Basic DC Circuit/Kirchoff problem with 2 power suppliesby wellcoughed Tags: basic, circuit or kirchoff, power, supplies 

#1
Dec1412, 11:06 AM

P: 11

Uploaded with ImageShack.us Hi I can't get my head round Kirchoffs law. I've been alright with series and parallel circuits with one voltage supply, but I've been given a problem with two voltage supplies. Would anyone be able to explain how I would get volt drop, current and power dissipated across these resistors please? I also don't understand why in some cases you add the emf's and sometimes you take them away from each other Ive been given this as an explanation to Kirchoffs Voltage law but it's just gobbledegook to me :/ "In any closed loop in a network, the algebraic sum of the voltage drops (i.e. products of current and resistance) taken around the loop is equal to the resultant emf acting in that loop" Many thanks if you can help :) 



#2
Dec1412, 12:07 PM

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P: 11,401

To explain by way of analogy, think of Kirchhoff's voltage law (KVL) like a walk on a hilly terrain that ends at the same location that it starts. Clearly the sum of all the "ups" has to to equal the sum of all the "downs" along the closed path.
The same is true of KVL: If you "walk" around a closed path in a circuit the sum of the potential changes must be zero. What causes potential changes? A battery will cause a rise or a drop depending on which direction you "walk" through it. A resistor will cause a change in potential that is set by the direction of the current that flows through it; if you walk through the resistor against the flow of current, you see a potential rise. If you walk with the current, you see a potential drop. So, label your circuit with currents shown for each branch. Don't worry about choosing the correct direction beforehand  the math will sort things out; if your guess happened to be "incorrect", then the math will give you a negative value for that current, meaning it's flowing in the opposite direction to your assumption. Next label each component with "+" and "" at its ends to show the direction of potential changes which occur in that component. For batteries it's fixed regardless of current direction. For resistors, a drop occurs in the (assumed) direction of the current flow. Next, identify the closed paths ("circuits") you wish to "walk" around. Usually this means recognizing the obvious loops in the circuit. You only need to find enough loops so that each component is "visited" at least once. Then, for each loop, "walk" around the loop and write down the potential changes as you pass through each component. For resistors, potential change have magnitude I*R and you get the sign from the +/ labels that you made previously. When you arrive back where you started, set the sum equal to zero. This procedure will result in an equation for each loop. Solve the set of equations to find the current values. 



#3
Dec1812, 11:37 AM

P: 11

Uploaded with ImageShack.us I've been asked to find the power dissipated across R3. I'm just wondering if I'm right in my method so far? Loop One: 6v + R1(I1) + R3(I1I2) = 0 I1 = 1.0471698Amps Loop Two: 1.5V + R2(I2) + R3(I2I1)= 0 I2 = 0.71698112 Amps But here's where it gets sticky... To find the current across I3 I've done I1I2 = I3 1.047Amps  0.717 Amps = 0.33Amps I3 = 0.33Amps Voltage Across I3 = I3 x R3 Voltage Across I3 = 0.33 x 15ohms Voltage = 4.95V Power dissipated across R3 = I3 x V3 Power dissipated across R3 = 0.33amps x 4.95V Power dissipated across R3 = 1.6335Watts But the confusion I have is the middle resistor is physically connected (+) to () to the 1.5V battery on the loop two side. Would this have an effect on the R3 volt drop that I haven't calculated for? Thanks if you can help :) 



#4
Dec1812, 12:05 PM

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Basic DC Circuit/Kirchoff problem with 2 power supplies 



#5
Dec1812, 12:22 PM

P: 11

so the method I've used in finding the power dissipated across R3 is correct?
And I've solved it? :) 



#6
Dec1812, 12:29 PM

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P: 11,401

By the way, we usually say that voltages are "across" components, currents flow "through" components, and power is dissipated "by" components Doesn't affect your answers though 



#7
Dec1812, 12:37 PM

P: 11

Thanks for taking the time to help me :)
Much appreciated 


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