---
title: "Long line voltage and current at intermediate point"
subject: "Power System Analysis I"
chapter: 5
key: "semester_4/power_system_analysis_i:long-line-intermediate-point"
source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/long-line-intermediate-point
site: BCT Engineering
---

# Long line voltage and current at intermediate point

Power System Analysis I · Chapter 5: Transmission Line Modeling

- Asked in: 2070 Bhadra, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/long-line-intermediate-point
- Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-5

## Question

For a long transmission line, derive the expression relating voltage and current at an intermediate point of the line and at the receiving end of the line.

## Short answer

For a point at distance x from the receiving end, voltage and current are V_x = V_R cosh gamma x + Z_c I_R sinh gamma x and I_x = (V_R/Z_c) sinh gamma x + I_R cosh gamma x, where gamma = sqrt(zy), Zc = sqrt(z/y). This exact long-line solution accounts for distributed parameters.

## Answer

In a long line the parameters are distributed, so voltage and current vary from point to point. Let $x$ be the distance of the intermediate point from the **receiving end**.

```text
 IS      x <------ measured from receiving end
 o-->--z.dx--+--z.dx--+-- ... --z.dx--+--> IR
             |        |               |
 VS        y.dx     y.dx            y.dx    VR
             |        |               |
 o-----------+--------+-- ... --------+----o
 sending                          receiving
```

#### Derivation
For an element $dx$ at distance $x$ from the receiving end, with $z = r + j\omega L$ and $y = g + j\omega C$ per km:

$$
\frac{dV}{dx} = zI, \qquad \frac{dI}{dx} = yV
$$

Differentiating the first and substituting the second:

$$
\frac{d^2V}{dx^2} = zy\,V = \gamma^2V, \qquad \gamma = \sqrt{zy} = \alpha + j\beta
$$

The general solution and the current (from $I = \frac{1}{z}\frac{dV}{dx}$) are:

$$
\begin{aligned}
V &= K_1e^{\gamma x} + K_2e^{-\gamma x} \\
I &= \frac{1}{Z_c}\left(K_1e^{\gamma x} - K_2e^{-\gamma x}\right), \quad Z_c = \sqrt{z/y}
\end{aligned}
$$

At $x = 0$: $V = V_R$ and $I = I_R$, so $K_1 + K_2 = V_R$ and $K_1 - K_2 = Z_cI_R$:

$$
K_1 = \frac{V_R + Z_cI_R}{2}, \qquad K_2 = \frac{V_R - Z_cI_R}{2}
$$

Substituting:

$$
\begin{aligned}
V_x &= \frac{V_R + Z_cI_R}{2}e^{\gamma x} + \frac{V_R - Z_cI_R}{2}e^{-\gamma x} \\
I_x &= \frac{V_R/Z_c + I_R}{2}e^{\gamma x} - \frac{V_R/Z_c - I_R}{2}e^{-\gamma x}
\end{aligned}
$$

Using $\cosh\gamma x = \frac{e^{\gamma x} + e^{-\gamma x}}{2}$ and $\sinh\gamma x = \frac{e^{\gamma x} - e^{-\gamma x}}{2}$:

$$
\begin{aligned}
V_x &= V_R\cosh\gamma x + Z_cI_R\sinh\gamma x \\
I_x &= \frac{V_R}{Z_c}\sinh\gamma x + I_R\cosh\gamma x
\end{aligned}
$$

In matrix form:

$$
\begin{bmatrix} V_x \\ I_x \end{bmatrix} = \begin{bmatrix} \cosh\gamma x & Z_c\sinh\gamma x \\ \frac{1}{Z_c}\sinh\gamma x & \cosh\gamma x \end{bmatrix}\begin{bmatrix} V_R \\ I_R \end{bmatrix}
$$

#### Notes
- If the point is given as distance $d$ from the **sending end**, use $x = l - d$.
- Putting $x = l$ gives the sending-end relations $V_S = V_R\cosh\gamma l + Z_cI_R\sinh\gamma l$ and $I_S = \frac{V_R}{Z_c}\sinh\gamma l + I_R\cosh\gamma l$.
- The terms with $e^{\gamma x}$ form the incident wave and those with $e^{-\gamma x}$ the reflected wave. Their sum changes with $x$, so $V_x$ and $I_x$ are different at every point.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which equations form the starting point for an element dx at distance x from the receiving end?
   A. dV/dx = zI and dI/dx = yV
   B. dV/dx = yI and dI/dx = zV
   C. V = zI and I = yV
   D. dV/dx = I/z and dI/dx = V/y
   Answer: A. The voltage change per length is the series drop zI and the current change is the shunt current yV. (Key point: Starting differential equations)

2. What second-order equation results and what is its solution?
   A. d²V/dx² = γ²V, V = K1 e^(γx) + K2 e^(-γx)
   B. d²V/dx² = -γ²V, V = K1 cos γx + K2 sin γx
   C. d²V/dx² = zV, V = K e^(zx)
   D. d²V/dx² = γV, V = K1 + K2 x
   Answer: A. Differentiating dV/dx = zI and using dI/dx = yV gives d²V/dx² = zyV = γ²V. (Key point: Wave equation and general solution)

3. How are K1 and K2 found?
   A. At x = 0, V = VR and I = IR, giving K1 = (VR + Zc IR)/2 and K2 = (VR - Zc IR)/2
   B. At x = l, V = 0
   C. K1 = K2 = VR
   D. K1 = VR Zc, K2 = IR
   Answer: A. At the receiving end (x = 0), K1 + K2 = VR and K1 - K2 = Zc IR. (Key point: Boundary conditions)

4. What is the final relation at distance x from the receiving end?
   A. Vx = VR cosh γx + Zc IR sinh γx; Ix = (VR/Zc) sinh γx + IR cosh γx
   B. Vx = VR sinh γx + Zc IR cosh γx
   C. Vx = VR cosh γx + IR sinh γx / Zc
   D. Vx = VR e^(-γx)
   Answer: A. Combining the exponentials into cosh and sinh gives this; setting x = l gives the sending-end relations. (Key point: Final result)

## Related in Power System Analysis I

- [Long line sending end voltage derivation](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Along-line-sending-voltage-derivation) — Power System Analysis I, chapter 5
- [Receiving end complex power and power circle](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areceiving-power-circle) — Power System Analysis I, chapter 6
- [Receiving end power in ABCD terms and maximum power](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Areceiving-power-abcd-max) — Power System Analysis I, chapter 6
- [Distributed parameters and current along a long line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Adistributed-parameters-long-line-current) — Power System Analysis I, chapter 5
- [280 km, 60 Hz line: nominal-π vs long line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A280km-60hz-nominal-pi-long) — Power System Analysis I, chapter 5, asked 2×
- [Receiving and sending end powers from ABCD, 132/134 kV](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aabcd-power-132kv-134kv) — Power System Analysis I, chapter 6

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- [Receiving-end voltage infinite at 1500 km](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Alossless-line-1500km-infinite-voltage) — High Voltage Engineering, chapter 3
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