what is p2 the power dissipated in resistor r2

anban

Preparation Statement

We are tending 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 complete resistance of the circuit is 3.54 ohms... I saved this by solving (1/Rtot) = (1/12)+(1/5) .

The total ongoing 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 state that the current going through R1 is named I1 and the electric current releas direct R2 is named I2. Using V = IR, I can and then say that 10 = (R1*I1) + (R2*I2). I can also suppose that I1 + I2 = 2.82 A.

I also have it away that P= V^2 / R. I can deduce that the add u power dissipated is (10)^2/3.54, which is 28.2 W.

If the totality baron dissipated is the sum of apiece 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 degraded through a single resistor, especially since they have different resistances. Any guidance on where to go close?

JayneDoe

Founded along 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
\final stage{cases}
Since [itex]I_1[/itex] and [itex]I_2[/itex] are the only unknown quantities, you give the axe well 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 up-to-date, then you can use that relationship to look the top executive dissipated in the individual resistor(s).

gneill

Supported the work on 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 region quantities, you can well 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 launch the incumbent current, then you can use that relationship to calculate the power sporting in the item-by-item resistor(s).

The resistors are in parallel so they will both take up the same potential (10V) across them. That renders your first equality incorrect.

JayneDoe

Ah, you're flop. Serves ME right for non thinking.

In that pillowcase, since you know the electromotive force is the same across both resistors, you know that [itex] 10 = R_1I_1 = R_2I_2 [/itex]. If you supercede my prototypical equating with this, then I opine that the rest of my base is still fine.

EDIT: However, now that I think about it, my method seems unnecessarily drawn-out. Since we know the voltage crosswise both resistors, doesn't information technology answer to use the equation [itex]P = \frac{V^2}{R} [/itex]?

gneill

Homework Command

We are inclined 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 resistance?

Prep Equations

V=IR
P=(I^2)R

There are two more than adroit expressions for power. Practise you bang what they are?

The Essay at a Solution

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

The tot current of the circuit is 2.82 A, which I base by victimization V=IR, Beaver State 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 and through R2 is named I2. Using V = Inland Revenue, I can then pronounce that 10 = (R1*I1) + (R2*I2). I can also say that I1 + I2 = 2.82 A.

You rear say the forward, only not the first. The resistors are in parallel, which means that they some partake the same potential difference. The potentials across them do not add, they are equal.

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

Hunky-dory. That uses extraordinary of the other convenient expressions for big businessman.

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

Why non use the same expression for force, P = V²/R, that you used in front, but for the individual resistors? You know the potential crosswise each...

anban

Thanks for all the responses.

I see how 10 = (R1*I1) + (R2*I2) is not valid present because of the loop rule out. An electron can only if down cardinal resistor.

Now, if I use P = V^2 /R, then I know that the power dissipated is 8.33 W for the 12 ohm resistance 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 undergo another small question.

Does the 5 Ohm resistor (the smaller resistor of the two) dissipate many power because it has a large current going finished it, and therefore gets hotter?

gneill

Thanks for each the responses.
I look how 10 = (R1*I1) + (R2*I2) is not valid here because of the loop rule. An negatron can lonesome follow through one resistance.

Now, if I use P = V^2 /R, past 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 and then each individual resistance. Is that it? It's that easy?!

Yup. It's that easy

If this is the case, I throw some other small interrogative.
Does the 5 ohm resistor (the smaller underground of the two) dissipate many power because it has a bigger present-day going through it, and therefore gets hotter?

Yes. The potential drop is the same but the underway is high. The other "W. C. Handy" expression for power is P = VI.

what is p2 the power dissipated in resistor r2

Power dissipated by a resistance in parallel of latitude

Preparation Statement

Homework Equations

The Attempt at a Solution

Answers and Replies

Homework Command

Prep Equations

The Essay at a Solution

Cognate Togs on Power dissipated by a resistor in parallel

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