I read in how stuff works that:
"When voltage is too high, an MOV can conduct a lot of current to eliminate the extra voltage.
As soon as the extra current is diverted into the MOV and to ground, the voltage in the hot line returns to a normal level."
I'm not sure though why the extra voltage doesnt reach eventually to the appliance itself.
The reason that if V(A) equals 10KV, V(B) wouldn't reach 10KV, is due to the fact that the reflection coefficient in point A is:
Γ(A) = (0 - Zc) / (0 + Zc) = -1 ; (Zc is the characteristic impedance of the line)
and therefore the propogating voltage and the regressing voltage cancel each other?
"When voltage is too high, an MOV can conduct a lot of current to eliminate the extra voltage.
As soon as the extra current is diverted into the MOV and to ground, the voltage in the hot line returns to a normal level."
I'm not sure though why the extra voltage doesnt reach eventually to the appliance itself.
The reason that if V(A) equals 10KV, V(B) wouldn't reach 10KV, is due to the fact that the reflection coefficient in point A is:
Γ(A) = (0 - Zc) / (0 + Zc) = -1 ; (Zc is the characteristic impedance of the line)
and therefore the propogating voltage and the regressing voltage cancel each other?