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A battery is connected to a potentiometer and a balance point is obtained at `84 cm` along the wire. When its terminals are connected by a `5 Omega` resistor, the balance point changes to `70 cm`. Calculate the internal resistance of the cell.A. `4 Omega`B. `2 Omega`C. `5 Omega`D. `1 Omega`

Answer» Correct Answer - D
`r = (l_(1) - l_(2))/(l_(2))R = (84 - 70)/(70) xx 5 = 1 Omega`
`1 = (l_(1) - l_(3))/(l_(3)) xx 4`, put `l_(1) = 84 cm` get `l_(3) = 67.2 cm`


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