Question

Figure Q26.10 shows a $3V$battery with metal wires attached to each end. What are the potential differences $∆V_{12}=V_2-V_1,∆V_{23}=V_3-V_2,$$∆V_{34}=V_4-V_3,\mathrm{and}∆V_{41}=V_1-V_4$?

Short answer

The required potential differences are :

$$\begin{gathered} \mathbf{∆}{\mathit{V}}_{\mathbf{12}}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{}\mathit{V} \\ \mathbf{∆}{\mathit{V}}_{\mathbf{23}}\mathbf{}\mathbf{=}\mathbf{3}\mathbf{}\mathit{V} \\ \mathbf{∆}{\mathit{V}}_{\mathbf{34}}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{}\mathit{V} \\ \mathbf{∆}{\mathit{V}}_{\mathbf{14}}\mathbf{}\mathbf{=}\mathbf{3}\mathbf{}\mathit{V} \end{gathered}$$

Step-by-step solution

: Given information and formula used   

Given circuit :

Theory used :

Equipotential or isopotential refers to a region in space where every point in it is at the same potential. This usually refers to a scalar potential

Defining ∆V12,∆V23,∆V34 and ∆V41

If no current is flowing, conducting wires are equipotential. The top wire is at the same potential as the battery's top, and the bottom wire is at the same potential as the battery's bottom.

As a result,

$$\begin{gathered} \mathbf{∆}{\mathit{V}}_{\mathbf{12}}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{}\mathit{V} \\ \mathbf{∆}{\mathit{V}}_{\mathbf{23}}\mathbf{}\mathbf{=}\mathbf{3}\mathbf{}\mathit{V} \\ \mathbf{∆}{\mathit{V}}_{\mathbf{34}}\mathbf{}\mathbf{=}\mathbf{0}\mathbf{}\mathit{V} \\ \mathbf{∆}{\mathit{V}}_{\mathbf{14}}\mathbf{}\mathbf{=}\mathbf{3}\mathbf{}\mathit{V} \end{gathered}$$,