Question

In a system of 132 kV, the circuit phase to ground capacitance is 0.01 $\mu F$ and the transformer inductance succeeding the circuit breaker is 5 H. If the circuit breaker breaks a magnetizing current of 10 A flowing in the transformer, the voltage (in kV) appearing across the poles of circuit breaker is

• A. $\frac{100}{\sqrt{5}}$
• B. 100
• C. 132
• D. $100\sqrt{5}$

Answer 1

The correct option is D $100\sqrt{5}$

Voltage across the circuit breaker contacts after the interruption of 10 A current

$V=i\sqrt{\frac{L}{C}}=10\sqrt{\frac{5}{0.01\times {10}^{-6}}}$
$=10\sqrt{5}\times {10}^4$
$=100\sqrt{5}$ kV