--- title: "Distributed parameters and current along a long line" subject: "Power System Analysis I" chapter: 5 key: "semester_4/power_system_analysis_i:distributed-parameters-long-line-current" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/distributed-parameters-long-line-current site: BCT Engineering --- # Distributed parameters and current along a long line Power System Analysis I · Chapter 5: Transmission Line Modeling - Asked in: 2076 Bhadra, 2+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/distributed-parameters-long-line-current - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-5 ## Question What is the reason for long transmission lines being represented by uniformly distributed line parameters? With necessary mathematical derivation, show that the current in a long line is different at different points. ## Short answer Long lines are represented by distributed parameters because charging current varies continuously along the line; lumped models fail for lengths>250 km. Solving dV/dx=zI, dI/dx=yV gives I(x)=I_R coshγx+(V_R/Zc) sinhγx, proving current differs with distance x. At no-load, current is zero at receiving end but non-zero elsewhere. ## Answer #### Why long lines use uniformly distributed parameters - Resistance, inductance and capacitance are not located at one point; every metre of conductor has its own series $r$, $L$ and shunt $C$ (and leakage $g$). - In short and medium lines the total charging current is small, so lumping $C$ at one or two points gives little error. - In a long line (above about 250 km) the charging current is large and **the current changes continuously along the line**: each section draws its own charging current. The voltage therefore also changes continuously. - The line length becomes a noticeable fraction of the wavelength ($\lambda \approx 6000$ km at 50 Hz), so travelling-wave effects (phase shift, attenuation, reflection) appear. These can only be represented by distributed parameters. - Lumped models then give wrong $V_S$, $I_S$, regulation and Ferranti rise, so an exact solution of the line's differential equations is used. #### Current is different at different points Let $z = r + j\omega L$ (Ω/km) and $y = g + j\omega C$ (S/km). Take an element $dx$ at distance $x$ 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 ``` Series drop and shunt current in the element: $$ \frac{dV}{dx} = zI, \qquad \frac{dI}{dx} = yV $$ Differentiating the second and substituting the first: $$ \frac{d^2I}{dx^2} = y\frac{dV}{dx} = yz\,I = \gamma^2 I, \qquad \gamma = \sqrt{zy} = \alpha + j\beta $$ General solution: $$ I(x) = K_1e^{\gamma x} + K_2e^{-\gamma x} $$ Then $V = \frac{1}{y}\frac{dI}{dx} = Z_c\left(K_1e^{\gamma x} - K_2e^{-\gamma x}\right)$, with $Z_c = \sqrt{z/y}$. Boundary conditions at $x = 0$: $I = I_R$, $V = V_R$: $$ K_1 + K_2 = I_R, \quad K_1 - K_2 = \frac{V_R}{Z_c} \;\Rightarrow\; K_1 = \frac{1}{2}\left(I_R + \frac{V_R}{Z_c}\right),\; K_2 = \frac{1}{2}\left(I_R - \frac{V_R}{Z_c}\right) $$ $$ \begin{aligned} I(x) &= \frac{1}{2}\left(I_R + \frac{V_R}{Z_c}\right)e^{\gamma x} + \frac{1}{2}\left(I_R - \frac{V_R}{Z_c}\right)e^{-\gamma x} \\ I(x) &= I_R\cosh\gamma x + \frac{V_R}{Z_c}\sinh\gamma x \end{aligned} $$ #### Interpretation - $I(x)$ depends on $x$ through $\cosh\gamma x$ and $\sinh\gamma x$, which change in magnitude and angle along the line. So the current is different at every point. - The term $\frac{V_R}{Z_c}\sinh\gamma x$ is the accumulated charging current; it is zero at the receiving end and largest at the sending end. - In wave form, the current is the sum of an incident wave ($e^{\gamma x}$) and a reflected wave ($e^{-\gamma x}$), whose magnitudes vary as $e^{\pm\alpha x}$ and phases as $\pm\beta x$ along the line. - At no load ($I_R = 0$), $I(x) = \frac{V_R}{Z_c}\sinh\gamma x$, which is zero at the receiving end but not at the sending end. - At $x = l$: $I_S = \frac{V_R}{Z_c}\sinh\gamma l + I_R\cosh\gamma l$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why must long lines use uniformly distributed parameters? A. Because resistance is zero in long lines B. Because charging current varies continuously along the line and travelling-wave effects appear C. Because lumped models are exact for long lines D. Because the voltage is constant along the line Answer: B. Each section draws its own charging current, so V and I change continuously along the line. (Key point: Reason for distributed parameters) 2. For a line element dx, the basic equations are: A. dV/dx = yI, dI/dx = zV B. dV/dx = zI, dI/dx = yV C. dV/dx = zV, dI/dx = yI D. dV/dx = z/I, dI/dx = y/V Answer: B. Series drop per length is zI and shunt current per length is yV. (Key point: Differential equations) 3. Eliminating V gives d^2I/dx^2 = γ^2 I with γ equal to: A. z/y B. sqrt(z/y) C. zy D. sqrt(zy) Answer: D. γ = sqrt(zy) = α + jβ is the propagation constant; sqrt(z/y) is Zc. (Key point: Wave equation) 4. After applying boundary conditions at the receiving end (x = 0): A. I(x) = IR cosh γx + (VR/Zc) sinh γx B. I(x) = IR C. I(x) = IR sinh γx + VR cosh γx D. I(x) = VR / Zc Answer: A. I depends on x through cosh and sinh, so current differs at every point. (Key point: Result for I(x)) 5. At no load (IR = 0), I(x) = (VR/Zc) sinh γx. This is: A. Largest at the receiving end B. Zero everywhere C. Equal at all points D. Zero at the receiving end and largest at the sending end Answer: D. It is the accumulated charging current, growing with x. 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