Respuesta :
1. [tex]I_2 = 0.14 I_1[/tex]
Explanation:
We have:
[tex]V_1 = 15.7 V[/tex] voltage in the primary coil
[tex]V_2 = 110 V[/tex] voltage in the secondary coil
The efficiency of the transformer is 100%: this means that the power in the primary coil and in the secondary coil are equal
[tex]P_1 = P_2\\V_1 I_1 = V_2 I_2[/tex]
where I1 and I2 are the currents in the two coils. Re-arranging the equation, we find
[tex]\frac{I_2}{I_1}=\frac{V_1}{V_2}=\frac{15.7 V}{110 V}=0.14[/tex]
which means that the current in the secondary coil is 14% of the value of the current in the primary coil.
2. 5.7 V
We can solve the problem by using the transformer equation:
[tex]\frac{N_p}{N_s}=\frac{V_p}{V_s}[/tex]
where:
Np = 400 is the number of turns in the primary coil
Ns = 19 is the number of turns in the secondary coil
Vp = 120 V is the voltage in the primary coil
Vs = ? is the voltage in the secondary coil
Re-arranging the formula and substituting the numbers, we find:
[tex]V_s = V_p \frac{N_s}{N_p}=(120 V)\frac{19}{400}=5.7 V[/tex]
1. Current is defined as the rate of flow of charge in a coil. The current in the secondary coil will be 0.14 times the current in the primary coil.
2. The voltage in the secondary coil will be 5.7 V.
What is defined as power in a transformer?
Power is defined as the product of voltage difference between the end of the coil and current flow through the coil. Its unit is the watt.it is given by
power = voltage difference × electric current
P=VI
V₁ is the voltage in the primary coil=15.7 V
V₂ is the voltage in the secondary coil=110 V
The efficiency given in the solution is 100%. So that power in the primary and secondary coil is the same. So that
P₁=P₂
V₁I₁=V₂I₂
15.7I₁=110I₂
I₂=0.14I₁
Hence the current in the secondary coil will be 0.14 times the current in the primary coil.
2.
Np is the number of turns in the primary coil = 400
Ns is the number of turns in the secondary coil= 19
Vp is the voltage in the primary coil = 120 V
Vs is the voltage in the secondary coil =?
According to transformer formula;
[tex]\frac{N_p}{N_s} =\frac{V_p}{V_s} \\\\\rm V_p=\frac{N_p}{N_s}\times{V_s} \\\\ \rm V_p=\frac{400}{19}\times{120}\\\\\rm V_p=5.7\;V[/tex]
Hence the voltage in the secondary coil will be 5.7 V.
To learn more about the power in transformer refer to the link;
https://brainly.com/question/13887561
