Chapter 28, Problem 8PP

$\begin{array}{l}\begin{array}{l}{\text{E}}_{\text{P}}277\text{ V}\hfill \\ {\text{I}}_{\text{P}}\text{\_\_\_\_\_}\hfill \\ \begin{array}{l}{\text{N}}_{\text{P}}\text{ 350 Turns}\\ {\text{E}}_{\text{S1}}480\text{ V}\\ {\text{I}}_{\text{S1}}\_\_\_\_\_\\ {\text{N}}_{\text{S1}}\_\_\_\_\_\end{array}\hfill \\ \begin{array}{l}\text{Ratio 1:}\\ {\text{R}}_{1}200\Omega \end{array}\hfill \\ {\text{E}}_{\text{s}}\text{208 V}\hfill \\ {\text{I}}_{\text{S}2}\text{ \_\_\_\_\_}\hfill \\ \begin{array}{l}{\text{N}}_{\text{s2}}\_\_\_\_\_\\ \text{Ratio 2:}\\ {\text{R}}_{2}60\Omega \\ {\text{E}}_{\text{S3}}120\text{ V}\\ {\text{I}}_{\text{S3}}\_\_\_\_\_\\ {\text{N}}_{\text{S3}}\_\_\_\_\_\\ \text{Ratio 3:}\end{array}\hfill \end{array}\\ {\text{R}}_3\text{ 24 }\Omega \end{array}$

To determine

The missing values in the given table.

Answer to Problem 8PP

EP = 277 V	ES1 = 480 V	ES2 = 208 V	ES3 = 120 V
IP = 8.93 A	IS1 = 2.4 A	IS2 = 3.47 A	IS3 = 5 A
NP = 350 turns	NS1 = 606 turns	NS = 263 turns	NS = 152 turns
	Ratio 1 =1:1.73	Ratio 2 =1.33:1	Ratio 3 = 2.30:1
	R1 =200 Ω	R2 = 60 Ω	R3 = 24 Ω

Explanation of Solution

The transformer in the fig 27-17 contains one primary winding and three secondary windings.

The primary is connected to 277 V AC and contains 350 turns of wire.

One secondary has an output voltage of 480 volts and a load resistance of 200 Ω.

Second secondary has an output voltage of 208 volts and a load resistance of 60 Ω.

Third secondary has an output voltage of 120 volts and a load impedance of 24 Ω.

The turns ratio of the first secondary can be found by dividing the smaller voltage into the larger:

$Ratio1=\frac{E_{S1}}{E_P}=\frac{480}{277}=1.73$

The turns ratio of the first secondary is re written as,

$Ratio1=1:1.73$

The current flow in the first secondary can be calculated using Ohm’s law:

$I_{S1}=\frac{E_{S1}}{R_1}=\frac{480}{200}=2.4\text{ A}$

The amount of primary current needed to supply this secondary winding can be found using the turns ratio. As this primary has less voltage, it requires more current:

$\begin{array}{l}{I}_{P1}={\text{ I}}_{S1}\times turnsRatio\\ =\text{ 2}\text{.4}\times \text{1}\text{.73}\\ \text{ = 4}\text{.15 A }\end{array}$

The number of turns of wire in the first secondary winding is found using the turns ratio. Because this secondary has a higher voltage than the primary, it must have more turns of wire:

$\begin{array}{l}{N}_{S1}=N_P\times turnsRatio\\ =\text{ 350}\times 1.73\\ \text{ = 606 turns }\end{array}$

The turns ratio of the second secondary winding is found by dividing the higher voltage by the lower:

$Ratio\text{ 2}=\frac{E_{S2}}{E_P}=\frac{208}{277}=\frac{1}{1.33}$

The turns ratio of the second secondary is re written as,

$Ratio\text{ 2}=1.33:1$

The amount of current flow in this secondary can be determined using Ohm’s law:

$I_{S2}=\frac{E_{S2}}{R_2}=\frac{208}{60}=3.47\text{ A}$

The amount of primary current needed to supply this secondary winding can be found using the turns ratio. As this primary has more voltage, it requires less current:

$\begin{array}{l}{I}_{P2}=\frac{{\text{I}}_{S2}}{turnsRatio}\\ =\frac{\text{3}\text{.47}}{1.33}\\ \text{ = 2}\text{.61 A }\end{array}$

Because the voltage of this secondary is lesser than the primary, it has less turns of wire than the primary. The number of turns of this secondary is found using the turns ratio:

$\begin{array}{l}{N}_{S2}=\frac{N_P}{turnsRatio}\\ =\frac{\text{350}}{1.33}\\ \text{ = 263 turns }\end{array}$

The turns ratio of the third secondary winding is calculated in the same way as the other two. The larger voltage is divided by the smaller:

$Ratio3=\frac{E_{S3}}{E_P}=\frac{120}{277}=\frac{1}{2.30}$

The turns ratio of the third secondary is re written as,

$Ratio\text{ 3}=2.30:1$

The secondary current is found using Ohm’s law:

$I_{S3}=\frac{E_{S3}}{R_3}=\frac{120}{24}=5\text{ A}$

The amount of primary current needed to supply this secondary winding can be found using the turns ratio. As this primary has more voltage, it requires less current:

$\begin{array}{l}{I}_{P3}=\frac{{\text{I}}_{S3}}{turnsRatio}\\ =\frac{\text{5}}{2.30}\\ \text{ = 2}\text{.17 A }\end{array}$

Because the voltage of this secondary is lesser than the primary, it has less turns of wire than the primary. The number of turns of this third secondary is found using the turns ratio:

$\begin{array}{l}{N}_{S3}=\frac{N_P}{turnsRatio}\\ =\frac{350}{2.30}\\ \text{ = 152 turns }\end{array}$

The primary must supply current to each of the three secondary windings. Therefore, the total amount of primary current is the sum of the currents required to supply each secondary:

$\begin{array}{l}{I}_P=I_{P1}+I_{P2}+I_{P3}\\ =4.15+2.61+2.17\\ =8.93\text{ A}\\ \end{array}$