91Ó°ÊÓ

The given data can be listed below,

• The potential difference between point A and B is,${\mathrm{V}}_{\mathrm{B}}−{\mathrm{V}}_{\mathrm{A}}=12\text{ V}$
• The potential difference between point B and C is,${\mathrm{V}}_{\mathrm{C}}−{\mathrm{V}}_{\mathrm{B}}=−5\text{ V}$
• The potential difference between point D and C is ,$V_D−V_C=−15\text{ V}$

Voltage is the term used to describe the electric potential. The law of conservation states that the sum of the potential difference at all the edges in a closed path is equal to the zero.

According to the law of conservation, total potential difference should be zero on the path ABCD, which can be given as follows,

$\begin{array}{c}\mathrm{Δ³Õ}=({\mathrm{V}}_{\mathrm{B}}−{\mathrm{V}}_{\mathrm{A}})+({\mathrm{V}}_{\mathrm{C}}−{\mathrm{V}}_{\mathrm{B}})+({\mathrm{V}}_{\mathrm{D}}−{\mathrm{V}}_{\mathrm{C}})+({\mathrm{V}}_{\mathrm{A}}−{\mathrm{V}}_{\mathrm{D}})\\ 0=({\mathrm{V}}_{\mathrm{B}}−{\mathrm{V}}_{\mathrm{A}})+({\mathrm{V}}_{\mathrm{C}}−{\mathrm{V}}_{\mathrm{B}})+({\mathrm{V}}_{\mathrm{D}}−{\mathrm{V}}_{\mathrm{C}})+({\mathrm{V}}_{\mathrm{A}}−{\mathrm{V}}_{\mathrm{D}})\end{array}$

Here,$V_B−V_A$is the potential difference between point A and B,$V_C−V_B$is the potential difference between point B and C,$V_D−V_C$ is the potential difference between point B and C, and $V_A−V_D$is the potential difference between point A and point D.

Rearrange the above expression to get the expression for the potential difference between point A and D.

$V_A−V_D=0−(V_B−V_A)−(V_C−V_B)−(V_D−V_C)$

Substitute all the values in the above expression.

$\begin{array}{c}{V}_A−V_D=−12\text{ V}−(−5\text{ V})−(−15\text{ V})\\ =8\text{ V}\end{array}$

Thus, the potential difference between point A and D is$8\text{ V}$ .