--- title: "Wave front retardation by terminal capacitance" subject: "High Voltage Engineering" chapter: 3 key: "semester_8/high_voltage_engineering:capacitance-termination-wavefront" source: https://www.bctengineer.com/q/semester_8/high_voltage_engineering/capacitance-termination-wavefront site: BCT Engineering --- # Wave front retardation by terminal capacitance High Voltage Engineering · Chapter 3: Overvoltages in Power System - Asked in: 2072 Magh, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/high_voltage_engineering/capacitance-termination-wavefront - Whole chapter: https://www.bctengineer.com/md/semester_8/high_voltage_engineering/chapter-3 ## Question An infinite rectangular wave on a line having a surge impedance of 500 Ω strikes a transmission line terminated with a capacitance of 0.004 μF. Calculate the extent to which the wave front is retarded. ## Short answer Time constant ZC = 500 times 0.004 uF = 2 us. Capacitor voltage rises exponentially toward 2V; it reaches the incident value V after 1.386 us and 90% of final value after 4.61 us. The rectangular wave front is retarded to a few microseconds. ## Answer An infinite rectangular wave $V$ arriving at a capacitance $C$ at the end of a line of surge impedance $Z$ sees the capacitor as a short circuit at first. The capacitor then charges towards $2V$ through $Z$: $$ V_C(t) = 2V\left(1 - e^{-t/ZC}\right) $$ So the vertical front of the incident wave becomes an **exponential front**: the wave is retarded (sloped). The amount of retardation is set by the time constant $ZC$. #### Time constant $$ \tau = ZC = 500 \times 0.004\times10^{-6} = 2\times10^{-6}\ \text{s} = 2\ \mu\text{s} $$ #### Measures of retardation - Time to reach the incident value $V$ (half the final $2V$): $$ 2V(1 - e^{-t/\tau}) = V \;\Rightarrow\; t = \tau\ln 2 = 2 \times 0.693 = 1.386\ \mu\text{s} $$ - Time to reach 90% of the final value $2V$: $$ t = \tau\ln 10 = 2 \times 2.303 = 4.61\ \mu\text{s} $$ - Practically full value (about 99%) after $5\tau = 10\ \mu$s. | $t$ (μs) | $V_C / 2V$ | |---|---| | 1.386 | 0.50 | | 2 | 0.632 | | 4.61 | 0.90 | | 10 | 0.993 | ```text V_C 2V | ___________ | __-- V | _--' | / | / |/________________________ t 0 1.39 4.61 us ``` A rectangular (zero rise time) wave thus becomes a wave with a front of a few microseconds. This reduces the steepness ($dV/dt$) of the surge, which protects the winding insulation of terminal equipment. **Answer:** time constant $ZC$ = **2 μs**. The front is retarded so that the voltage reaches the incident value after **1.39 μs** and 90% of its final value ($2V$) after **4.61 μs**. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How does the capacitor at the line end behave at the instant the wave arrives? A. Open circuit B. Short circuit, then charges towards 2V through Z C. Short circuit permanently D. Same as a resistor Z Answer: B. An uncharged capacitor is a short at t = 0, then VC = 2V(1 - e^(-t/ZC)). (Key point: Capacitor behaviour) 2. For Z = 500 Ω and C = 0.004 μF the time constant is: A. 0.2 μs B. 20 μs C. 2 μs D. 2 ms Answer: C. 500 x 0.004e-6 = 2e-6 s. (Key point: Time constant) 3. The time to reach the incident value V (half of 2V) is: A. τ ln 2 = 1.39 μs B. τ ln 10 = 4.61 μs C. τ = 2 μs D. 5τ = 10 μs Answer: A. 2V(1 - e^(-t/τ)) = V gives e^(-t/τ) = 0.5, so t = τ ln 2. (Key point: Time to reach V) 4. Why is the retardation useful? A. It raises the surge magnitude B. It removes the wave C. It reduces dV/dt and so the stress on winding insulation D. It speeds up propagation Answer: C. A sloped front reduces steepness, protecting the first turns of terminal equipment. 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