Am I properly identifying the resistor terminal voltage?

Question

The frequency applied to a circuit of voltage 120 V with a real coil and a resistor has a value of 50 Hz. The resistance of the resistor is 10 $\Omega$. The voltage at the resistor terminals $u_1=60V$. The voltage at the coil's terminals is $u_2=90V$.

$$\nu=50\text{ Hz}$$
$$R=10\ {\Omega }$$ $$U=120\text{ V}$$
$$u_1=60\text{ V - resistor terminal voltage}$$
$$u_2=90\text{ V - coil terminal voltage}$$

Find:

• the intensity of the current $I$.
• the parameters of the coil.

I think by "parameters of the coil", it is meant the resistance and the impedance of the coil.

$$I=?$$
$$L=?$$
$$R_L=?$$

I've been trying a bit, but I am quite poor at physics. This problem is suggested in a book, I want to prepare for a testpaper.

What I've been thinking of is to calculate $$\cos\phi=\frac{U_r }{U}=\frac{60}{120}=\frac{1}{2}$$ $$\implies \phi=\pi/3$$ But I am not sure if the resistor terminal voltage is the same thing with $U_r$. Is this right?

Answer 1

Ok , it was a nice one.! Let's consider the real inductor as a pure inductor plus resistance pair.

the value of I(rms) can be get using the Pure resistor outside this pair,

U=I(rms)*R2 ; r2=10;

I(rms)=6;

Now potential drop across real inductor Ulr (90) , across pure inductor Ul and across resistance Ur are pythagorean triplets.

Ulr^2=Ul^2+Ur^2;

We get

225=R^2+(50L)^2 ; I=6A;

Now , see circuit as a whole;

potential drop is ==> sqrt[(10+R)^2 + (50L)^2 ] 6 ; ie. ZI=120

we get => 100 + R^2 + 2R + (50L)^2 = 400;

use the two equations to get R and L