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A potentiometer wire of length 1 m has a resistance of 100Omega. It is connected toa 6V battery in series with a resistance of 5Omega. Determine the emf of the primary cell which gives a balance point at 40cm. |
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Answer» SOLUTION :Here , l=1m `R_1=10 Omega , V=6V , R_2=5Omega` Current flowing in potentiometer wire, `I=V/(R_1+R_2)=6/(10+5)=6/15`=0.4 A Potential drop across the potentiometer wire V = IR = 0.4 × 10 = 4V Potential gradient, `K=(V')/1=4/1`=4V/m EMF of the primary cell = KI = 4 × 0.4 = 1.6 V |
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