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Find the energy in a circuit, delivered to resistors, by battery, etc 
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#1
May1112, 01:30 PM

P: 297

1. The problem statement, all variables and given/known data
The circuit shown in the figure below is connected for 2.10 min. (Assume R1 = 6.00 Ω, R2 = 1.60 Ω, and V = 16.0 V.) (a) Determine the current in each branch of the circuit. *already did* (b) Find the energy delivered by each battery. (c) Find the energy delivered to each resistor. (d) Identify the type of energy storage transformation that occurs in the operation of the circuit. (e) Find the total amount of energy transformed into internal energy in the resistors. 2. Relevant equations *Not even sure* Summation of current through junction = 0 Summation of potential difference in loop = 0 ΔV = IR U = ΔV / q ? ΔU + ΔEint = 0 ? 3. The attempt at a solution Just need help with parts after A, not sure where to start with regards to them. I already solved for part A with Kirchoff's rules, the currents are: I (left wire / branch) = 1.254 A down I (middle wire / branch) = 0.58 A down I (right wire / branch) = 1.842 A up any help is appreciated. 


#2
May1112, 02:20 PM

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P: 11,621

What is the formula for the energy provided by a voltage source V with a current I flowing OUT of its positive terminal?
What is the formula for the energy dissipated by a resistance with a current I flowing though it? 


#3
May1112, 06:42 PM

P: 297

and P = IΔV But I can't figure out how it relates to work, and how to apply it here. 


#4
May1112, 09:38 PM

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P: 11,621

Find the energy in a circuit, delivered to resistors, by battery, etc



#5
May1112, 10:48 PM

P: 297

I am not even sure how the current / which current applies to what battery... I think it is just for the same wire / branch though (ie only 0.58A applying to just the 4V battery)
I take it though that you calculate power and then multiply by the time? Is this right? Also forgot P = (ΔV)^{2} / R For the middle branch would I simply plug in P = (4V)^{2} / 6 ohms or P = (0.58A)^{2}(6 ohms) and then multiply either by the time (126 seconds in this case) ? 


#6
May1112, 10:55 PM

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P: 11,621

And yes, when you multiply power (J/s) by time (s) you end up with total energy delivered or dissipated over that time. 


#7
May1112, 11:31 PM

P: 297

Did (0.58A)^2 multiplied by 6 ohms and then multiplied by 126 seconds. Do I have to account for the current in the left resistor as well or anything? Or do I use the P = IΔV formula instead for the energy delivered by the battery, and then multiply by time again? 


#8
May1212, 06:05 AM

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#9
May1212, 03:24 PM

P: 297

How about for the middle resistors individually? You said P = I^{2}R can be used for individual resistors, so for the middle resistors can I just use the current 0.58A and then each resistance separately for R and then multiply by time? Is the current through each resistor in the middle branch negative like in the battery, since the current is flowing toward the battery? Or do I have to use potential? 


#10
May1212, 06:02 PM

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#11
May1212, 10:03 PM

P: 297

WebAssign is just weird....
says "You appear to be calculating this correctly using an incorrect value for the current through the 4V battery" yet says the magnitude of the current downwards in the middle is correct (0.58A). Or is the current actually different from that of the branch when using P = (I^2)(R) ? Because as far as I remember, the current through a single wire with anything connected in series (doesn't matter if they are batteries, resistors, or capacitors) is the same value. Got 2 of the individual resistors right doing what you said though. All of the energy is also transformed into internal energy so you just add them in the end. You helped me a lot anyway, thanks gneill! 


#12
May1312, 07:57 AM

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Two things occur to me. One, the program may be being picky about accuracy in this case. You are using 0.58 A for the branch current, but should keep a couple more decimal places for purposes of calculation and round only the answers to the required significant figures. Two, did you submit the value with the correct sign? 


#13
May1412, 01:33 AM

P: 297

Yeah that's what I was thinking to be honest.
I got all of them right now... had to look at the currents again. The current in the middle was actually 0.5877 A not just 0.58 A Next time I won't round off so much, I just did it because I was in a hurry and didn't feel like writing down that many more digits / numbers. Again, thanks for the help. 


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