Disconnect the resistor R from the circuit, and set the disconnected upper and lower ends as nodes A and B respectively.

The external variables of the voltage source E are two parallel branches, the series impedances of C and L are Xc and XL respectively, and the negative pole of the voltage source is node N, then:

Guan (phasor) = e× jxl/(jxl-jxc) = XL× e/(XL-xc);

Ubn (phasor) = e× (-jxc)/(jxl-jxc) =-xc× e/(XL-xc).

So: Uoc (phasor) =Uab (phasor) =Uan-Ubn=(XL+Xc)E/(XL-Xc).

Then the voltage source is short-circuited, and the circuit is changed to two LC parallel circuits in series, and the Thevenin equivalent impedance is obtained as follows: zeq = zab = jxl× (-jxc)/(jxl-jxc)+jxl× (-jxc) =-j2lxc/(XL-xc).

According to Thevenin theorem: v (phasor) =Uoc (phasor) × r/(r+zeq) = (XL+xc) × r× e/(XL-xc)/[-j2xlxc/(XL-xc)] = (jxl+jxc) × r× e/.

So: V/E=(jXL+jXc)R/(2XLXc).