Chapter 37, Problem 22A

The transformers are all powered with 100 W, and all have 100 turns on the primary. The number of turns on each secondary varies as shown.

a. Rank the voltage output of the secondaries from greatest to least.

b. Rank the current in the secondaries from greatest to least.

c. Rank the power output in the secondaries from greatest to least.

To determine

To arrange: the values of secondary voltage in decreasing order.

Answer to Problem 22A

The decreasing order of the secondary voltage is :

${\mathrm{V}}_{B,s}>V_{C,s}>V_{A,s}$

Explanation of Solution

Given:

The primary voltage for all systems, ${\mathrm{V}}_P=240V$

The primary turns for all systems, ${\mathrm{N}}_P=100$

The secondary conditions for all system given below:

System A:

The secondary turns, ${\mathrm{N}}_{A,s}=50$

System B:

The secondary turns, ${\mathrm{N}}_{B,s}=200$

System C:

The secondary turns, ${\mathrm{N}}_{C,s}=100$

Concept Used:

The transformer relation is given as:

$\begin{array}{l}\frac{V_P}{N_P}=\frac{V_S}{N_S}\\ N_P=numberofprimaryturns\\ N_S=numberofsecondaryturns\\ V_P=primaryvoltage\\ V_S=secondaryvoltage\end{array}$

Calculation:

For system A:

$\begin{array}{l}{V}_P=240V\\ N_P=100\\ N_S=50\end{array}$

The voltage produced is calculated as:

$\begin{array}{l}\frac{240}{100}=\frac{V_{S,A}}{50}\\ V_{S,A}=120V\mathrm{...............................................................................}(1)\end{array}$

For system B:

$\begin{array}{l}{V}_P=240V\\ N_P=100\\ N_S=200\end{array}$

The voltage produced is calculated as:

$\begin{array}{l}\frac{240}{100}=\frac{V_{S,B}}{200}\\ V_{S,B}=480V\mathrm{................................................................}(2)\end{array}$

For system C:

$\begin{array}{l}{V}_P=240V\\ N_P=100\\ N_S=200\end{array}$

The voltage produced is calculated as:

$\begin{array}{l}\frac{240}{100}=\frac{V_{S,C}}{100}\\ V_{S,B}=240V\mathrm{............................................................................}(3)\end{array}$

From (1) , (2) & (3) we get:

${\mathrm{V}}_{S,B}>V_{S,C}>V_{S,A}$

Conclusion:

The secondary voltage values greatest to least:

${\mathrm{V}}_{S,B}>V_{S,C}>V_{S,A}$

To determine

To arrange: the reading of the current in decreasing order.

Answer to Problem 22A

The decreasing order of the current is :

${\mathrm{I}}_{S,A}>I_{S,C}>I_{S,B}$

Explanation of Solution

Given:

The primary voltage for all systems, ${\mathrm{V}}_P=240V$

The primary turns for all systems, ${\mathrm{N}}_P=100$

The power for each system, $\mathrm{P}=100W$

The secondary conditions for all system given below:

System A:

The secondary turns,

${\mathrm{N}}_{A,s}=50$

System B:

The secondary turns, ${\mathrm{N}}_{B,s}=200$

System C:

The secondary turns, ${\mathrm{N}}_{C,s}=100$

Concept Used:

The power delivered to the transformer:

  $\begin{array}{l}{P}_{in}=VI\\ I=\frac{P_{in}}{V}\\ P=inputpower,W\\ V=secondaryvoltage,V\\ I=current,A\end{array}$

Calculation:

For system A:

$\begin{array}{l}{P}_{in}=100W\\ V=120V\end{array}$

The current is calculated as:

  ${\mathrm{I}}_A=\frac{100}{120}=0.83A\mathrm{................................................................}(4)$

For system B:

$\begin{array}{l}{P}_{in}=100W\\ V=480V\end{array}$

The current is calculated as:

${\mathrm{I}}_B=\frac{100}{480}=0.2075A\mathrm{................................................................}(5)$

For system C:

$\begin{array}{l}{P}_{in}=100W\\ V=240V\end{array}$

The current is calculated as:

${\mathrm{I}}_B=\frac{100}{240}=0.415A\mathrm{................................................................}(5)$

From (4) , (5) & (6) we get:

${\mathrm{I}}_{S,A}>I_{S,C}>I_{S,B}$

Conclusion:

The current readings from greatest to least:

${\mathrm{I}}_{S,A}>I_{S,C}>I_{S,B}$

To determine

To arrange: the output power in decreasing order.

Answer to Problem 22A

The decreasing order of the output power is :

${\mathrm{P}}_{out,A}=P_{out,B}=P_{out,C}$

Explanation of Solution

Introduction:

The power is defined as the rate of energy transfer.

As per energy conservation the total energy will be conserved. So the power in the primary coils is equal to the power in the secondary coils.

${\mathrm{P}}_{in}=P_{out}=100W$

Conclusion:

The output power will be same in all systems.

${\mathrm{P}}_{out,A}=P_{out,B}=P_{out,C}$