Voltage and current waves on short-circuited line

Consider a lossless line of surge impedance $Z_0$ and travel time $\tau$. It is short-circuited at the far end and switched on at $t = 0$ to an ideal DC (step) source $E$ of zero internal impedance.

Reflection coefficients

Successive reflections

• $0 < t < \tau$: a wave $+E$ with current $I = E/Z_0$ travels forward.
• At $t = \tau$ (short end): the voltage reflects as $-E$, so the net voltage is $E - E = 0$. The current reflects as $+E/Z_0$, so the net current is $2E/Z_0$.
• At $t = 2\tau$ (source end): the $-E$ wave reflects as $+E$, and the source voltage is restored to $E$. The current adds another $E/Z_0$, giving $3E/Z_0$.
• At $t = 3\tau$: the second $+E$ wave is reflected at the short as $-E$, so the line voltage again falls to zero. The current becomes $4E/Z_0$.

 V |  E __    __    __
   |   |  |  |  |  |  |
   | 0 |  |__|  |__|  |__   (periodic: E, 0, E, 0)
   +----------------------> t

 I |               ____
   |          ____|
   |     ____|            (rises E/Z0 each tau)
   |____|
   +----------------------> t

Interpretation

• The voltage at any point keeps switching between $E$ and $0$. It reduces to zero periodically, because each reflection from the short or from the source reverses its sign.
• The current never reverses sign. Each reflection adds another $E/Z_0$. After $n$ transits the current is $nE/Z_0$, so as $t \to \infty$ the current $\to \infty$ on a lossless line.

Steady-state (phasor) view

For AC, a short-circuited line has $V(x) = jZ_0 I_r\sin\beta x$ and $I(x) = I_r\cos\beta x$ ($x$ measured from the short). With $V_s$ fixed:

$$I_r = \frac{V_s}{jZ_0\sin\beta l}$$

When $\beta l = n\pi$, $I_r \to \infty$, while the voltage has nodes (zeros) every half wavelength.

This is the same result: on a short-circuited lossless line the voltage periodically falls to zero and the current builds up without limit. In a practical line, resistance and source impedance limit the current to the short-circuit value.