Electric circuit with antenna
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This is getting ridiculous. You keep posting these drawings that are barely legible. The scribbling is bad enough, but the uncropped images are unacceptable. The image you originally posted below was 2500 x 1667 pixels, even though the drawing occupied only a small piece of that. I cropped it to just around the drawing, which resulted in 300 x 476 pixels. That can be displayed natively in a post here (up to 640 pixels wide is apparently the limit), so the result isn't shrunk to oblivion like your original. It still looks like it was drawn by a second grader with crayons, but is more readable.
I fixed it for you this one time. I'm not going to do that again. Next time your question will simply be closed. Don't be so lazy. Stop disrespecting the volunteers here you are asking for free help.
And no, saying you don't know how to crop images, use MathJax, create legible schematics, or anything else to produce a properly readable post is no excuse. Do it right, or you don't get to do it here.
Suppose we have this circuit with an antenna.In order for the antenna to radiate EM fields electrons must accumulate to and from the edge of the antenna.
But how does that happen since an antenna is an open circuit?And how do we apply Kirchoff's laws in that case?
1 answer
Antennas are not lumped-parameter systems. They may be open circuits or shorts at DC, but this changes as the frequency goes up. They aren't just a bunch of wires anymore at the intended operating frequency.
Take a simple dipole as example. It's just two disconnected wires at DC, but that doesn't tell you anything useful about what happens at the optimum radiation frequency. If you apply a steady voltage, one wire becomes positive, and the other negative. However, if you keep changing that voltage, then charges must move back and forth.
If you change the voltage at the right rate so that it matches the speed of propagation in the wire, then you're exiting resonance. Charges slosh back and forth smoothly from one end to the other. Just as the charges get to one end, the driving voltage is changed so that they need to slosh to the other end again. You end up with a voltage profile of half a sine wave. The current profile is also half a sine wave, but with a 90° phase shift.
The concept is similar to moving your hand back and forth in a bathtub of water. If you do it at the right frequency, a little back and forth with your hand in the middle causes a lot of back and forth of all the water in the bathtub. Each time your hand goes back and forth, it adds a little energy to the system. That energy goes into increasing the amplitude of the water's movement. The wave peaks at each end get higher, and the volume of water that sloshes past the middle each cycle gets larger.
The height of the water is like the voltage on the dipole, and volume that moves each cycle is like the current.
The dipole has an additional trick that the bathtub doesn't. The charges sloshing back and forth is a current, which creates a magnetic field. The voltage going up and down creates an electric field. When these are matched at the right phase and relative amplitudes, then energy is propagated into space.
Since energy is radiated from the antenna, it has to be replaced if a signal is transmitted continuously. This appears to the driving circuit as the antenna having a resistive component. In fact, at the optimum radiation frequency, the antenna ideally looks like a pure resistance to the driving circuit. That resistance happens to be 75 Ω for a simple half wave dipole. Other antennas have other characteristic impedances.
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