--- title: "Arrester current and resistance for 750 kV line" subject: "High Voltage Engineering" chapter: 4 key: "semester_8/high_voltage_engineering:arrester-750kv-z3000" source: https://www.bctengineer.com/q/semester_8/high_voltage_engineering/arrester-750kv-z3000 site: BCT Engineering --- # Arrester current and resistance for 750 kV line High Voltage Engineering · Chapter 4: Insulation Coordination - Asked in: 2077 Chaitra, 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/arrester-750kv-z3000 - Whole chapter: https://www.bctengineer.com/md/semester_8/high_voltage_engineering/chapter-4 ## Question For a 750 kV line, take Vw = 3000 kV, crest, travelling on the line and Vp = 1700 kV. The line surge impedance is Z = 3000 ohms. Calculate and discuss (i) the current flowing in the line before reaching the arrester, (ii) the current through the arrester, and (iii) the value of arrester resistance for this condition, and (iv) reflection and refraction coefficients for voltage and current. ## Short answer Assuming a realistic surge impedance of 300 Ω, incident current is 10 kA. Arrester current = (2×3000–1700)/300 = 14.33 kA; arrester resistance = 1700/14.33 = 118.6 Ω. Voltage reflection coefficient = –0.433, refraction = 0.567; current reflection = 0.433, refraction = 1.433. ## Answer **Note on the data:** a surge impedance of 3000 Ω is not realistic for an overhead line (typical values are 300–500 Ω). The same problem in other papers uses $Z = 300\ \Omega$, so it is solved here with $Z = 300\ \Omega$. Results for 3000 Ω are given at the end. The reflection and refraction coefficients are the same in both cases. **Data:** $V_w = 3000$ kV, $V_p = 1700$ kV (arrester protective level), $Z = 300\ \Omega$. When the wave reaches the arrester (line end), the Thevenin equivalent is $2V_w$ behind $Z$, and the arrester holds the voltage at $V_p$. #### (i) Current in the line before reaching the arrester $$ I_w = \frac{V_w}{Z} = \frac{3000}{300} = 10\ \text{kA} $$ #### (ii) Current through the arrester $$ \begin{aligned} I_a &= \frac{2V_w - V_p}{Z} = \frac{2 \times 3000 - 1700}{300} \\ &= \frac{4300}{300} = 14.33\ \text{kA} \end{aligned} $$ #### (iii) Arrester resistance $$ R = \frac{V_p}{I_a} = \frac{1700}{14.33} = 118.6\ \Omega $$ #### (iv) Reflection and refraction coefficients Treat the arrester as a termination $R = 118.6\ \Omega$: $$ \begin{aligned} \text{Voltage reflection: } \Gamma_v &= \frac{R - Z}{R + Z} = \frac{118.6 - 300}{418.6} = -0.4333 \\ \text{Voltage refraction: } T_v &= \frac{2R}{R + Z} = \frac{V_p}{V_w} = 0.5667 \\ \text{Current reflection: } \Gamma_i &= -\Gamma_v = +0.4333 \\ \text{Current refraction: } T_i &= \frac{2Z}{R + Z} = 1.4333 \end{aligned} $$ Check: reflected voltage $= -0.4333 \times 3000 = -1300$ kV, and $3000 - 1300 = 1700$ kV ✓. Reflected current $= 0.4333 \times 10 = 4.33$ kA, and $10 + 4.33 = 14.33$ kA ✓. | Quantity | $Z = 300\ \Omega$ | $Z = 3000\ \Omega$ (as printed) | |---|---|---| | Line current $I_w$ | 10 kA | 1 kA | | Arrester current | 14.33 kA | 1.433 kA | | Arrester resistance | 118.6 Ω | 1186 Ω | | $\Gamma_v$ / $T_v$ | −0.433 / 0.567 | −0.433 / 0.567 | | $\Gamma_i$ / $T_i$ | 0.433 / 1.433 | 0.433 / 1.433 | #### Discussion - The arrester clamps the voltage at 1700 kV instead of letting it double to 6000 kV at the open end. A negative voltage reflection of −1300 kV travels back along the line. - The arrester current (14.3 kA) is larger than the line current (10 kA), because current reflection at a low-resistance termination is positive. - The arrester must be rated to carry this current and to absorb the energy. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How does the answer handle the given Z = 3000 ohm? A. It treats 3000 ohm as normal for overhead lines B. It notes it is unrealistic (typical 300-500 ohm) and solves with 300 ohm, also giving 3000 ohm results C. It rejects the problem as unsolvable D. It solves only for 3000 ohm Answer: B. The answer flags 3000 ohm as unrealistic, solves with 300 ohm and tabulates the 3000 ohm results too. (Key point: Realistic surge impedance) 2. Which expression gives the current through the arrester? A. Vw / Z B. (2Vw - Vp) / Z C. (Vw - Vp) / Z D. Vp / Z Answer: B. Thevenin equivalent is 2Vw behind Z and the arrester holds Vp, so Ia = (2Vw - Vp)/Z = 14.33 kA for 300 ohm. (Key point: Arrester current) 3. How is the arrester resistance found? A. R = Vw / Ia B. R = Z / 2 C. R = Vp / Ia D. R = 2Vw / Ia Answer: C. The arrester holds Vp while carrying Ia, so R = Vp / Ia = 1700 / 14.33 = 118.6 ohm. (Key point: Arrester resistance) 4. For the arrester termination (R = 118.6 ohm, Z = 300 ohm), what are the voltage reflection and refraction coefficients? A. +0.433 and 1.433 B. -0.433 and 0.567 C. -0.567 and 0.433 D. 0.567 and -0.433 Answer: B. Reflection = (R - Z)/(R + Z) = -0.433 and refraction = 2R/(R + Z) = 0.567; the current coefficients are +0.433 and 1.433. (Key point: Coefficients) ## Close variants set in other exams - [Arrester current and resistance (Z = 300 Ω)](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Aarrester-750kv-z300) — High Voltage Engineering, chapter 4, asked 2× ## Related in High Voltage Engineering - [Arrester current, resistance and rating on 132 kV line](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Aarrester-132kv-line-3000kv) — High Voltage Engineering, chapter 4 - [Arrester with 1000 kV BIL on 300 Ω line](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Aarrester-bil-1000kv-300-ohm) — High Voltage Engineering, chapter 4, asked 3× - [Arrester with 1000 kV BIL on 350 Ω line](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Aarrester-bil-1000kv-350-ohm) — High Voltage Engineering, chapter 4, asked 3× - [Surge impedance and velocity of 220 kV line](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Asurge-impedance-220kv-200km) — High Voltage Engineering, chapter 3, asked 2× - [Transformer BIL and arrester resistance for 500 kV line](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Abil-500kv-ehv-line-arrester) — High Voltage Engineering, chapter 4 - [Arrester separation for a 500 kV/μs surge](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Aarrester-distance-500kv-per-us) — High Voltage Engineering, chapter 4