Nope. They are not conducting current, so they amount to pieces of wire connected to an empty glass bottle.
R = V/I and current (I) would be zero. So R = V/0, which is infinite. SO put infinite impedance in parallel with whatever the other tube impedance might be, and they calculate out to whatever that other tube is. A not conducting tube might as well be the air.
R = V/I and current (I) would be zero. So R = V/0, which is infinite. SO put infinite impedance in parallel with whatever the other tube impedance might be, and they calculate out to whatever that other tube is. A not conducting tube might as well be the air.
Comment