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

In summary: R3 is then how are you supposed to find the equivalent??If you don't know what resistor R3 is, then you can't use the equation V = I*R. You would need to find the equivalent resistance for R3 first and then use that in the equation.
  • #1
ElegantSir
37
0

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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  • #2
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.
 
  • #3
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?
 
  • #4
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 ?
 
  • #5
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.
 
  • #6
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
 
  • #7
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??
 
  • #8
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.
 
  • #9
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?
 

1. What is the difference between R, P, V, and I in electric circuits?

R, P, V, and I are all important components in understanding and analyzing electric circuits. R stands for resistance, which is the measure of how much a material opposes the flow of electric current. P stands for power, which is the rate at which energy is transferred in a circuit. V stands for voltage, which is the difference in electric potential between two points in a circuit. I stands for current, which is the flow of electric charge through a circuit.

2. How do you calculate the total resistance of a series circuit?

The total resistance of a series circuit is calculated by adding up the individual resistances of each component in the circuit. This is because in a series circuit, the current flows through each component in succession, so the total resistance is equal to the sum of all the individual resistances.

3. Can you explain Ohm's Law and how it relates to electric circuits?

Ohm's Law states that the current through a conductor between two points is directly proportional to the voltage across the two points, and inversely proportional to the resistance between them. This means that as voltage increases, current will also increase, and as resistance increases, current will decrease. In electric circuits, Ohm's Law helps us understand the relationship between voltage, current, and resistance.

4. How do you calculate the power dissipated in a circuit?

The power dissipated in a circuit can be calculated using the formula P = VI, where P is power in watts, V is voltage in volts, and I is current in amps. This formula can also be rewritten as P = I²R or P = V²/R, depending on the information given in the problem. It is important to note that power is measured in watts, which is equal to joules per second.

5. What is the difference between a parallel circuit and a series circuit?

In a parallel circuit, the components are connected in such a way that there are multiple paths for the current to flow through. This means that the voltage across each component will be the same, but the current will be divided between the branches. In a series circuit, the components are connected in a single loop, so the same current flows through each component, but the voltage may be different across each component. Additionally, if one component breaks in a series circuit, the entire circuit will stop functioning, whereas in a parallel circuit, only the branch with the broken component will be affected.

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