Questions: A transformer has 500 turns of the primary winding and 10 turns of the secondary winding. a) Determine the secondary voltage if the secondary circuit is open and the primary voltage is 120 V. b) Determine the current in the primary and secondary winding, given that the secondary winding is connected to a resistance load 15 Ohms?

Solution Steps

To find the secondary voltage (\(V_s\)) when the secondary circuit is open, we use the transformer turns ratio formula:

\[ \frac{V_p}{V_s} = \frac{N_p}{N_s} \]

where:

• \(V_p\) is the primary voltage (120 V)
• \(N_p\) is the number of turns in the primary winding (500)
• \(N_s\) is the number of turns in the secondary winding (10)

Rearranging the formula to solve for \(V_s\):

\[ V_s = V_p \cdot \frac{N_s}{N_p} \]

Substituting the given values:

\[ V_s = 120 \, \text{V} \cdot \frac{10}{500} = 120 \, \text{V} \cdot 0.02 = 2.4 \, \text{V} \]

Given that the secondary winding is connected to a resistance load of \(15 \, \Omega\), we can use Ohm's Law to find the secondary current (\(I_s\)):

\[ I_s = \frac{V_s}{R} \]

Substituting the known values:

\[ I_s = \frac{2.4 \, \text{V}}{15 \, \Omega} = 0.16 \, \text{A} \]

To find the primary current (\(I_p\)), we use the power equivalence in an ideal transformer, where the power in the primary winding equals the power in the secondary winding:

\[ V_p \cdot I_p = V_s \cdot I_s \]

Rearranging to solve for \(I_p\):

\[ I_p = \frac{V_s \cdot I_s}{V_p} \]

Substituting the known values:

\[ I_p = \frac{2.4 \, \text{V} \cdot 0.16 \, \text{A}}{120 \, \text{V}} = \frac{0.384 \, \text{W}}{120 \, \text{V}} = 0.0032 \, \text{A} \]

Final Answer