A 6 volt battery of negligible internal resistance is connected across a uniform wire AB of length 100 cm. The positive terminal of another battery of emf 4V and internal resistance 1 Ω Ω is joined to the point A as shown in figure. Take the potential at B to be zero.
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Sol. V A = 6 V, V C = 2V Ans.
E = x λ
⇒ 4 = λ × 
⇒ λ =
cm Ans.
In secondary circuit current is zero Ans.
6 V, 6 – 7.5 = – 1.5 V, no such point D exists Ans.
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