Chapter 3.2, Problem 6RQ

A step-up transformer has a primary current of 32 A and an applied voltage of 240 V. The secondary coil has a current of 2 A. Assuming ideal transformer conditions, calculate the following:
a. Power input of the primary winding coil
b. Power output of the secondary winding coil
c. Secondary coil winding voltage
d. Turns ratio

To determine

(a)

The power input of primary winding coil assuming ideal transformer condition.

Answer to Problem 6RQ

The power input of primary winding coil assuming ideal transformer condition is $7680\text{W}$.

Explanation of Solution

Given information:

The primary winding current is $32\text{A}$.

The secondary winding current is $2\text{A}$.

The voltage applied to primary is $240\text{V}$.

Write the expression for the power input of the primary winding coil assuming ideal transformer condition.

$P_P=V_P{I}_P$ ...... (I)

Here, the voltage in the primary winding is $V_P$ and the current in the primary winding is $I_P$.

Calculation:

Susbtitute $32\text{A}$ for $I_P$ and $240\text{V}$ for $V_P$ in equation (I).

$\begin{array}{c}{P}_P=\left(240\text{V}\right)\left(32\text{A}\right)\\ =7680\text{V}\cdot \text{A}\left(\frac{1\text{W}}{1\text{V}\cdot \text{A}}\right)\\ =7680\text{W}\end{array}$

Conclusion:

Therefore, the power input of primary winding coil assuming ideal transformer condition is $7680\text{W}$.

To determine

(b)

The power output of the secondary winding coil assuming ideal transformer condition.

Answer to Problem 6RQ

The power output of secondary winding coil assuming ideal transformer condition is $7680\text{W}$.

Explanation of Solution

Given information:

The primary winding current is $32\text{A}$.

The secondary winding current is $2\text{A}$.

The voltage applied to primary is $240\text{V}$.

The primary power and secondary power in an ideal transformer are always equal

$P_P=P_S$ ...... (II)

Here, the primary power is $P_P$ and secondary power is $P_S$.

Calculation:

Susbtitute $7680\text{W}$ for $P_S$ in equation (II).

$P_P=7680\text{W}$

Conclusion:

Therefore, the power output of secondary winding coil assuming ideal transformer condition is $7680\text{W}$.

To determine

(c)

The secondary coil winding voltage assuming ideal transformer condition.

Answer to Problem 6RQ

The secondary coil winding voltage assuming ideal transformer condition is $3840\text{V}$.

Explanation of Solution

Given information:

The primary winding current is $32\text{A}$.

The secondary winding current is $2\text{A}$.

The voltage applied to primary is $240\text{V}$.

Write the expression for the secondary coil winding voltage assuming ideal transformer condition.

$V_S=\frac{P_S}{I_S}$ ...... (III)

Here, the current in the secondary coil is $I_S$.

Calculation:

Substitute $7680\text{W}$ for $P_S$ and $2\text{A}$ for $I_S$ in equation (III)

$\begin{array}{c}{V}_S=\frac{7680\text{W}}{2\text{A}}\\ =3840\text{V}\end{array}$

Conclusion:

Therefore, the secondary coil winding voltage assuming ideal transformer condition is $3840\text{V}$.

To determine

(d)

The turns ratio assuming ideal transformer condition.

Answer to Problem 6RQ

The turns ratio assuming ideal transformer condition is $1:16$.

Explanation of Solution

Given information:

The primary winding current is $32\text{A}$.

The secondary winding current is $2\text{A}$.

The voltage applied to primary is $240\text{V}$.

The turns ratio in a transformer is the ratio of the number of turns in the primary winding to number of turns in the secondary winding.

$TR=\frac{I_S}{I_P}$ ....... (IV)

Here, the turns ratio is $TR$.

Calculation:

Susbtitute $32\text{A}$ for $I_P$ and $2\text{A}$ for $I_S$ in equation (IV)

$\begin{array}{l}TR=\frac{2\text{A}}{32\text{A}}\\ TR=\frac{1}{16}\\ TR=1:16\end{array}$

Conclusion:

Therefore, the turns ratio assuming ideal transformer condition is $1:16$.