom R1. From node C to node D however it is flowing is some hellish fashion they don't ask you about really.

But the one thing you know about the total flow from C to D is that the current isn't not adding up to what enters and what exits, because it must add up.

Hence the current flow between C and D must be the current flowing from the Voltage source.

This would then be V/Requivalent (the answer you found in the first part.)

The Req you mention, is that just for loop ACDB? or for the whole circuit? The one I tried was the whole circuit so maybe that's where I went wrong

Incidentally, I get a slightly different number for the first part.

I get 84.7 Ω

Hmm, well the Computer accepted my answer for part one, but it does give a bit of error room.

Well I just maxed out all my tries, but thanks anyway. I'd still like to understand how to do it for future reference. my last try was 125/84.7 which didnt work :(.

The next question was all about loops and I couldn't get it either, even following the book step by step. stupid profs giving stuff they haven't taught us yet.

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Well, to find the current flowing from C to D, you must find the potential difference (read: voltage) across these two points to apply Ohm's law. You have to replace the given circuit with an equivalent (has the same full resistance but is done with different resistors) Do you know how to do it for this kind circuit?

EDIT: Cut off the source and the resistors connected between A and C, B and D, leaving just the part which shows all possible ways of the current flowing from C to D. Now compute the total resistance for it to flow from C to D through anywhere but R2 (the shortest way). Then apply the Kirchoff's law. You have the total current and the current flowing NOT through R2 from C to D, so it's just a subtraction left.

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kbaumen is correct. I misread what was asked. I didn't see you wanted the current just through the R2 between CD.

That makes it a simple current divider problem.

The current flowing into C is split between the R2||Req the rest of the mess to the right.

Figure the Req of everything to the right and then determine the portion of the current flowing just through R2. (Current through R2 and through Req is equal to the total coming from the voltage source.)

Sorry for any confusion my misreading it may have caused.

## 1. What is the total resistance in this circuit?

The total resistance in this circuit can be calculated using the formula R = R1 + R2 + R3. Plugging in the given values, we get R = 17.0 + 84.0 + 24.0 = 125 ohms.

## 2. What is the current flowing through the circuit?

To determine the current flowing through the circuit, we can use Ohm's Law which states that I = V/R, where I is the current, V is the voltage, and R is the resistance. Plugging in the given values, we get I = 125V/125 ohms = 1 amp.

## 3. What is the voltage drop across each resistor?

The voltage drop across each resistor can be calculated using Ohm's Law again. The voltage drop across R1 is V1 = I * R1 = 1 amp * 17.0 ohms = 17 volts. Similarly, V2 = I * R2 = 1 amp * 84.0 ohms = 84 volts and V3 = I * R3 = 1 amp * 24.0 ohms = 24 volts.

## 4. What is the power dissipated by each resistor?

The power dissipated by each resistor can be calculated using the formula P = I^2 * R, where P is the power, I is the current, and R is the resistance. Plugging in the values, we get P1 = (1 amp)^2 * 17.0 ohms = 17 watts, P2 = (1 amp)^2 * 84.0 ohms = 84 watts, and P3 = (1 amp)^2 * 24.0 ohms = 24 watts.

## 5. How can I simplify this circuit?

This circuit can be simplified using the equivalent resistance formula for resistors in parallel, which is 1/R = 1/R1 + 1/R2 + 1/R3. Plugging in the values, we get 1/R = 1/17.0 + 1/84.0 + 1/24.0 = 0.0833 + 0.0119 + 0.0417 = 0.1369. Therefore, R = 1/0.1369 = 7.3 ohms. So, the circuit can be simplified to a single resistor of 7.3 ohms in series with the voltage source.

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