Chapter 6, Problem 15PEB

The step-down transformer in a local neighborhood reduces the voltage from a 7,200 V line to 120 V. (a) If there are 125 loops on the secondary, how many are on the primary coil? (b) What current does the transformer draw from the line if the current in the secondary is 36 A? (c) What are the power input and output?

To determine

The number of loops at the primary side of a step-down transformer.

Answer to Problem 15PEB

Solution:

The number of loops at the primary side is $7500$.

Explanation of Solution

Given data:

A step-down transformer with 125 loops on the secondary side. The voltage at the primary side is $7200\text{ V}$ and the voltage at the secondary side is $120\text{ V}$. The current on secondary side is $36\text{ A}$.

Formula used:

The expression that relates the number of loops and the voltage is written as,

$\frac{V_p}{N_p}=\frac{V_s}{N_s}$

Here, $V_p$ is the primary voltage, $N_p$ are the number of loops at the primary side, $V_s$ is the secondary voltage and $N_s$ are the number of loops at the secondary side.

Also, power and current can be determined by the formula, therefore

$V_p{I}_p=V_s{I}_s$

Here, $I_p\text{ and }{I}_s$ are current at primary and secondary side, respectively.

Explanation:

Consider the expression,

$\frac{V_p}{N_p}=\frac{V_s}{N_s}$

Rearrange to get the secondary voltage $V_s$.

$N_p=N_s\frac{V_p}{V_s}$

Substitute $125$ for $N_s$, $7200\text{ V}$ for $V_p$ and $120\text{ V}$ for $V_s$.

$\begin{array}{c}{N}_p=N_s\frac{V_p}{V_s}\\ =125\times \frac{7200\text{ V}}{120\text{ V}}\\ =7500\end{array}$

Conclusion:

Hence, the number of loops ate the primary side is $7500$.

To determine

The primary current in a step down transformer.

Answer to Problem 15PEB

Solution:

The primary current is $0.6\text{ A}$.

Explanation of Solution

Given data:

A step-down transformer with 125 loops on the secondary side. The voltage at the primary side is $7200\text{ V}$ and the voltage at the secondary side is $120\text{ V}$. The current on secondary side is $36\text{ A}$.

Formula used:

The expression for power in a transformer is written as,

$V_p{I}_p=V_s{I}_s$

Here, $I_p\text{ and }{I}_s$ are current at primary and secondary side, respectively.

Explanation:

Consider the expression for power in the transformer.

$V_p{I}_p=V_s{I}_s$

Rearrange the power expression to get the primary current $I_p$.

$I_p=\frac{V_s{I}_s}{V_p}$

Substitute $120\text{ V}$ for $V_s$, $36\text{ A}$ for $I_s$ and $7200\text{ V}$ for $V_p$.

$\begin{array}{c}{I}_p=\frac{V_s{I}_s}{V_p}\\ =\frac{120\text{ V}\times 36\text{ A}}{7200\text{ V}}\\ =0.6\text{ A}\end{array}$

Conclusion:

Hence, the primary current is $0.6\text{ A}$.

To determine

The power of the transformer at the input and the output sides.

Answer to Problem 15PEB

Solution: The power at the input and the output side is same and is equal to $4320\text{ W}$.

Explanation of Solution

Given data:

A step-down transformer with 125 loops on the secondary side. The voltage at the primary side is $7200\text{ V}$ and the voltage at the secondary side is $120\text{ V}$. The current on secondary side is $36\text{ A}$.

Formula used:

The expression for power at the input side is written as,

$P{i}_{}=V_p{I}_p$

Here, $P_i$ is the input power, $V_p$ is the primary voltage and $I_p$ is the primary current.

The expression for power at the output side is written as,

$P{o}_{}=V_s{I}_s$

Here, $P_o$ is the output power, $V_s$ is the secondary voltage and $I_s$ is the secondary current.

Explanation:

Consider the expression for the input power.

$P{i}_{}=V_p{I}_p$

Substitute $7200\text{ V}$ for $V_p$ and $0.6\text{ A}$ for $I_p$.

$\begin{array}{c}{P}_i=V_p{I}_p\\ =7200\text{ V}\times 0.6\text{ A}\\ =4320\text{ W}\end{array}$

Similarly, consider the expression for the output power.

$P{o}_{}=V_s{I}_s$

Substitute $120\text{ V}$ for $V_s$ and $36\text{ A}$ for $I_s$.

$\begin{array}{c}{P}_o=V_s{I}_s\\ =120\text{ V}\times 36\text{ A}\\ =4320\text{ W}\end{array}$

Conclusion:

Hence, the power at the input and the output side is same and is equal to $4320\text{}\text{ W}$.