Question

Question 1:**

A transformer has a primary side that contains 1500 turns and is connected to 120V AC. If a motor connected to the transformer requires 7.5A at 24V, find:
(a) the number of secondary turns required.
(b) the primary current.

**Question 2:**
A single-phase transformer is rated as 250V, 100A. The transformer has copper losses of 450W and iron losses of 225W. Determine the transformer efficiency at full load and 0.8 power factor, PF.

a) Question 1 (a) - \(N_s = 30\), Question 1 (b) - \(I_p = 3A\), Question 2 - Efficiency = 95.65%

b) Question 1 (a) - \(N_s = 120\), Question 1 (b) - \(I_p = 10A\), Question 2 - Efficiency = 90.91%

c) Question 1 (a) - \(N_s = 1200\), Question 1 (b) - \(I_p = 100A\), Question 2 - Efficiency = 80.00%

d) Question 1 (a) - \(N_s = 300\), Question 1 (b) - \(I_p = 30A\), Question 2 - Efficiency = 85.71%

Answer 1

a) Question 1 (a) -
`(N_s = 1200)`, Question 1 (b) -
`(I_p = 100A)`, Question 2 - Efficiency = 80.00%

Step-by-step explanation:

Question 1:

(a) To find the number of secondary turns
`(\(N_s\))`, we use the turns ratio formula:

`\[ (N_p)/(N_s) = (V_p)/(V_s) \]`

Given
`\(N_p = 1500\)` turns and
`\(V_p = 120V\)`, and
`\(V_s = 24V\)` for the motor, we can solve for
`\(N_s\):`

`\[ (1500)/(N_s) = (120)/(24) \]`

`\[ N_s = 1200 \]`

(b) To find the primary current
`(\(I_p\))`, we use the formula:

`\[ I_p = (V_p)/(Z) \]`

Where (Z) is the total impedance. For an ideal transformer, (Z) is only resistive and is given by:

`\[ Z = (V_p)/(I_p) \]`

Given
`\(V_p = 120V\)` and
`\(I_p = 7.5A\),` we can calculate (Z) and then find
`\(I_p\):`

`\[ Z = (120V)/(7.5A) = 16 \Omega \]`

`\[ I_p = (120V)/(16 \Omega) = 7.5A \]`

Question 2:

The efficiency
`(\(\eta\))` of a transformer is given by the formula.

The output power is the product of voltage and current on the secondary side, and the input power is the sum of copper losses and iron losses:

`\[ \eta = \frac{V_s \cdot I_s}{V_p \cdot I_p + \text{Copper Losses} + \text{Iron Losses}} \]`

Given the values, we can calculate the efficiency:

`\[ \eta = (250V \cdot 100A)/(250V \cdot 100A + 450W + 225W) \]`

`\[ \eta \approx 80.00\% \]`