Question

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 (32-E28). 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)?

Answer 1

(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
$$\begin{gathered} \frac{V_{\mathrm{AD}}}{V_{\mathrm{BD}}}=\frac{l_{\mathrm{AD}}}{l_{\mathrm{DB}}} \\ \Rightarrow \frac{4}{2}=\frac{l_{\mathrm{AD}}}{100-l_{\mathrm{AD}}} \\ \Rightarrow 4\left(100-l_{\mathrm{AD}}\right)=2l_{\mathrm{AD}} \\ \Rightarrow 6l_{\mathrm{AD}}=400 \\ \Rightarrow l_{\mathrm{AD}}=\frac{400}{6}=66.7\mathrm{cm} \end{gathered}$$

(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.