--- title: "L and C of double circuit, 0.75 cm radius" subject: "Power System Analysis I" chapter: 4 key: "semester_4/power_system_analysis_i:double-circuit-l-c-0-75cm" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/double-circuit-l-c-0-75cm site: BCT Engineering --- # L and C of double circuit, 0.75 cm radius Power System Analysis I · Chapter 4: Line Parameter Calculations - Asked in: 2076 Bhadra, 8 marks - Video: [Inductance of three phase double circuit lines](https://www.youtube.com/watch?v=zo9HhiMnYeI) — Surumi Shafi - 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/double-circuit-l-c-0-75cm - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-4 ## Question A 3-Φ double circuit line is arranged as shown in figure below. The conductors are transposed. The radius of each conductor is 0.75 cm. Phase sequence is abc. Find the inductance and capacitance per phase per km. [Figure: vertical double-circuit arrangement; top row a and c' 4 m apart; middle row b and b' 5.5 m apart; bottom row c and a' 4 m apart; vertical spacing 3 m between rows.] ## Short answer Compute equivalent mutual GMD and self GMD for each phase group using r' for inductance and r for capacitance. L = 0.622 mH/km per phase, C = 18.64 nF/km per phase to neutral. ## Answer For a transposed double-circuit line, the two conductors of each phase act as a two-conductor group. We find the equivalent mutual GMD $D_m$ and the self-GMD $D_s$ of each phase group. ```text a o------ 4 m ------o c' -- 3 m b o------- 5.5 m -----o b' -- 3 m c o------ 4 m ------o a' -- ``` **Data:** $r = 0.75$ cm $= 0.0075$ m, $r' = 0.7788 \times 0.0075 = 0.005841$ m. Horizontal offset of b from a is $(5.5-4)/2 = 0.75$ m. #### Distances | Pair | Distance (m) | |---|---| | ab, bc, a'b', b'c' | $\sqrt{3^2+0.75^2} = 3.0923$ | | ab', a'b, bc', b'c | $\sqrt{3^2+4.75^2} = 5.6181$ | | ac, a'c' | 6 | | ac', a'c | 4 | | aa', cc' | $\sqrt{6^2+4^2} = 7.2111$ | | bb' | 5.5 | #### Mutual GMD $$ \begin{aligned} D_{AB} &= (3.0923 \times 5.6181 \times 5.6181 \times 3.0923)^{1/4} = 4.1681\ \text{m} \\ D_{BC} &= 4.1681\ \text{m} \\ D_{CA} &= (6 \times 4 \times 4 \times 6)^{1/4} = 4.8990\ \text{m} \\ D_m &= \sqrt[3]{4.1681 \times 4.1681 \times 4.8990} = 4.3987\ \text{m} \end{aligned} $$ #### Inductance Self GMD of each phase using $r'$: $$ \begin{aligned} D_{sA} &= \sqrt{r'D_{aa'}} = \sqrt{0.005841 \times 7.2111} = 0.20523\ \text{m} \\ D_{sB} &= \sqrt{0.005841 \times 5.5} = 0.17924\ \text{m}, \quad D_{sC} = 0.20523\ \text{m} \\ D_{sL} &= \sqrt[3]{0.20523 \times 0.17924 \times 0.20523} = 0.19617\ \text{m} \\ L &= 2\times10^{-7}\ln\frac{4.3987}{0.19617} = 2\times10^{-7}\times3.1101 \\ &= 6.220\times10^{-7}\ \text{H/m} = 0.622\ \text{mH/km} \end{aligned} $$ #### Capacitance Self GMD using actual radius $r$: $$ \begin{aligned} D_{sA} &= \sqrt{0.0075 \times 7.2111} = 0.23256\ \text{m} \\ D_{sB} &= \sqrt{0.0075 \times 5.5} = 0.20310\ \text{m}, \quad D_{sC} = 0.23256\ \text{m} \\ D_{sC,eq} &= \sqrt[3]{0.23256 \times 0.20310 \times 0.23256} = 0.22229\ \text{m} \\ C &= \frac{2\pi\varepsilon_0}{\ln(4.3987/0.22229)} = \frac{55.63\times10^{-12}}{2.9851} \\ &= 18.64\times10^{-12}\ \text{F/m} = 18.64\ \text{nF/km} \end{aligned} $$ **Answer:** $L = 0.622$ mH/km per phase; $C = 18.64$ nF/km (0.01864 µF/km) per phase to neutral. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In this double-circuit problem, which radius is used for the self-GMD in the inductance calculation? A. r = 0.0075 m B. r' = 0.7788 x 0.0075 m C. 2r D. sqrt(r) Answer: B. Inductance uses r' to include internal flux; capacitance uses the actual r. (Key point: Inductance vs capacitance radius) 2. How is D_sA found for phase A (conductors a and a' at distance 7.2111 m)? A. sqrt(r x D_aa') (with r' for inductance) B. (r x D_aa')^(1/4) C. D_aa'/2 D. r + D_aa' Answer: A. For a 2-conductor group n^2 = 4 terms give D_s = sqrt(r x D_aa'). (Key point: Self GMD of phase A) 3. How is the overall D_m of the transposed line obtained? A. Cube root of D_AB x D_BC x D_CA, each a 4th-root GMD B. Average of the three distances C. The largest of D_AB, D_BC, D_CA D. Square root of D_AB x D_CA Answer: A. D_m = (D_AB D_BC D_CA)^(1/3) = 4.3987 m, each D being the geometric mean of four distances. (Key point: Mutual GMD) 4. Which pair of results is correct? A. L = 0.622 mH/km, C = 18.64 nF/km B. L = 6.22 mH/km, C = 18.64 nF/km C. L = 0.622 mH/km, C = 1.864 nF/km D. L = 18.64 mH/km, C = 0.622 nF/km Answer: A. L = 2e-7 ln(4.3987/0.19617) = 0.622 mH/km and C = 55.63e-12/2.9851 = 18.64 nF/km. (Key point: Result comparison) ## Related in Power System Analysis I - [Inductance of transposed double circuit, 0.9 cm radius](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Adouble-circuit-inductance-0-9cm) — Power System Analysis I, chapter 4 - [Capacitance of double circuit line, 6 mm radius](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Adouble-circuit-capacitance-6mm) — Power System Analysis I, chapter 4 - [Inductance and capacitance of 4-conductor bundle line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aquad-bundle-220kv-l-c) — Power System Analysis I, chapter 4 - [Capacitance of double circuit line, 9 mm radius](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Adouble-circuit-capacitance-9mm) — Power System Analysis I, chapter 4 - [Inductance of composite-conductor single-phase line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Asingle-phase-composite-conductors) — Power System Analysis I, chapter 4 - [Inductance of 3-conductor bundled line, 6.1 m spacing](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Ainductance-three-subconductor-bundle-6-1m) — Power System Analysis I, chapter 4 ## Related in other subjects - [Capacitance matrices, untransposed and transposed (15 m)](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Acapacitance-matrix-transposed-15m) — High Voltage Engineering, chapter 5, asked 3× - [Capacitance and charge matrices, transposed and untransposed](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Acapacitance-charge-matrix-midspan) — High Voltage Engineering, chapter 5 - [Inductance of 420 kV line, transposed and untransposed](https://www.bctengineer.com/question.md?key=semester_8%2Fhigh_voltage_engineering%3Ainductance-420kv-transposed) — High Voltage Engineering, chapter 5 - [Air clearances for 132 kV double circuit line](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aair-clearance-132kv-double-circuit) — Transmission and Distribution System Design, chapter 4