Concept explainers
Figure 4.34 shows double-circuit conductors' relative positions in segment I of transposition of a completely transposed three-phase overhead transmission line. The inductance is given by
Where
With mean distances defined by equivalent spacings
And
Now consider a 345-kV, three-phase, double-circuit line with phase-conductor’s GMR of 0.0588 ft and the horizontal conductor configuration shown in Figure 4.35.
- Determine the inductance per meter per phase in Henries (H).
- Calculate the inductance of just one circuit and then divide by 2 to obtain the inductance of the double circuit.
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Power System Analysis & Design
- I need help with this problem and an explanation of the solution for the image described below. (Introduction to Signals and Systems)arrow_forwardFind Rth at the open terminals using a 1V test source.arrow_forwardI need help with this problem and an explanation of the solution for the image described below. (Introduction to Signals and Systems)arrow_forward
- Find Rth at the open terminals using 1V test source.arrow_forwardHow many atoms are there in a simple cubic unit cell? in a bcc unit cell? in a fcc unit cell? in the unit cell characterizing the diamond lattice?arrow_forwardConsider the homogeneous RLC circuit (no voltage source) shown in the diagram below. Before the switch is closed, the capacitor has an initial charge go and the circuit has an initial current go- R 9(1) i(t)↓ After the switches closes, current flows through the circuit and the capacitor begins to discharge. The equation that describes the total voltage in the loop comes from Kirchoff's voltage law: L di(t) + Ri(t)+(0) = 0, (1) where i(t) and q(t) are the current and capacitor charge as a function of time, L is the inductance, R is the resistance, and C is the capacitance. Using the fact that the current equals the rate of change of the capacitor charge, and dividing by L, we can write the following homogeneous (no input source) differential equation for the charge on the capacitor: 4(1) +29(1)+w79(1)=0, ཀྱི where a= R 2L and The solution to this second order linear differential equation can be written as: 9(1) =Aent - Beat, where (3) (4) (5) A= (81+20)90 +90 (82+20)90 +90 and B= (6)…arrow_forward
- Consider the homogeneous RLC circuit (no voltage source) shown in the diagram below. Before the switch is closed, the capacitor has an initial charge go and the circuit has an initial current go. R w i(t) q(t) C н After the switches closes, current flows through the circuit and the capacitor begins to discharge. The equation that describes the total voltage in the loop comes from Kirchoff's voltage law: di(t) L + Ri(t) + (t) = 0, dt (1) where i(t) and q(t) are the current and capacitor charge as a function of time, L is the inductance, R is the resistance, and C is the capacitance. Using the fact that the current equals the rate of change of the capacitor charge, and dividing by L, we can write the following homogeneous (no input source) differential equation for the charge on the capacitor: ä(t)+2ag(t)+wg(t) = 0, (2) where R a 2L and w₁ = C LC The solution to this second order linear differential equation can be written as: where 81= q(t) = Ae³¹- Bel 82 = (3) (4) (5)arrow_forwardI need help with this problem and an explanation of the solution for the image described below. (Introduction to Signals and Systems)arrow_forwardFind Rth at open terminals using a 1V test source.arrow_forward
- Power System Analysis and Design (MindTap Course ...Electrical EngineeringISBN:9781305632134Author:J. Duncan Glover, Thomas Overbye, Mulukutla S. SarmaPublisher:Cengage Learning