How Do You Calculate the Power Supplied by Batteries A and B?

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Homework Help Overview

The discussion revolves around calculating the power supplied by two batteries in a circuit, specifically focusing on the equations related to power (P = iV, P = i²R, and P = V²/R) and their application in determining the contributions of each battery.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the different equations for power and express confusion about their applications. There is an exploration of how to determine the power supplied by each battery based on current and voltage values.

Discussion Status

Some participants have provided guidance on using the appropriate power equation depending on the components involved. There is an ongoing exploration of how to apply these equations to the specific circuit in question, with attention to the direction of current flow in relation to the batteries.

Contextual Notes

Participants note the need for component values to verify calculations and emphasize the importance of understanding current direction when determining whether a battery is supplying or absorbing power.

electrohau5
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Homework Statement



Calculate the power Pa and Pb supplied by batteries A and B respectively

(see attached diagram)

Homework Equations



P=iV, P=i^2R, and P=V^2/R

The Attempt at a Solution



I already found i1, i2 and i3 to be 5.714, 1.143, and -4.571 respectively.

I am confused over the difference between P=iV, P=i^2R, and P=V^2/R
so I am not sure what to do next.
 

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You could use I^2R to work out the power dissipated in each resistor BUT that wouldn't necessarily tell you which battery supplied it or how much (at least not without further calculations).

The power delivered by a battery is simply the battery voltage multiplied by the battery current.
 
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electrohau5 said:

Homework Statement



Calculate the power Pa and Pb supplied by batteries A and B respectively

(see attached diagram)

Homework Equations



P=iV, P=i^2R, and P=V^2/R

The Attempt at a Solution



I already found i1, i2 and i3 to be 5.714, 1.143, and -4.571 respectively.
You'll have to supply component values if we're to check your work.
I am confused over the difference between P=iV, P=i^2R, and P=V^2/R
so I am not sure what to do next.
Each of the expressions will yield the same value for power if you know the current, voltage, and resistance of the given component. For components without a specified resistance (such as ideal voltage supplies or current supplies) you would choose the P = IV expression, having obtained the voltage and current for the device.
 
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"I am confused over the difference between P=iV, P=i^2R, and P=V^2/R"
Start with P = VI and recall that V = IR
Substitute IR for V in P = VI and you get P = I^2R
From V = IR, I = V/R Substitute V/R for I and you get P = V^2/R
 
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gneill said:
You'll have to supply component values if we're to check your work.

Each of the expressions will yield the same value for power if you know the current, voltage, and resistance of the given component. For components without a specified resistance (such as ideal voltage supplies or current supplies) you would choose the P = IV expression, having obtained the voltage and current for the device.

So would I in P=IV be I1, since it goes through each battery, and would V be the emf of each battery?
 
electrohau5 said:
So would I in P=IV be I1, since it goes through each battery, and would V be the emf of each battery?

Right. Be careful with the current direction; current flowing out of the + terminal of the battery means the battery is supplying power to the circuit, while current flowing into the + terminal means the battery is sinking (absorbing) power.
 
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What gneil said. In this circuit both batteries are supplying power, however you might come across other problems involving two batteries where the direction of the current means one battery is being charged by the other. In that case the power delivered by the battery being charged is negative.
 
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thanks guys, I think I got this now
 

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