Electric Circuits Homework: R, P, V, I Solutions

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Homework Help Overview

The discussion revolves around an electric circuits problem involving resistors, voltage, current, and power calculations. Participants are exploring the relationships between these quantities in a circuit with multiple resistors, specifically focusing on how voltage and current behave across different components.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to determine the voltage and current values that carry over to subsequent resistors in a circuit. There is a focus on calculating resistance using Ohm's law and understanding the implications of voltage drops across resistors in parallel.

Discussion Status

Some participants have provided guidance on using equivalent resistance in calculations, while others are questioning the assumptions made about voltage and current in the circuit. There is an ongoing exploration of how to correctly apply the principles of circuit analysis without reaching a definitive conclusion.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the information available or the methods they are allowed to use. There is a noted uncertainty regarding the definitions and calculations related to equivalent resistance and the behavior of current and voltage in the circuit.

ElegantSir
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Homework Statement


The question is given in the thumbnail...

Homework Equations


R = resistance P = power V = voltage I = current
[/B]
R = V / I
P = V * I
P = I * R
P = V^2 / R

The Attempt at a Solution


The attempt at a solution is also given in the thumbnail...

I know that the total voltage is 3V and after resistor 1: 3V - 1.8V that there will be 1.2 volts remaining. So I'm assuming from here (since I was given the current for resistor 2) that the 1.2 volts carries over to resistor 3, allowing me to determine the resistance r = 1.2v / 0.06a = 20.

Here is where I'm stuck. With resistor 4, how do I know whether the volts or the current is the value that carries over? I know it has to be one of those two values seeing as how resistor 4 gives you its resistance, so if I were to add the voltage or current to it, I could calculate everything I need. So the question stands, is it the 1.2 volts that carries over to resistor 4, or the 0.06 amps?
 

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ElegantSir said:

Homework Statement


The question is given in the thumbnail...

Homework Equations


R = resistance P = power V = voltage I = current
[/B]
R = V / I
P = V * I
P = I * R
P = V^2 / R

The Attempt at a Solution


The attempt at a solution is also given in the thumbnail...

I know that the total voltage is 3V and after resistor 1: 3V - 1.8V that there will be 1.2 volts remaining. So I'm assuming from here (since I was given the current for resistor 2) that the 1.2 volts carries over to resistor 3, allowing me to determine the resistance r = 1.2v / 0.06a = 20.

Here is where I'm stuck. With resistor 4, how do I know whether the volts or the current is the value that carries over? I know it has to be one of those two values seeing as how resistor 4 gives you its resistance, so if I were to add the voltage or current to it, I could calculate everything I need. So the question stands, is it the 1.2 volts that carries over to resistor 4, or the 0.06 amps?
The total current going into the part of the circuit with parallel resistors must equal the total current coming out. The voltage drop in each parallel branch must be the same.
 
SteamKing said:
The total current going into the part of the circuit with parallel resistors must equal the total current coming out. The voltage drop in each parallel branch must be the same.
Ahh okay. So for the last two columns of the chart:

Resistor 3: 20Ω, 0.06A, 1.2V
Resistor 4: 12Ω, 0.06A, 0.6V

And then the power would be Ptotal = Vtotal * I total
P = 3V * 0.30A
P = 0.90 watts
Therefore the power of the circuit would be 0.90 watts. I believe this is correct, yes?
 
ElegantSir said:
Ahh okay. So for the last two columns of the chart:

Resistor 3: 20Ω, 0.06A, 1.2V
Resistor 4: 12Ω, 0.06A, 0.6V

And then the power would be Ptotal = Vtotal * I total
P = 3V * 0.30A
P = 0.90 watts
Therefore the power of the circuit would be 0.90 watts. I believe this is correct, yes?
How did you get R3 = 20Ω ?
 
SteamKing said:
How did you get R3 = 20Ω ?
I know that the amperage had to be 0.06A because the circuit had a total of 0.3A and resistor 2 had 0.24A going through it, so 0.3 - 0.24 = 0.06A for R3. Also after R1, the remaining voltage was 1.2V which was carried to both R2 and R3. I used 1.2V / 0.06A to get a Ω of 20.

Im only assuming that this was correct.
 
ElegantSir said:
I know that the amperage had to be 0.06A because the circuit had a total of 0.3A and resistor 2 had 0.24A going through it, so 0.3 - 0.24 = 0.06A for R3. Also after R1, the remaining voltage was 1.2V which was carried to both R2 and R3. I used 1.2V / 0.06A to get a Ω of 20.

Im only assuming that this was correct.

In the branch with R3 and R4, you must use the equivalent resistance Req in the equation V = I*R.

For this branch, Req = R3 + R4 = R3 + 12Ω
 
SteamKing said:
In the branch with R3 and R4, you must use the equivalent resistance Req in the equation V = I*R.

For this branch, Req = R3 + R4 = R3 + 12Ω
But if you don't know what R3 is then how are you supposed to find the equivalent??
 
ElegantSir said:
But if you don't know what R3 is then how are you supposed to find the equivalent??
You know what the current I is and you know that the voltage drop in each branch is the same. You can calculate the voltage drop in the branch which contains R2 and use this voltage drop to calculate R3, as described above.
 
SteamKing said:
You know what the current I is and you know that the voltage drop in each branch is the same. You can calculate the voltage drop in the branch which contains R2 and use this voltage drop to calculate R3, as described above.
Im not quite following...because if I do that i'd be getting the exact same answer:

The amperage in R2 is 0.24, and its resistance is 5. That means that 0.24 * 5 = 1.2 volts. Which would mean that 1,2 volts also goes into R3. In R3 if the voltage is 1.2 and the amperage is 0.06, you would plug them in. R = V / I R = 1.2volts / 0.06amps resistance = 20.

Im not sure where you were going with the R3 + R4 business...I didnt learn that at all during my lesson. But how would this NOT be the answer?
 
  • #10
ElegantSir said:
Im not quite following...because if I do that i'd be getting the exact same answer:

The amperage in R2 is 0.24, and its resistance is 5. That means that 0.24 * 5 = 1.2 volts. Which would mean that 1,2 volts also goes into R3. In R3 if the voltage is 1.2 and the amperage is 0.06, you would plug them in. R = V / I R = 1.2volts / 0.06amps resistance = 20.

Im not sure where you were going with the R3 + R4 business...I didnt learn that at all during my lesson. But how would this NOT be the answer?
You're not following because the R you are calculating is the Req of the branch.

Remember, Req = 20Ω = R3 + R4, and R4 = 12Ω. So what is R3 ?
 
  • #11
SteamKing said:
You're not following because the R you are calculating is the Req of the branch.

Remember, Req = 20Ω = R3 + R4, and R4 = 12Ω. So what is R3 ?
So R3 would be 8 then...?
 
  • #12
ElegantSir said:
So R3 would be 8 then...?
Correct, but more accurately 8Ω.
 
  • #13
SteamKing said:
Correct, but more accurately 8Ω.
I have the final answer but I am afraid I am not understanding how I got there. You said that Req was equal to 20...but where did that number even come from?
 

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