Respuesta :
Answer:
Therefore,
a) [tex]\dfrac{N_{1}}{N_{2}}=\dfrac{10}{1}[/tex]
b) I₂ = 2.55 Ampere
c) Power = 30.6 Watts
d) R₁ = 470.58 ohms.
Explanation:
Given:
Let the primary voltage be,
[tex]V_{1}=120\ V[/tex]
Secondary Voltage,
[tex]V_{2}=12\ V[/tex]
Secondary Resistance,
[tex]R_{2}=4.7\ ohms[/tex]
To Find:
a) [tex]\dfrac{N_{1}}{N_{2}}=?[/tex]
b) I₂ = ?
c) Power = ?
d) R₁ = ?
Solution:
The ratio of primary to secondary turns of the transformer, is the ratio of primary to secondary Voltage and is given by,
[tex]\dfrac{N_{1}}{N_{2}}=\dfrac{V_{1}}{V_{2}}[/tex]
Substituting the values we get
[tex]\dfrac{N_{1}}{N_{2}}=\dfrac{120}{12}=\dfrac{10}{1}[/tex]
For, rms current for the secondary supply,
[tex]I_{2}=\dfrac{V_{2}}{R_{2}}[/tex]
Substituting the values we get
[tex]I_{2}=\dfrac{12}{4.7}=2.55\ Ampere[/tex]
Now for average power at load,
[tex]Power=V_{2}\times I_{2}[/tex]
Substituting the values we get
[tex]Power=12\times 2.55=30.6\ Watts[/tex]
Now for resistance connected directly across the source line Power will remain same,
[tex]Power=\dfrac{(V_{1})^{2}}{R_{1}}[/tex]
Substituting the values we get
[tex]R_{1}=\dfrac{(V_{1})^{2}}{Power}=\dfrac{14400}{30.6}=470.58\ ohms[/tex]
Therefore,
a) [tex]\dfrac{N_{1}}{N_{2}}=\dfrac{10}{1}[/tex]
b) I₂ = 2.55 Ampere
c) Power = 30.6 Watts
d) R₁ = 470.58 ohms.
