Chapter 8, Problem 2P

To determine

The missing values.

Explanation of Solution

The given values are shown in below table:

Primary (P)		Secondary (S)		Load (L)
${\text{E}}_{\text{P}\left(\text{P}\right)}$		${\text{E}}_{\text{P}\left(\text{S}\right)}$		${\text{E}}_{\text{P}\left(\text{L}\right)}$
${\text{I}}_{\text{P}\left(\text{P}\right)}$		${\text{I}}_{\text{P}\left(\text{S}\right)}$		${\text{I}}_{\text{P}\left(\text{L}\right)}$
${\text{E}}_{\text{L}\left(\text{P}\right)}$	7200	${\text{E}}_{\text{L}\left(\text{S}\right)}$	240	${\text{E}}_{\text{L}\left(\text{L}\right)}$
${\text{I}}_{\text{L}\left(\text{P}\right)}$		${\text{I}}_{\text{L}\left(\text{s}\right)}$		${\text{I}}_{\text{L}\left(\text{L}\right)}$
Ratio		$\text{Z}$	$4\Omega$

Refer to the circuit shown in Figure 8-30.

The primary windings of three single phase transformer are connected in wye connection. In wye connection, the line voltage is equal to $\sqrt{3}$ times phase voltage.

  $\begin{array}{c}{\text{E}}_{\text{L}\left(\text{P}\right)}=\sqrt{3}{\text{E}}_{\text{P}\left(\text{P}\right)}\\ {\text{E}}_{\text{P}\left(\text{P}\right)}=\frac{{\text{E}}_{\text{L}\left(\text{P}\right)}}{\sqrt{3}}\\ =\frac{7200}{\sqrt{3}}\\ =4156.92\end{array}$

The secondary windings of three single phase transformer are connected in delta connection. In delta connection, the line voltage is equal to phase voltage.

  $\begin{array}{l}{\text{E}}_{\text{P}\left(\text{S}\right)}={\text{E}}_{\text{L}\left(\text{S}\right)}\\ {\text{E}}_{\text{P}\left(\text{S}\right)}=240\end{array}$

Calculate the turns ratio of primary and secondary winding.

  $\begin{array}{c}\text{Turns ratio}=\frac{{\text{E}}_{\text{P}\left(\text{P}\right)}}{{\text{E}}_{\text{P}\left(\text{S}\right)}}\\ =\frac{4156.92}{240}\\ =\frac{17.3205}{1}\\ \approx \frac{17.32}{1}\end{array}$

Thus, the turns ratio $\left({\text{E}}_{\text{P}\left(\text{P}\right)}:{\text{E}}_{\text{P}\left(\text{S}\right)}\right)$ is $17.32:1$.

Since, the secondary winding (S) is directly connected to the load (L). Thus, the line voltage of secondary winding and load are equal.

  ${\text{E}}_{\text{L}\left(\text{S}\right)}={\text{E}}_{\text{L}\left(\text{L}\right)}=240$

Here, the load uses wye connection. In wye connection, the line voltage is equal to $\sqrt{3}$ times phase voltage.

  $\begin{array}{c}{\text{E}}_{\text{L}\left(\text{L}\right)}=\sqrt{3}{\text{E}}_{\text{P}\left(\text{L}\right)}\\ {\text{E}}_{\text{P}\left(\text{L}\right)}=\frac{{\text{E}}_{\text{L}\left(\text{L}\right)}}{\sqrt{3}}\\ =\frac{240}{\sqrt{3}}\\ =138.56\end{array}$

Calculate the phase current of the load $\left({\text{I}}_{\text{P}\left(\text{L}\right)}\right)$.

  $\begin{array}{c}{\text{I}}_{\text{P}\left(\text{L}\right)}=\frac{{\text{E}}_{\text{P}\left(\text{L}\right)}}{\text{Z}}\\ =\frac{138.56}{4\Omega }\\ =34.64\end{array}$

In wye connection, the phase current $\left({\text{I}}_{\text{P}}\right)$ and line current $\left({\text{I}}_{\text{L}}\right)$ are equal.

  $\begin{array}{l}{\text{I}}_{\text{L}\left(\text{L}\right)}={\text{I}}_{\text{P}\left(\text{L}\right)}\\ {\text{I}}_{\text{L}\left(\text{L}\right)}=34.64\end{array}$

Since, the secondary winding (S) is directly connected to the load (L). Thus, the line current of secondary winding and load are equal.

  ${\text{I}}_{\text{L}\left(\text{L}\right)}={\text{I}}_{\text{L}\left(\text{S}\right)}=34.64$

Here, the secondary winding uses delta connection. In delta connection, the line current is equal to $\sqrt{3}$ times of phase current.

  $\begin{array}{c}{\text{I}}_{\text{L}\left(\text{S}\right)}=\sqrt{3}{\text{I}}_{\text{P}\left(\text{S}\right)}\\ {\text{I}}_{\text{P}\left(\text{S}\right)}=\frac{{\text{I}}_{\text{L}\left(\text{S}\right)}}{\sqrt{3}}\\ =\frac{34.64}{\sqrt{3}}\\ =19.999\\ \approx 20\end{array}$

Calculate the phase current in primary winding $\left({\text{I}}_{\text{P}\left(\text{P}\right)}\right)$.

  $\begin{array}{c}\frac{{\text{I}}_{\text{P}\left(\text{P}\right)}}{{\text{I}}_{\text{P}\left(\text{S}\right)}}=\frac{{\text{E}}_{\text{P}\left(\text{S}\right)}}{{\text{E}}_{\text{P}\left(\text{P}\right)}}\\ {\text{I}}_{\text{P}\left(\text{P}\right)}=\frac{{\text{E}}_{\text{P}\left(\text{S}\right)}}{{\text{E}}_{\text{P}\left(\text{P}\right)}}\times {\text{I}}_{\text{P}\left(\text{S}\right)}\\ =\frac{1}{17.32}\times 20\\ =1.15\end{array}$

Since, the primary winding uses wye connection. In wye connection, the line current is equal to phase current.

  $\begin{array}{l}{\text{I}}_{\text{L}\left(\text{P}\right)}={\text{I}}_{\text{P}\left(\text{P}\right)}\\ {\text{I}}_{\text{L}\left(\text{P}\right)}=1.15\end{array}$

Thus, the all missing values are calculated and shown in below table:

Primary (P)		Secondary (S)		Load (L)
${\text{E}}_{\text{P}\left(\text{P}\right)}$	4156.92	${\text{E}}_{\text{P}\left(\text{S}\right)}$	440	${\text{E}}_{\text{P}\left(\text{L}\right)}$	138.56
${\text{I}}_{\text{P}\left(\text{P}\right)}$	1.15	${\text{I}}_{\text{P}\left(\text{S}\right)}$	20	${\text{I}}_{\text{P}\left(\text{L}\right)}$	34.64
${\text{E}}_{\text{L}\left(\text{P}\right)}$	7200	${\text{E}}_{\text{L}\left(\text{S}\right)}$	240	${\text{E}}_{\text{L}\left(\text{L}\right)}$	240
${\text{I}}_{\text{L}\left(\text{P}\right)}$	1.15	${\text{I}}_{\text{L}\left(\text{S}\right)}$	34.64	${\text{I}}_{\text{L}\left(\text{L}\right)}$	34.64
Ratio	17.32:1	$\text{Z}$	$4\Omega$