Inductance in Single Conductor Power Transmission Line
Reason of Transmission Line InductanceGenerally, electric power is transmitted through the transmission line with AC high voltage and current. High valued alternating current while flowing through the conductor sets up magnetic flux of high strength with alternating nature. This high valued alternating magnetic flux makes a linkage with other adjacent conductors parallel to the main conductor. Flux linkage in a conductor happens internally and externally. Internally flux linkage is due to self-current and externally flux linkage due to external flux. Now the term inductance is closely related to the flux linkage, denoted by λ. Suppose a coil with N number of turn is linked by flux Φ due to current I, then, But for transmission line N = 1. We have to calculate only the value of flux Φ, and hence, we can get the transmission line inductance.
Calculation of Inductance of Single Conductor
Calculation of Internal Inductance due to Internal Magnetic Flux of a ConductorSuppose a conductor is carrying current I through its length l, x is the internal variable radius of the conductor and r is the original radius of the conductor. Now the cross-sectional area with respect to radius x is πx2 square – unit and current Ix is flowing through this cross-sectional area. So the value of Ix can be expressed in term of original conductor current I and cross-sectional area πr2 square – unit
Now consider small thickness dx with the 1m length of the conductor, where Hx is the magnetizing force due to current Ix around the area πx2. And magnetic flux density Bx = μHx, where μ is the permeability of this conductor. Again, µ = µ0µr. If it is considered that the relative permeability of this conductor µr = 1, then µ = µ0. Hence, here Bx = μ0 Hx. dφ for small strip dx is expressed by
Here entire cross-sectional area of the conductor does not enclose the above expressed flux. The ratio of the cross sectional area inside the circle of radius x to the total cross section of the conductor can be thought about as fractional turn that links the flux. Therefore the flux linkage is
Now, the total flux linkage for the conductor of 1m length with radius r is given by Hence, the internal inductance is