Answer to Question #116910 in Electric Circuits for havefun7741

Question

[img]https://upload.cc/i1/2020/05/09/uYgzwX.jpg[/img]




R=4

Answer 1

Solution

From the function for voltage we see that the angular frequency is 314 rad/s. Therefore, we can calculate the magnitude of current:

$I=\frac{V}{Z}=\frac{V}{\sqrt{R^2+(\omega L)^2}}=\\ \space\\ =\frac{380}{\sqrt{5^2+(100\pi\cdot6\cdot10^{-3})^2}}=71.1\text{ A}.$

Therefore, since all elements are linear, the current as a function of time is

$i(t)=71.1\text{ sin}(100\pi t).$

The phase shift between voltage and current:

$\phi=\text{atan}\frac{X_L}{R}=\text{atan}\frac{100\pi\cdot0.006}{5}=20.7^\circ.$

The average power consumed by the resistor:

$P=\frac{UI}{2}\text{ cos}\phi=\frac{380\cdot71.1}{2}\text{ cos}20.7^\circ=12700\text{ W}.$

Solution

The voltage generated in the secondary winding:

$V_2=(V-I_1R_1)\frac{N_2}{N_1}.\\ \space\\ I_1=\frac{N_2}{N_1}I_2.\\ \space\\ I_2=\frac{V_2}{R_2}, \\\space\\\rightarrow I_1=\frac{N_2V_2}{N_1R_2.}$

Substituting this into the first equation, obtain V₂:

$V_2=V\frac{R_2N_1N_2}{R_2N_1^2+R_1N_2^2}.$

Calculate the power:

$P_2=\frac{V_2^2}{R_2}=R_2\bigg(\frac{VN_1N_2}{R_2N_1^2+R_1N_2^2}\bigg)^2=40\text{ W}.$