Power dissipated by a resistor in parallel

In summary, the power dissipated in the 12 ohm resistor is 28.2 W and the power dissipated in the 5 ohm resistor is 20 W.
  • #1
anban
20
0

Homework Statement



We are given a circuit as follows: V= 10 V and two resistors are in parallel. R1 is 12 ohms and R2 is 5 ohms. What is the power dissipated in the 12 ohm resistor?


Homework Equations



V=IR
P=(I^2)R

The Attempt at a Solution



I know that the total resistance of the circuit is 3.54 ohms... I found this by solving (1/Rtot) = (1/12)+(1/5) .

The total current of the circuit is 2.82 A, which I found by using V=IR, or 10 = (3.54) I. I will call this current I3.

I will just say that the current going through R1 is named I1 and the current going through R2 is named I2. Using V = IR, I can then say that 10 = (R1*I1) + (R2*I2). I can also say that I1 + I2 = 2.82 A.

I also know that P= V^2 / R. I can deduce that the total power dissipated is (10)^2/3.54, which is 28.2 W.

If the total power dissipated is the sum of each resistor's power dissipation, then 28.2 = (I1^2)(R1) + (I2^2)(R2).

I have a lot of pieces but I still am not sure how to figure out the power dissipated through a single resistor, especially since they have different resistances. Any guidance on where to go next?
 
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  • #2
Based on the work you've already done, you have the system of equations
\begin{cases}
10 = R_1I_1 + R_2I_2 \\
2.82 = I_1 + I_2
\end{cases}
Since [itex]I_1[/itex] and [itex]I_2[/itex] are the only unknown quantities, you can easily solve this system for either one.

You've stated both that [itex]P = \frac{V^2}{R}[/itex] and V = IR - this tells you that [itex] P = I^2R [/itex]. Once you have found the necessary current, then you can use that relationship to calculate the power dissipated in the individual resistor(s).
 
  • #3
JayneDoe said:
Based on the work you've already done, you have the system of equations
\begin{cases}
10 = R_1I_1 + R_2I_2 \\
2.82 = I_1 + I_2
\end{cases}
Since [itex]I_1[/itex] and [itex]I_2[/itex] are the only unknown quantities, you can easily solve this system for either one.

You've stated both that [itex]P = \frac{V^2}{R}[/itex] and V = IR - this tells you that [itex] P = I^2R [/itex]. Once you have found the necessary current, then you can use that relationship to calculate the power dissipated in the individual resistor(s).

The resistors are in parallel so they will both have the same potential (10V) across them. That renders your first equation incorrect.
 
  • #4
Ah, you're right. Serves me right for not thinking.

In that case, since you know the voltage is the same across both resistors, you know that [itex] 10 = R_1I_1 = R_2I_2 [/itex]. If you replace my first equation with this, then I think that the rest of my post is still fine.

EDIT: However, now that I think about it, my method seems unnecessarily long. Since we know the voltage across both resistors, doesn't it suffice to use the equation [itex]P = \frac{V^2}{R} [/itex]?
 
  • #5
anban said:

Homework Statement



We are given a circuit as follows: V= 10 V and two resistors are in parallel. R1 is 12 ohms and R2 is 5 ohms. What is the power dissipated in the 12 ohm resistor?


Homework Equations



V=IR
P=(I^2)R
There are two more handy expressions for power. Do you know what they are?

The Attempt at a Solution



I know that the total resistance of the circuit is 3.54 ohms... I found this by solving (1/Rtot) = (1/12)+(1/5) .

The total current of the circuit is 2.82 A, which I found by using V=IR, or 10 = (3.54) I. I will call this current I3.

I will just say that the current going through R1 is named I1 and the current going through R2 is named I2. Using V = IR, I can then say that 10 = (R1*I1) + (R2*I2). I can also say that I1 + I2 = 2.82 A.
You can say the second, but not the first. The resistors are in parallel, which means that they both share the same potential difference. The potentials across them do not add, they are equal.
I also know that P= V^2 / R. I can deduce that the total power dissipated is (10)^2/3.54, which is 28.2 W.
Okay. That uses one of the other handy expressions for power.
If the total power dissipated is the sum of each resistor's power dissipation, then 28.2 = (I1^2)(R1) + (I2^2)(R2).
Why not use the same expression for power, P = V2/R, that you used before, but for the individual resistors? You know the potential across each...
 
  • #6
Thanks for all the responses.

I see how 10 = (R1*I1) + (R2*I2) is not valid here because of the loop rule. An electron can only go through one resistor.

Now, if I use P = V^2 /R, then I know that the power dissipated is 8.33 W for the 12 ohm resistor and 20 W for the 5 ohm resistor. I got this by plugging in V= 10 V and then each individual resistance. Is that it? It's that easy?!

If this is the case, I have another small question.

Does the 5 ohm resistor (the smaller resistance of the two) dissipate more power because it has a bigger current going through it, and therefore gets hotter?
 
  • #7
anban said:
Thanks for all the responses.

I see how 10 = (R1*I1) + (R2*I2) is not valid here because of the loop rule. An electron can only go through one resistor.

Now, if I use P = V^2 /R, then I know that the power dissipated is 8.33 W for the 12 ohm resistor and 20 W for the 5 ohm resistor. I got this by plugging in V= 10 V and then each individual resistance. Is that it? It's that easy?!
Yup. It's that easy :smile:
If this is the case, I have another small question.

Does the 5 ohm resistor (the smaller resistance of the two) dissipate more power because it has a bigger current going through it, and therefore gets hotter?

Yes. The potential drop is the same but the current is higher. The other "handy" expression for power is P = VI.
 

Related to Power dissipated by a resistor in parallel

What is power dissipation?

Power dissipation is the rate at which energy is converted into heat within a circuit component, such as a resistor. It is typically measured in watts (W) and is an important factor to consider when designing and analyzing electronic systems.

How is power dissipated by a resistor in parallel calculated?

The power dissipated by a resistor in parallel can be calculated using the formula P = V^2/R, where P is power in watts, V is voltage in volts, and R is resistance in ohms. This formula shows that the power dissipation is proportional to the square of the voltage and inversely proportional to the resistance.

What factors affect power dissipation in a parallel resistor circuit?

The main factors that affect power dissipation in a parallel resistor circuit are the voltage across the resistor, the resistance of the resistor, and the ambient temperature. Higher voltages and lower resistances will result in higher power dissipation, while higher temperatures can decrease power dissipation.

How does power dissipation impact the performance of a circuit?

Power dissipation can impact the performance of a circuit in several ways. It can cause the temperature of the circuit to increase, potentially leading to overheating and damage. It can also affect the efficiency of the circuit by converting some of the input energy into heat instead of usable output energy. Additionally, excessive power dissipation can result in voltage drops and affect the accuracy of the circuit's measurements.

What are some methods for reducing power dissipation in a parallel resistor circuit?

There are several methods for reducing power dissipation in a parallel resistor circuit. One approach is to use resistors with higher power ratings, which can handle larger amounts of power without overheating. Another method is to use resistors with lower resistance values, as this will decrease the power dissipation. Additionally, implementing cooling methods and proper circuit design can also help reduce power dissipation.

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