PHY2049ch27B%282-17-10%29

PHY2049ch27B%282-17-10%29 - Multiloop Circuits Multiloop...

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Multi–loop Circuits Multi–loop circuits may contain multiple power supplies so we first consider multiple power supplies in a single loop. Many devices that you use require multiple batteries. Generally this is because they need a higher voltage than is provided by a single battery. Power supplies in series, oriented so that they face the same direction, boost the potential difference between the ends. Consider: + 1 ξ 2 ξ 3 ξ R a b With the potential at point a taken to be V = 0 , Kirchhoff’s loop rule gives, 123 R V0 ξ+ξ+ξ− =
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+ 1 ξ 2 ξ 3 ξ R a b 123 R V ξ +ξ +ξ = Evidently the potential difference between a and b is the sum of the supply potentials. (if each is a 1.5 V battery the net potential between a and b is 4.5 V ). A power supply effectively takes current in at it’s negative terminal passes the charge as an internal current to its positive terminal, raising the potential of that charge to its rated potential difference and passes that charge out of the positive terminal with the same current . ξ The current through a single loop can only go in one direction. Here because each supply is in the same direction they all work to boost the total current.
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Suppose now that the 3 rd supply is flipped around , 123 iR ξ+ξ+ξ= + 1 ξ 2 ξ 3 ξ R a b The current is found using Ohm’s law to replace R Vi R = i R ξ +ξ +ξ = Then, Where i is clockwise, & greater than if there were only one supply. 1 ξ 2 ξ 3 ξ R a b Now the loop rule gives, iR ξ +ξ −ξ = i R ξ+ξ−ξ =
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+ 1 ξ 2 ξ 3 ξ R a b 123 i R ξ +ξ −ξ = The current remains clockwise so long as but if the current is zero while if the current is actually reversed to circulate counter-clockwise. 12 3 ξ +ξ >ξ 3 ξ+ξ =ξ 3 ξ+ξ<ξ Where would this situation of opposing batteries come up? Battery 3 might be a rechargeable battery being recharged. The current flowing through it “backwards” reverses the chemical processes from when it supplies power, charging it back up. In multi–loop, circuits some loops may contain “reversed” power supplies to perform the circuits intended function.
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Homework hint problem 27.6 + 1 ξ 2 ξ 3 ξ R a b Since only potential differences are relevant we’ve generally chosen the potential at some point of the circuit to be zero, but this need not be the case. We could chose the potential at some point to be any fixed value. Suppose the potential here at point a is V = 100 V . Then starting at a the loop rule now gives, 123 100V iR 100V +ξ +ξ −ξ − = iR 0 ξ+ξ−ξ− = Which is the same as if we’d used V a = 0 b V 100V = +ξ +ξ −ξ But note that now,
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To show how multi–loop problems are solved, we’ll work out the problem we solved last class by series–parallel reduction. ξ 1 R 3 R 2 R 3 i 2 i 1 i In addition to the resistors we label the currents through each leg of the circuit where they could be different with a different symbol.
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PHY2049ch27B%282-17-10%29 - Multiloop Circuits Multiloop...

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