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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 $u_2=90V$.

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


  • 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'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?

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Hi Bujanca, and welcome to Physics Stack Exchange! This site is for conceptual questions about physics, not general homework help - which means the first part of your question is fine here, but the second part may not be. I edited your question a bit to fix it up. Still, though, I'm not sure if you've provided enough information to get a proper answer. Could you quote the complete problem statement in your question? – David Z Mar 12 '13 at 20:48
@DavidZaslavsky thanks a lot for help. I hope that now it does fit the site requirements :) – Bujanca Mihai Mar 12 '13 at 21:10
Yep, that should definitely help! I edited it again only to "spruce up" the formatting: it's kind of conventional to quote the problem statement using blockquote syntax, >. Hopefully I got it right ;-) – David Z Mar 12 '13 at 21:20
@DavidZaslavsky yes, you did. Thank you for help and time, i appreciate – Bujanca Mihai Mar 12 '13 at 21:24

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

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

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


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


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. Z*I=120

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

use the two equations to get R and L

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