Surge at junction of 600 Ω/200 Ω lines

At a junction with two branch lines, the branches act as two impedances in parallel for the incoming surge. The transmitted (refracted) voltage is the same on both branches.

Data: $V = 200$ kV, $Z_1 = 500\ \Omega$, $Z_2 = 600\ \Omega$, $Z_3 = 200\ \Omega$.

Equivalent impedance of the branches

$$Z_p = \frac{Z_2 Z_3}{Z_2 + Z_3} = \frac{600 \times 200}{800} = 150\ \Omega$$

Transmitted voltage

$$\begin{aligned} V_t &= \frac{2 Z_p}{Z_1 + Z_p} V = \frac{2 \times 150}{500 + 150} \times 200 \\ &= 92.31\ \text{kV} \end{aligned}$$

This same voltage travels into both branch lines.

Transmitted currents

$$\begin{aligned} I_2 &= \frac{V_t}{Z_2} = \frac{92.31}{600} = 0.1538\ \text{kA} = 153.8\ \text{A} \\ I_3 &= \frac{V_t}{Z_3} = \frac{92.31}{200} = 0.4615\ \text{kA} = 461.5\ \text{A} \end{aligned}$$

Check with the reflected wave

• Reflected voltage: $V_r = V_t - V = 92.31 - 200 = -107.69$ kV
• Incident current: $I = 200/500 = 400$ A; reflected current: $I_r = -V_r/Z_1 = +215.4$ A
• Current in line 1 at the junction: $400 + 215.4 = 615.4$ A $= I_2 + I_3 = 153.8 + 461.5$ ✓

 Z1=500 ohm          +--- Z2=600: 92.31 kV, 153.8 A
 200 kV ---->  J ----+
                     +--- Z3=200: 92.31 kV, 461.5 A

Answer: transmitted voltage = 92.31 kV in each branch; currents = 153.8 A (600 Ω line) and 461.5 A (200 Ω line).