Question

A transformer circuit breaker is equipped with 500/5 A C.T, connected to an induction type over current relays. The relays have 125% plug setting and 0.4 time setting. If a 3-$\varphi$ fault current of 7500 A flows from C.T. and relays follow the characteristics table given below at TMS = 1, then actual operating time of the relay will be

$\begin{array}{ccccc}\text{PSM}& 4& 8& 12& 16\\ \text{OPERATING TIME (SEC.}& 5& 3& 2.8& 2.2\end{array}$

• A. 1.12 sec
• B. 1.01 sec
• C. 2.81 sec
• D. 2.24 sec

Answer 1

The correct option is A 1.12 sec

$\text{The secondary current,}{I}_s=\frac{I_p}{n}$

$\text{Where,}n=\frac{500}{5}$

$I_s=7500\times \frac{5}{500}=75A$

$\text{Relay current setting = 125\% of 5A = 1.25}\times \text{5 = 6.25 A}$

$\text{Plug setting multiplier (PSM)}=\frac{75}{6.25}=12$

$\text{Using data in characteristic table is}$

$\text{Operating time corresponding to PSM = 12 is 2.8 sec (at TMS = 1)}$

$\text{Operating time of relay = 2.8 sec}\times \text{0.4 = 1.12 sec.}$