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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. |
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Answer» SOLUTION :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`. |
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