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www.GOALias.blogspot.comCurrentElectricitySo, we have ε/V = l 1/l 2But, ε = I (r + R) and V = IR. This givesε/V = (r+R )/RFrom Eq. [3.94(a)] and [3.94(b)] we have[3.94(a)][3.94(b)](R+r)/R = l 1/l 2⎛lr R l⎞1= ⎜ –1⎝⎟2 ⎠(3.95)Using Eq. (3.95) we can find the internal resistance of a given cell.The potentiometer has the advantage that it draws no current fromthe voltage source being measured. As such it is unaffected by the internalresistance of the source.Example 3.10 A resistance of R Ω draws current from apotentiometer. The potentiometer has a total resistance R 0Ω(Fig. 3.29). A voltage V is supplied to the potentiometer. Derive anexpression for the voltage across R when the sliding contact is in themiddle of the potentiometer.FIGURE 3.29Solution While the slide is in the middle of the potentiometer onlyhalf of its resistance (R 0/2) will be between the points A and B. Hence,the total resistance between A and B, say, R 1, will be given by thefollowing expression:1 1 1= +R1 R ( R0/2)R RR01= R 0+ 2 RThe total resistance between A and C will be sum of resistance betweenA and B and B and C, i.e., R 1+ R 0/2∴ The current flowing through the potentiometer will beV 2VI = =R + R /2 2R + R1 0 1 0The voltage V 1taken from the potentiometer will be the product ofcurrent I and resistance R 1,⎛ 2V⎞V 1= I R 1= ⎜ × R1⎝2R+ R ⎟⎠1 0EXAMPLE 3.10123