Correct Answer - Option 3 : 88.54 kV
Concept:
When low inductive current is being interrupted and the arc quenching force of the circuit breaker is more than necessary to interrupt a low magnitude of current, the current will be interrupted before its natural zero instant.
In this situation, the energy stored in the magnetic field appears in the form of high voltage across the stray capacitance which will cause restriking of the arc.
The energy stored in the magnetic field is \(\frac{1}{2}L{i^2}\) where i is the instantaneous value of the current which is interrupted.
This will appear in the form of electrostatic energy equal to \(\frac{1}{2}C{v^2}\).
Where V is called the prospective value of the voltage.
As these two energies are equal, they can be related as
\(\frac{1}{2}L{i^2} = \frac{1}{2}C{v^2}\)
\( \Rightarrow v = i\;\sqrt {\frac{L}{C}}\)
Calculation:
Given that, magnetizing current (I) = 7 A.
Inductance (L) = 8 H
Capacitance (C) = 0.05 × 10-6 F
Prospective voltage, \(V = i\sqrt {\frac{L}{C}} \)
\(= 7\;\sqrt {\frac{{8}}{{0.05 \times {{10}^{ - 6}}}}} \)
= 88.54 kV
