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Surge on bifurcated 150 Ω and 90 Ω lines

Short answer

The two outgoing lines act in parallel, equivalent surge impedance 56.25 ohm. Transmitted voltage is 2*56.25/(300+56.25)*150 = 47.37 kV on both branches. Branch currents are 315.8 A in 150 ohm and 526.3 A in 90 ohm; reflected wave is -102.63 kV.

An overhead line of 300Ω surge impedance bifurcates into two lines of 150Ω and 90Ω surge impedances respectively. If a step wave of 150 kV is launched on the line, estimate the magnitude of waves transmitted on the bifurcated lines.

At the bifurcation point, the two outgoing lines are in parallel for the incoming wave. The same transmitted voltage travels on both.

Data: V=150V = 150 kV, Z1=300 ΩZ_1 = 300\ \Omega, Z2=150 ΩZ_2 = 150\ \Omega, Z3=90 ΩZ_3 = 90\ \Omega.

Parallel impedance

Zp=150×90150+90=56.25 ΩZ_p = \frac{150 \times 90}{150 + 90} = 56.25\ \Omega

Transmitted voltage

Vt=2ZpZ1+ZpV=2×56.25300+56.25×150=0.3158×150=47.37 kV\begin{aligned} V_t &= \frac{2Z_p}{Z_1 + Z_p}V = \frac{2 \times 56.25}{300 + 56.25} \times 150 \\ &= 0.3158 \times 150 = 47.37\ \text{kV} \end{aligned}

Transmitted currents

I2=47.37150=0.3158 kA=315.8 AI3=47.3790=0.5263 kA=526.3 A\begin{aligned} I_2 &= \frac{47.37}{150} = 0.3158\ \text{kA} = 315.8\ \text{A} \\ I_3 &= \frac{47.37}{90} = 0.5263\ \text{kA} = 526.3\ \text{A} \end{aligned}

Reflected wave and check

  • Vr=Vt−V=47.37−150=−102.63V_r = V_t - V = 47.37 - 150 = -102.63 kV
  • Incident current =150/300=500= 150/300 = 500 A; reflected current =102.63/300=342.1= 102.63/300 = 342.1 A
  • Line current at junction =500+342.1=842.1= 500 + 342.1 = 842.1 A =315.8+526.3= 315.8 + 526.3 ✓
 Z1=300 ohm            +--- 150 ohm: 47.37 kV, 315.8 A
 150 kV ----->  J -----+
                       +---  90 ohm: 47.37 kV, 526.3 A

Answer: a 47.37 kV wave travels on each branch, with currents 315.8 A (150 Ω line) and 526.3 A (90 Ω line). A −102.6 kV wave is reflected back.

✓ Answer checked by Claude Opus 5.5 (high effort)

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