Question

A source of ac volatge `V = V_(0) cos omega t` delivers enegry yo a consummer by means of along straight coaxial cable with negligible active resistance. The current in the circuit varies as `I = I_(0) cos omegat - varphi`. Find the time-averaged enegry flux through the cross-section of the cable. The sheath is thin.

Answer 1

As in the previous problem
`E_(r) = (V_(0)cos omegat)/(rIn(r_(2))/(r_(1))` and `H_(theta) = (I_(0)cos(omegat - varphi))/(2pir)`
Hence time averged power flux (along the `z` axis) `= (1)/(2)V_(0)I_(0)cos varphi`
On using `lt cos omega t cos (omegat - varphi) gt = (1)/(2)cos varphi`.