## Modelling a Wire With Damaged Insulation

If the insulation on a wire becomes damaged, current may be lost through the damaged insulation. We can model this current leakage using the diagram below.To aid the analysis, divide the section of wire into distinct sections as shown.

in section A

\[i\]

is the current flowing into and out of terminal 1 and I is the current flowing into and out of terminal 2. Let \[v\]

be the voltage between the input terminals and let \[V\]

be the voltage difference between the output terminals. A current \[i\]

flowing through a resistance \[r\]

experience a voltage drop \[ir\]

hence \[V=v-ir\]

. The current along the bottom of the diagram is constant - \[I=i\]

.We can represent the input - section A - with the vector

\[\mathbf{w}= \begin{pmatrix}v\\i\end{pmatrix}\]

and the output with the vector \[\mathbf{W}= \begin{pmatrix}V\\I\end{pmatrix}\]

. We can represent the input and output in matrix form as \[\mathbf{V}= A \mathbf{v}, \: A= \left( \begin{array}{ccc} 1 & -r \\ 0 & 1 \end{array} \right)\]

.Now analyse section B. The input to section B is the output from section A so

\[v=V\]

. A current \[V/R\]

passes through the resistance \[R\]

and the input current at B is less by this amount so \[I=i-V/R\]

. We can represent section B by the matrix system \[\mathbf{V}= B \mathbf{v}, \: B= \left( \begin{array}{ccc} 1 & 0 \\ -1/R+ & 1 \end{array} \right)\]

.The input and output of the circuit representing the wire is

\[\begin{equation} \begin{aligned} \mathbf{OUTPUT} &= (AB)^3 \mathbf{INPUT} \\ &= ( \left( \begin{array}{ccc} 1 & -r \\ 0 & 1 \end{array} \right) \left( \begin{array}{ccc} 1 & 0 \\ -1/R & 1 \end{array} \right) )^3 \\ &= \left( \begin{array}{ccc} 1+r/R & -r \\ -1/R+ & 1 \end{array} \right)^3 \mathbf{INPUT} \end{aligned} \end{equation}\]