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*Sigh*. I'm embarrassed to ask this question because it seems like a silly question. But at the same time, if I don't ask, I don't know how else I'll find out the answer.
I have some documentation for work that says, "The relays are 110 V AC coils..." Believe it or not, that snippet is all you need. The actual context isn't important. I've read about relays, seen photos, and seen one in real life. From what I understand, they are remote switches -- electromechanical devices consisting of a solenoid that actuates an armature magnetically. If the relay is normally open, then the armature closes a contact, making an electrical connection.
Here is my question: for this application, isn't a constant magnetic field required? Especially if the contact is supposed to remain closed as long as there is current in the coil? If so, isn't DC required? Sure, I understand the applications of AC coils in situations in which one wishes to induce an EMF (e.g. in a transformer). But, I don't understand how a relay with an AC coil would work. Just to make sure I wasn't totally off, I reviewed solenoids:
The magnetic field of of our tightly-wound, ideal, infinite solenoid with n turns per unit length and a steady current I flowing through it is given by:
[tex] \mathbf{B} = \mu_0 n I \mathbf{\hat{z}} [/tex]...[1]
inside the solenoid. z is the longitudinal coordinate. Now, when I tried to think of what would happen in a solenoid supplied with an AC voltage, I got muddled up. If the voltage supplied was of the form:
[tex] v_s(t) = V_0 \cos(\omega t + \phi) [/tex]
then at first, the solenoid will try to draw a current:
[tex] i(t) = I_0 \cos(\omega t + \phi) [/tex]
with the relationship between I0 and V0 depending on the resistive (real) part of the coil's impedance, RL I.e.:
[tex]I_0 = \frac{V_0}{R_L} [/tex]
But whoah! The presence of a time-varying current in the coil and the corresponding changing magnetic flux through the coil induces an opposing EMF given by:
[tex] v_L(t) = L\frac{di(t)}{dt} [/tex]...[2]
So, since the EMF is always opposing, I'm guessing that the current at any instant is reduced, in the following manner:
[tex] i_L = \frac{v_s - v_L}{R_L} [/tex]...[3]
But wait! From [2], this alteration in the current alters vL. Which alters iL. Ad infinitum? ARRRRRRRRRRRGHH! I can't figure it out. Maybe one of you can explain it to me. But you know what? For the moment, I don't care. I'm just going to assume that in the steady state, the current through a solenoid supplied with an AC voltage can be expressed in the form:
[tex] i(t) = I \cos(\omega t + \phi) [/tex]
In which case (getting back to the issue at hand), from [1] the magnetic field in the coil would be given by:
[tex] \mathbf{B} = \mu_0 n i(t) \mathbf{\hat{z}} [/tex]
[tex] = \mu_0 n I \cos(\omega t + \phi) \mathbf{\hat{z}} [/tex]
The magnetic field also varies sinusoidally, and so it keeps switching direction, back and forth. That doesn't strike me as being very useful for actuating a switch! So what gives?
I have some documentation for work that says, "The relays are 110 V AC coils..." Believe it or not, that snippet is all you need. The actual context isn't important. I've read about relays, seen photos, and seen one in real life. From what I understand, they are remote switches -- electromechanical devices consisting of a solenoid that actuates an armature magnetically. If the relay is normally open, then the armature closes a contact, making an electrical connection.
Here is my question: for this application, isn't a constant magnetic field required? Especially if the contact is supposed to remain closed as long as there is current in the coil? If so, isn't DC required? Sure, I understand the applications of AC coils in situations in which one wishes to induce an EMF (e.g. in a transformer). But, I don't understand how a relay with an AC coil would work. Just to make sure I wasn't totally off, I reviewed solenoids:
The magnetic field of of our tightly-wound, ideal, infinite solenoid with n turns per unit length and a steady current I flowing through it is given by:
[tex] \mathbf{B} = \mu_0 n I \mathbf{\hat{z}} [/tex]...[1]
inside the solenoid. z is the longitudinal coordinate. Now, when I tried to think of what would happen in a solenoid supplied with an AC voltage, I got muddled up. If the voltage supplied was of the form:
[tex] v_s(t) = V_0 \cos(\omega t + \phi) [/tex]
then at first, the solenoid will try to draw a current:
[tex] i(t) = I_0 \cos(\omega t + \phi) [/tex]
with the relationship between I0 and V0 depending on the resistive (real) part of the coil's impedance, RL I.e.:
[tex]I_0 = \frac{V_0}{R_L} [/tex]
But whoah! The presence of a time-varying current in the coil and the corresponding changing magnetic flux through the coil induces an opposing EMF given by:
[tex] v_L(t) = L\frac{di(t)}{dt} [/tex]...[2]
So, since the EMF is always opposing, I'm guessing that the current at any instant is reduced, in the following manner:
[tex] i_L = \frac{v_s - v_L}{R_L} [/tex]...[3]
But wait! From [2], this alteration in the current alters vL. Which alters iL. Ad infinitum? ARRRRRRRRRRRGHH! I can't figure it out. Maybe one of you can explain it to me. But you know what? For the moment, I don't care. I'm just going to assume that in the steady state, the current through a solenoid supplied with an AC voltage can be expressed in the form:
[tex] i(t) = I \cos(\omega t + \phi) [/tex]
In which case (getting back to the issue at hand), from [1] the magnetic field in the coil would be given by:
[tex] \mathbf{B} = \mu_0 n i(t) \mathbf{\hat{z}} [/tex]
[tex] = \mu_0 n I \cos(\omega t + \phi) \mathbf{\hat{z}} [/tex]
The magnetic field also varies sinusoidally, and so it keeps switching direction, back and forth. That doesn't strike me as being very useful for actuating a switch! So what gives?
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