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Emf and internal resistance of a battery 
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#1
Oct2508, 01:54 AM

P: 267

When switch S in the figure is open, the voltmeter V of the battery reads 3.07 V. When the switch is closed, the voltmeter reading drops to 2.95 V, and the ammeter A reads 1.70 A. Assume that the two meters are ideal, so they do not affect the circuit.
Find the value of the emf. (see picture posted below) the equation would be EIrIR or E= I(r+R) E=Ir however I am totally confused on how to find the internal resistance value,r, when the switch is open and the value of R when the switch is close. Any help would be appreciated. Thanks. I apprecaite the help. Stephen 


#2
Oct2508, 09:04 AM

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Write another equation describing the given voltage reading across the battery terminals when the switch is closed. 


#3
Oct2508, 03:22 PM

P: 267

I have two equations there. Are they right? What would the emf be? the 3.07V or 2.95V? And how would you find the r and R?
Thanks Stephen 


#4
Oct2508, 05:16 PM

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Emf and internal resistance of a battery



#5
Oct2508, 07:12 PM

P: 267

So use 3.07 for E in E=Ir and 1.7 for I thus r=3.07/1.7=1.81? Then to find the emf you do the voltage 3.07 + (1.7)(1.81)?Then to find R would you use 2.95=I(r+R)=(1.7)(1.81+R)????
Are these right? 


#6
Oct2508, 07:15 PM

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#7
Oct2508, 08:24 PM

P: 267

The terminal voltage across the battery (which is what the voltmeter measures) is the EMF minus the voltage drop across the internal resistance (which is given by Ir).
Well doesn't this mean that the emf = the voltage measured by voltmeter +the voltage drop, Ir? and to find r you use E=Ir or 3.07=(1.7)*r? If this is not right I need a little bit more info? The concept is not sinking in. 


#8
Oct2608, 07:27 AM

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And the current (I) depends on whether the switch is open or closed. 


#9
Oct2608, 05:13 PM

P: 267

From your equation EMF = Measured voltage +Ir
right? And since the switch is not closed what would the current be when the voltmeter measures 3.07V? And then how would you find r from the equation E=Ir? Would you use the 3.07V for E and the current found above for I? If this is not right, would you please guide me to the right direction. Thanks. 


#10
Oct2608, 06:16 PM

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#11
Oct2608, 06:58 PM

P: 267

I have no idea my professor only went over closed circuits. Since the circuit is not closed, I do not see how a current exists. But since the voltmeter is giving a measurement there is obviously a current. Need a starting step.
V=Ir or I = V/r, but since we do not know r, I do not see how we can solve it. 


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