In section 10.5, I claimed that if you had an infinitely long coaxial cable and

Question

In section 10.5, I claimed that if you had an infinitely long coaxial cable and applied a constant voltage V to the left end of it, you would need to supply a constant current I = V/Z, where Z is characteristic impedance of the coax, in order to charge up the capacitance between the inner and outer conductors, as the wavefront of V propagates to the right. Although we have already made arguments that show this is correct in section 10.5, sometimes it is helpful to see things from more than one point of view. Starting from Q = CV, and taking into account the time dependence of the effective capacitance (as the wavefront propagates to the right), show that, indeed, I = V/Z.

Explanation / Answer

Ans:- if the infinitely long coaxial cable. Voltage V . so we are suppose Impedance(Z)  of the wire and the dielectric insulator (suppose 75 ohm coaxial cable) and no resistance of the coaxial cable. if the superconducting 75 ohm coaxial cable with 0 resistance coaxial cable but still has Impedance(Z) of 75 ohms.

so you apply a voltage (V) in coaxial cable, the starting current flows passing through the self inductance of the wire & charges the self capacitance of coaxial wire. if the starting current is calculating the ratio of the self inductance to the self capacitance for that function is called Impedance of the coaxial cable.

So that the current I=V / (R+Z0) so we are assume the bulb requirement of the current to produce the light. a step voltage passing from left to right charge up the self capacitance of the wire & the self inductance of the wire. so that the voltage is power to be zero. when that are reach the short circuit &  2nd step is reflected back in source. Because Z0 =Z , So no need further reflection. so steady state is increase the speed after 2 x time to travel along the coaxial wire. for example :- Light travels about 1ft in 1 nano second, & electricity speed 2/3 at this speed, it means ray passing in coaxial cable speed is 300 nano seconds for a 100 ft long cable. as per this

So the Voltage Vs =V × Z / (Z+Z0)   travels from left to right. so the current which flows is I =(V×Z / (Z+Z0) )Z.

the Impedance Z and output resistance Z0 are very much less than resistance R, so this current is greater than I=V / (R+Z0). Suppose we are taken the bulb. when the wave front passing bulb. so the Impedance Z is much more greater than resistance R. so that current is very lesser than I=V / (R+Z0). so we are know no bulbs light in the current or voltage wave front are reflected back from the short circuit. The bulb lights Z0=Z no reflection takes place when the voltage source and the steady state current I=V / (R+Z0) . Now flowing along the coaxial cable. so We can ignore the lamp resistance. because impedance Z0 is greater than Resistance R. so we write as I = V / Z .

So we prove that the time dependence of the effective capacitance as the wave front propagate to the right.

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