Question

Starting with (5.1.1) of the text, show that

$A=\frac{V_S{I}_S+V_R{I}_R}{V_R{I}_S+V_S{I}_R}$and$\mathrm{B}=\frac{{\mathrm{V}}_{\mathrm{S}}^2-{\mathrm{V}}_{\mathrm{R}}^2}{{\mathrm{V}}_{\mathrm{R}}{\mathrm{I}}_{\mathrm{S}}+{\mathrm{V}}_{\mathrm{S}}{\mathrm{I}}_{\mathrm{R}}}$

Short answer

The value of parameter A is $\frac{V_S{I}_S+V_R{I}_R}{V_R{I}_S+V_S{I}_R}$.

The value of parameter B is $\frac{V_S^2-V_R^2}{V_R{I}_S+V_S{I}_R}$.

Step-by-step solution

Write the given data from the question.

The given expression to find a solution of$A=\frac{V_S{I}_S+V_R{I}_R}{V_R{I}_S+V_S{I}_R}$.

The given expression to find a solution of $B=\frac{V_S^2-V_R^2}{V_R{I}_S+V_S{I}_R}$.

Determine the formula of given solution of Newton-Raphson method.

Write the formula of given solution of A by Newton-Raphson method.

${\mathbf{V}}_{\mathbf{S}}\mathbf{=}{\mathbf{AV}}_{\mathbf{R}}\mathbf{+}{\mathbf{BI}}_{\mathbf{R}}$..... (1)

Here, $V_S$is sending end voltage, $V_R$is receiving end voltage and $I_R$is receiving end current.

Write the formula of given solution of B by Newton-Raphson method.

$\mathbf{B}=\frac{{\mathbf{V}}_{\mathbf{S}}-{\mathbf{AV}}_{\mathbf{R}}}{{\mathbf{I}}_{\mathbf{R}}}$

Here, $V_S$is sending end voltage, $V_R$ is receiving end voltage and$I_R$ is receiving end current.

(a) Determine the given given expression to find a solution of  A and B.

Determine for a short-transmission line current.

$I_{\mathrm{S}}=C{V}_{\mathrm{R}}+D{I}_{\mathrm{R}}$...... (2)

$A=D$

Here, $I_S$is sending end current, $V_R$is receiving end voltage and $I_R$is receiving end current.

Multiply equation (1) by $I_S$on both the sides.

$V_{\mathrm{S}}{I}_{\mathrm{S}}=A{V}_{\mathrm{R}}{I}_{\mathrm{S}}+B{I}_{\mathrm{R}}{I}_{\mathrm{S}}$ …… (3)

Substitute equation (2) into equation (3).

localid="1655877357897" $\begin{array}{c}{V}_S{I}_S=A{V}_{\mathrm{R}}{I}_{\mathrm{S}}+B{I}_{\mathrm{R}}\left(C{V}_{\mathrm{R}}+D{I}_{\mathrm{R}}\right)\\ =A{V}_{\mathrm{R}}{I}_{\mathrm{S}}+B{I}_{\mathrm{R}}\left(C{V}_{\mathrm{R}}+A{I}_{\mathrm{R}}\right)\\ V_{\mathrm{S}}{I}_{\mathrm{S}}=A{V}_{\mathrm{R}}{I}_{\mathrm{S}}+BC{V}_{\mathrm{R}}{I}_{\mathrm{R}}+AB{I}_{\mathrm{R}}^2\end{array}$ …… (4)

Add $V_{\mathrm{R}}{I}_{\mathrm{R}}$on both the sides of equation (4).

$\begin{array}{l}{V}_{\mathrm{S}}{I}_{\mathrm{S}}+V_{\mathrm{R}}{I}_{\mathrm{R}}=A{V}_R{I}_{\mathrm{S}}+BC{V}_{\mathrm{R}}{I}_{\mathrm{R}}+AB{I}_{\mathrm{R}}^2+V_{\mathrm{R}}{I}_{\mathrm{R}}\\ V_{\mathrm{S}}{I}_{\mathrm{S}}+V_{\mathrm{R}}{I}_{\mathrm{R}}=A{V}_{\mathrm{R}}{I}_{\mathrm{S}}+\left(BC+1\right)V_{\mathrm{R}}{I}_{\mathrm{R}}+AB{I}_{\mathrm{R}}^2\end{array}$…… (5)

By using the following relation:

For short transmission line $A=D$.

Therefore,

$\begin{array}{l}{A}^2-BC=1\\ A^2=1+BC\end{array}$

Determine thegiven expression to find a solution of A.

Now substitute $A^2$for $1+BC$ into equation (5).

$\begin{array}{c}{V}_{\mathrm{S}}{I}_{\mathrm{S}}+V_{\mathrm{R}}{I}_{\mathrm{R}}=A{V}_{\mathrm{R}}{I}_{\mathrm{S}}+A^2{V}_{\mathrm{R}}{I}_{\mathrm{R}}+AB{I}_{\mathrm{R}}^2\\ =A\left(V_{\mathrm{R}}{I}_{\mathrm{S}}+A{V}_{\mathrm{R}}{I}_{\mathrm{R}}+B{I}_{\mathrm{R}}^2\right)\\ =A\left(V_{\mathrm{R}}{I}_{\mathrm{S}}+I_{\mathrm{R}}\left(A{V}_{\mathrm{R}}+B{I}_{\mathrm{R}}\right)\right)\\ =A\left(V_{\mathrm{R}}{I}_{\mathrm{S}}+I_{\mathrm{R}}{V}_{\mathrm{S}}\right)\end{array}$

Solve further as

$A=\frac{V_{\mathrm{S}}{I}_{\mathrm{S}}+V_{\mathrm{R}}{I}_{\mathrm{R}}}{V_{\mathrm{R}}{I}_{\mathrm{S}}+V_{\mathrm{S}}{I}_{\mathrm{R}}}$ …… (6)

Therefore, the expression for parameter A is $\frac{V_{\mathrm{S}}{I}_{\mathrm{S}}+V_{\mathrm{R}}{I}_{\mathrm{R}}}{V_{\mathrm{R}}{I}_{\mathrm{S}}+V_{\mathrm{S}}{I}_{\mathrm{R}}}$.

Rewrite the above equation (1).

$B=\frac{V_{\mathrm{S}}-A{V}_{\mathrm{R}}}{I_{\mathrm{R}}}$…… (7)

Now, substitute the value of from equation (6) into equation (7).

$\begin{array}{c}B=\frac{V_{\mathrm{S}}}{I_{\mathrm{R}}}-\frac{V_{\mathrm{R}}}{I_{\mathrm{R}}}\left(\frac{V_{\mathrm{S}}{I}_{\mathrm{S}}+V_{\mathrm{R}}{I}_{\mathrm{R}}}{V_{\mathrm{R}}{I}_{\mathrm{S}}+V_{\mathrm{S}}{I}_{\mathrm{R}}}\right)\\ =\frac{V_{\mathrm{S}}\left(V_{\mathrm{R}}{I}_{\mathrm{S}}+V_{\mathrm{S}}{I}_{\mathrm{R}}\right)-V_{\mathrm{R}}\left(V_{\mathrm{S}}{I}_{\mathrm{S}}+V_{\mathrm{R}}{I}_{\mathrm{R}}\right)}{I_R\left(V_{\mathrm{R}}{I}_{\mathrm{S}}+V_{\mathrm{S}}{I}_{\mathrm{R}}\right)}\\ =\frac{V_{\mathrm{R}}{V}_{\mathrm{S}}{I}_{\mathrm{S}}+V_{\mathrm{S}}^2{I}_{\mathrm{R}}-V_{\mathrm{S}}{V}_{\mathrm{R}}{I}_{\mathrm{S}}-V_{\mathrm{R}}^2{I}_{\mathrm{R}}}{I_{\mathrm{R}}\left(V_{\mathrm{R}}{I}_{\mathrm{S}}+V_{\mathrm{S}}{I}_{\mathrm{R}}\right)}\\ =\frac{I_{\mathrm{R}}\left(V_{\mathrm{S}}^2-V_{\mathrm{R}}^2\right)}{I_{\mathrm{R}}\left(V_{\mathrm{R}}{I}_{\mathrm{S}}+V_{\mathrm{S}}{I}_{\mathrm{R}}\right)}\end{array}$

Solve further as,

$B=\frac{V_{\mathrm{S}}^2-V_{\mathrm{R}}^2}{V_{\mathrm{R}}{I}_{\mathrm{S}}+V_{\mathrm{S}}{I}_{\mathrm{R}}}$

Therefore, the expression for parameter B is $\frac{V_{\mathrm{S}}^2-V_{\mathrm{R}}^2}{V_{\mathrm{R}}{I}_{\mathrm{S}}+V_{\mathrm{S}}{I}_{\mathrm{R}}}$.