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A 6 volt battery of negligible internal ...

A 6 volt battery of negligible internal resistance is connected across a potentiometer wire is AB of length 100 cm and uniform area of cross-section. The posistive terminal of another battery of emf 4 V and internal resistance `1 Omega` is joined to the point A as shown in figure. If we take potential at B to be zero, (a) What are the potentials at points A and C. (b) At which point D of the potentiometer C? ( C) If the points C and D are connected by a wire, What will be the current through itgt (d) if the 4 V battery is replaced by 7.5 V battery, what would be the answers of parts (a) and (b)?

Text Solution

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(a)Potential difference between A and B = 6 V. As potential at B = 0,
So the potential at A = 6V
Since the potential difference between points A and C = 4 V. Therefore, potential at C = 6 - 4 = 2 V
(b) Let the potential at point D be the same as at C. Then potential difference between A and D = potential between A and C = 4V.
The length AD of potentimeter wire
`=100/6 xx 4 = 66.7 cm`
(c ) IF points C and D are connceted by a wire, then current through this wire is zero because potential at C is equal to potential at D.
(d) If 4V battery is replaced by 7.5 V battery, then potential difference between A and C = 7.5 V. As point A is at 6V, the point C will be at - 1.5 V. Since the voltage of the battery to be balanced on potentiometer wire is more than fall of potential across the potentiometer wire, no balance point like D would exist on wire.
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