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A potentiometer wire of length 1 m has a resistance of `100Omega`. It is connected to a 6V battery in series with a resistance of `5Omega`. Determine the emf of the primary cell which gives a balance point at 40cm.A. 1.2 VB. 1.8 VC. 1.6 VD. 1.9 V |
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Answer» Correct Answer - C Given, length of wire, l = 1m = 100 cm Resistance, `R = 10 Omega` The emf of a battery, `E_(1)=6V` `R_(1)=5 Omega` `X = 40 cm` `therefore " "` Current, `I=(E_(1))/(R+R_(1))=(6)/(10+5)=(6)/(15)=(22)/(5)A` `V_(AB)=IR=(2)/(5)xx10=4 V` `therefore` The emf of the primary cell `= (V_(AB))/(l)xx x=(4)/(100)xx40 = 1.6 V` |
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