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Surge at junction of 800 Ω/200 Ω lines

Short answer

Transmitted voltage is 42.11 kV into both lines. Transmitted currents are 52.63 A in the 800 Ω line and 210.5 A in the 200 Ω line.

A surge of 100 kV travelling in a line of natural impedance 600 Ω arrives at a junction with two lines of impedances 800 Ω and 200 Ω respectively. Find the surge voltages and currents transmitted into each line.

At the junction the two outgoing lines appear in parallel to the incoming surge. Both outgoing lines carry the same transmitted voltage.

Data: V=100V = 100 kV, Z1=600 ΩZ_1 = 600\ \Omega, Z2=800 ΩZ_2 = 800\ \Omega, Z3=200 ΩZ_3 = 200\ \Omega.

Parallel impedance of the branches

Zp=800×200800+200=160 ΩZ_p = \frac{800 \times 200}{800 + 200} = 160\ \Omega

Transmitted voltage

Vt=2ZpZ1+ZpV=2×160600+160×100=42.11 kV\begin{aligned} V_t &= \frac{2 Z_p}{Z_1 + Z_p} V = \frac{2 \times 160}{600 + 160} \times 100 \\ &= 42.11\ \text{kV} \end{aligned}

Transmitted currents

I2=42.11800=0.05263 kA=52.63 AI3=42.11200=0.2105 kA=210.5 A\begin{aligned} I_2 &= \frac{42.11}{800} = 0.05263\ \text{kA} = 52.63\ \text{A} \\ I_3 &= \frac{42.11}{200} = 0.2105\ \text{kA} = 210.5\ \text{A} \end{aligned}

Check

  • Reflected voltage Vr=42.11−100=−57.89V_r = 42.11 - 100 = -57.89 kV
  • Incident current =100/600=166.7= 100/600 = 166.7 A; reflected current =57.89/600=+96.5= 57.89/600 = +96.5 A
  • Total current in line 1 =166.7+96.5=263.2= 166.7 + 96.5 = 263.2 A =52.63+210.5= 52.63 + 210.5 ✓
LineVoltage (kV)Current (A)
800 Ω42.1152.63
200 Ω42.11210.53

Answer: 42.11 kV in each branch; currents 52.63 A (800 Ω) and 210.5 A (200 Ω).

✓ Answer checked by Claude Opus 5.5 (high effort)

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