Surge on 400 Ω line terminated by 1000 Ω

Data: $V = 10$ kV, $Z = 400\ \Omega$, terminating resistance $R = 1000\ \Omega$.

(i) Incident current wave

$$I = \frac{V}{Z} = \frac{10\times10^3}{400} = 25\ \text{A}$$

(ii) Line terminated by $R = 1000\ \Omega$

Reflection coefficient:

$$\Gamma = \frac{R - Z}{R + Z} = \frac{1000 - 400}{1000 + 400} = 0.4286$$

Reflected voltage and current:

$$\begin{aligned} V_r &= \Gamma V = 0.4286 \times 10 = 4.286\ \text{kV} \\ I_r &= -\frac{V_r}{Z} = -\frac{4286}{400} = -10.71\ \text{A} \end{aligned}$$

Voltage and current in the resistance:

$$\begin{aligned} V_R &= V + V_r = 14.286\ \text{kV} \;\left(= \tfrac{2R}{R+Z}V\right) \\ I_R &= I + I_r = 25 - 10.71 = 14.29\ \text{A} \;\left(= V_R/R\right) \end{aligned}$$

Rate of energy dissipated in $R$:

$$P_R = V_R I_R = 14.286\ \text{kV} \times 14.286\ \text{A} = 204.1\ \text{kW}$$

Rate of energy reflection:

$$P_r = \frac{V_r^2}{Z} = \frac{(4286)^2}{400} = 45.92\ \text{kW}$$

Check: incident power $= V^2/Z = (10^4)^2/400 = 250$ kW, and $204.1 + 45.9 = 250$ kW ✓

Since $R > Z$, the reflected voltage is positive and the reflected current negative. 81.6% of the incident power is absorbed and 18.4% is reflected.

Answer: (i) 25 A. (ii) energy dissipated at 204.1 kW; reflected voltage 4.286 kV; reflected current −10.71 A; energy reflected at 45.9 kW.