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Voltage build-up at line–cable junction

Short answer

At the junction, the transmitted voltage is VJ = 2Z2/(Z1+Z2) times the incident surge. With 500 kV, Z1=500 Ω and Z2=60 Ω, VJ = (120/560)×500 = 107.14 kV. A reflected wave of -392.86 kV returns on the line.

A transmission line of 500 Ω surge impedance is connected to a cable of 60 Ω surge impedance at other end. If a surge of 500 kV travels along the line to the junction point. Find the voltage build-up at the junctions?

At the junction of the line (Z1Z_1) and the cable (Z2Z_2), the voltage that builds up equals the transmitted (refracted) wave:

VJ=2Z2Z1+Z2VV_J = \frac{2Z_2}{Z_1 + Z_2}V

Data: V=500V = 500 kV, Z1=500 ΩZ_1 = 500\ \Omega (line), Z2=60 ΩZ_2 = 60\ \Omega (cable).

Voltage at the junction

VJ=2×60500+60×500=0.2143×500=107.14 kV\begin{aligned} V_J &= \frac{2 \times 60}{500 + 60} \times 500 \\ &= 0.2143 \times 500 = 107.14\ \text{kV} \end{aligned}

Reflected wave on the line

Vr=VJ−V=107.14−500=−392.86 kVV_r = V_J - V = 107.14 - 500 = -392.86\ \text{kV}

The reflection coefficient is (60−500)/560=−0.786(60 - 500)/560 = -0.786. Most of the wave is reflected with negative sign, because the cable looks almost like a short circuit to the line.

Currents

  • Incident: I=500/500=1.0I = 500/500 = 1.0 kA
  • Transmitted into the cable: It=107.14/60=1.786I_t = 107.14/60 = 1.786 kA
  • Reflected: Ir=−Vr/Z1=392.86/500=0.786I_r = -V_r/Z_1 = 392.86/500 = 0.786 kA, and 1.0+0.786=1.7861.0 + 0.786 = 1.786 ✓
  line 500 ohm          cable 60 ohm
 500 kV ------>  J  ------> 107.14 kV
       <------ -392.86 kV

This is why a short cable between an overhead line and a substation reduces the surge voltage entering the equipment.

Answer: voltage built up at the junction = 107.14 kV.

✓ Answer checked by Claude Opus 5.5 (high effort)

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