InterviewSolution
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InterviewSolution
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Figure shows a potentiometer with a cell of 2.0 V and internal resistance `0.40 Omega` maintain a potential drop across the resistor wire AB. A standard cell which maintains a constant emf of 1.02 V (for very moderate currents upto a few mA) gives a balances point at 63.3 cm length of the wire. To ensure very low currents drawn from the standard cell, a very high resistance of `600 Omega` is put is series with it, which is shortedd close to the balance point. The standard cell is then replaced by a cell of unknown emf `epsilon` and the balance point found similarly, turns out to be at 82.3 cm length of the wire. What is the value `epsilon` ? |
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Answer» Here `epsilon = 1.02 V, L_(1) = 67.3cm, epsilon_(2) = epsilon = ?, L_(2) = 82.3 cm` Since `(epsilon_(2))/(epsilon_(1)) = (L_(2))/(L_(1)) :. Epsilon = (L_(2))/(L_(1)) xx epsilon_(1) = (82.3)/(67.3) xx 1.02 = 1.247 V` |
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