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 4 V and internal resistance 1 Ω is joined to the point A, as shown in the figure. Take the potential at B to be zero. (a) What are the potentials at the points A and C? (b) At which point D of the wire AB, the potential is equal to the potential at C? (c) If the points C and D are connected by a wire, what will be the current through it? (d) If the 4 V battery is replaced by a 7.5 V battery, what would be the answers of parts (a) and (b)?

#### Solution

(a) Potential difference across AB = Potential at A - Potential at B

Potential at B = 0 V

⇒ Potential at point A = Potential difference across AB = 6 V

⇒ Potential difference across AC = Potential at A - Potential at B

⇒ 4 = 6 - Potential at C

⇒ Potential at C = 2 V

(b) Given:-

Potential across AD = Potential across AC = 4 V

⇒ Potential across DB = 2 V

\[\frac{V_{AD}}{V_{BD}} = \frac{l_{AD}}{l_{DB}}\]

\[ \Rightarrow \frac{4}{2} = \frac{l_{AD}}{100 - l_{AD}}\]

\[ \Rightarrow 4\left( 100 - l_{AD} \right) = 2 l_{AD} \]

\[ \Rightarrow 6 l_{AD} = 400\]

\[ \Rightarrow l_{AD} = \frac{400}{6} = 66 . 7 cm\]

(c) When the points C and D are connected by a wire, current flowing through the wire will be zero because the points are at the same potential.

(d) Potential difference across AC = Potential at A - Potential at C

⇒ 7.5 = 6 - Potential at C

⇒ Potential at C = −1.5 V

Since the potential at C is negative now, this point will go beyond point B, which is at 0 V. Hence, no such point D will exist between the points A and B.