Question

An overcurrent relay set to operate at $\mathbf{10}\mathbf{A}$is connected to the CT in Figure 10.8 with a$\mathbf{500}\mathbf{:}\mathbf{5}$ CT ratio. Determine the minimum primary fault current that the relay will detect if the burden${\mathbf{Z}}_{\mathbf{B}}$ is (a)$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\Omega }$ , (b)$\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{\Omega }$ , and (c) $\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{\Omega }$.

Short answer

(a) The minimum primary fault current is $1004\text{\hspace{0.17em}A}$.

(b) The minimum primary fault current is$1008\text{\hspace{0.17em}A}$ .

(c) The minimum primary fault current is$1009.5\text{\hspace{0.17em}A}$ .

Step-by-step solution

Write the given data from the question.

Overcurrent relay setting, $I^{\text{'}}=10\text{\hspace{0.17em}A}$

CT ratio,$n=500:5$

The secondary resistance corresponding to CT ratio from the figure 10.8, $Z^{\text{'}}=0.242\Omega$.

Write the relay impedance.

(a)$Z_B=1\Omega$

(b)$Z_B=4\Omega$

(c)$Z_B=5\Omega$

Determine the equations to calculate the minimum primary fault current.

The equation to calculate the secondary voltage of the CT is given as follows.

${\mathbf{E}}^{\mathbf{\text{'}}}\mathbf{=}{\mathbf{I}}^{\mathbf{\text{'}}}\mathbf{(}{\mathbf{Z}}^{\mathbf{\text{'}}}\mathbf{+}{\mathbf{Z}}_{\mathbf{B}}\mathbf{)}$ …… (1)

The equation to calculate the minimum primary fault current is given as follows.

role="math" localid="1656430318326" $\mathbf{I}\mathbf{=}\mathbf{n}\mathbf{(}{\mathbf{I}}^{\mathbf{\text{'}}}\mathbf{+}{\mathbf{I}}_{\mathbf{c}}\mathbf{)}$ …… (2)

Here, ${\mathbf{I}}_{\mathbf{c}}$ is the exciting current.

Calculate the minimum primary fault current for ZB=1 Ω.

(a)

Calculate the secondary voltage of the CT,

Substitute $10\text{\hspace{0.17em}A}$for $I^{\text{'}}$, $0.242\Omega$for $Z^{\text{'}}$and $1\Omega$for $Z_B$into equation (1).

$\begin{array}{l}{E}^{\text{'}}=10(0.242+1)\\ E^{\text{'}}=10\times 1.242\\ E^{\text{'}}=12.42\text{\hspace{0.17em}V}\end{array}$

The exciting current corresponding to the secondary voltage of the CT$12.42\text{\hspace{0.17em}V}$ is$0.04\text{\hspace{0.17em}A}$.

$I_c=0.04\text{\hspace{0.17em}A}$

Calculate the minimum primary fault current.

Substitute$500:5$for$n$,$10\text{\hspace{0.17em}A}$ for$I^{\text{'}}$and$0.04\text{\hspace{0.17em}A}$for $I_c$into equation (2).

$\begin{array}{l}I=\frac{500}{5}(10+0.04)\\ I=100\times 10.04\\ I=1004\text{\hspace{0.17em}A}\end{array}$

Hence, the minimum primary fault current is$1004\text{\hspace{0.17em}A}$ .

Calculate the minimum primary fault current for ZB=4 Ω.

(b)

Calculate the secondary voltage of the CT,

Substitute$10A$for $I^{\text{'}}$,$0.242\Omega$for$Z^{\text{'}}$and$4\Omega$for$Z_B$into equation (1).

$\begin{array}{l}{E}^{\text{'}}=10(0.242+4)\\ E^{\text{'}}=10\times 4.242\\ E^{\text{'}}=42.42\text{\hspace{0.17em}V}\end{array}$

The exciting current corresponding to the secondary voltage of the CT$42.42\text{\hspace{0.17em}V}$ is$0.08\text{\hspace{0.17em}A}$.

$I_c=0.08\text{\hspace{0.17em}A}$

Calculate the minimum primary fault current.

Substitute$500:5$ for$n$ ,$10A$ for$I^{\text{'}}$ and$0.08\text{\hspace{0.17em}A}$ for$I_c$ into equation (2).

$\begin{array}{l}I=\frac{500}{5}(10+0.08)\\ I=100\times 10.08\\ I=1008\text{\hspace{0.17em}A}\end{array}$

Hence, the minimum primary fault current is $1008\text{\hspace{0.17em}A}$.

Calculate the minimum primary fault current for ZB=5 Ω.

(c)

Calculate the secondary voltage of the CT,

Substitute $10A$for $I^{\text{'}}$, $0.242\Omega$for $Z^{\text{'}}$and $5\Omega$for $Z_B$into equation (1).

$\begin{array}{l}{E}^{\text{'}}=10(0.242+5)\\ E^{\text{'}}=10\times 5.242\\ E^{\text{'}}=52.42\text{\hspace{0.17em}V}\end{array}$

The exciting current corresponding to the secondary voltage of the CT$52.42\text{\hspace{0.17em}V}$is$0.095\text{\hspace{0.17em}A}$.

$I_c=0.095\text{\hspace{0.17em}A}$

Calculate the minimum primary fault current.

Substitute$500:5$ for $n$, $10A$for$I^{\text{'}}$ and $0.095\text{\hspace{0.17em}A}$for $I_c$into equation (2).

$\begin{array}{l}I=\frac{500}{5}(10+0.095)\\ I=100\times 10.095\\ I=1009.5\text{\hspace{0.17em}A}\end{array}$

Hence, the minimum primary fault current is$1009.5\text{\hspace{0.17em}A}$ .