Cell in circuit is reversed

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Retsek
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#1
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#1
Three identical cells, each of internal resistance R, are connected in series with an external resistor of resistance R. The current in the external resistor is I. If one of the cells is reversed I'm the circuit, what is the new current in the external resistor? (In terms of I and R)
_____________
Can someone please help me with this question? I have no idea how to deal with reversed cells, I know the total resistance is going to be 4R and the current is constant throughout the entire circuit, but that's as far as I've gotten.
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Eimmanuel
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(Original post by Retsek)
Three identical cells, each of internal resistance R, are connected in series with an external resistor of resistance R. The current in the external resistor is I. If one of the cells is reversed I'm the circuit, what is the new current in the external resistor? (In terms of I and R)
_____________
Can someone please help me with this question? I have no idea how to deal with reversed cells, I know the total resistance is going to be 4R and the current is constant throughout the entire circuit, but that's as far as I've gotten.
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The cells are in opposition, so the p.d. across XY is zero.
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chemlegend27
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Does anyone have the method to figure this out? I'm stuck on it as well
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i_do_maths
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sum of p.d. in a closed loop must equal to the sum of emf. in this case the sum of emf is 0 ∴there is no p.d. in the circuit. However, the internal resistance still adds up, because they're in series, ∴if there's another cell/battery in the circuit this must be considered.
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Eimmanuel
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(Original post by chemlegend27)
Does anyone have the method to figure this out? I'm stuck on it as well

When the cells are connected in series, the combined e.m.f across the battery is equal to the absolute value of the algebraic sum of the voltage change across each of the individual cells.

You can think of emfs of the cells as vector when they are connected in series and then sum the vector arrows to get the emf of the cells connected in series. For example,

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So the emf of the cells connected in series is 1.5 V. Note that we ignore the minus sign.
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Bullyhunter76
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When the cells are connected in series, the combined e.m.f across the battery is equal to the absolute value of the algebraic sum of the voltage change across each of the individual cells.

You can think of emfs of the cells as vector when they are connected in series and then sum the vector arrows to get the emf of the cells connected in series. For example,

Image

So the emf of the cells connected in series is 1.5 V. Note that we ignore the minus sign.
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madhatter101
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#7
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If each cell has emf of 'v', initially the total current in circuit=total voltage/total resistance = 3v/4R.As people have explained before, reversing a cell will cause it and its neighbouring cell's voltage to effectively cancel out, but their resistances will remain affecting the circuit.So now the current = v/4RTherefore the current has divided by 3 as (3v/4R)/(v/4R)=1/3.Therefore the new current in terms of I(and R) is I/3.
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