Khi dong dien qua cuon day bang nua gia tri cuc dai thi nang luong tu truong la
$W_{t}=\frac{W}{4}$ voi $W$ la nang luong dien tu tong cong.
Luc nay nang luong dien truong o hai tu la nhu nhau va co gia tri la
$W_{tC1}=W_{tC2}=\frac{W-W_{t}}{2}=\frac{3W}{8}$
Khi mach nhanh chua tu $C_{2}$ ho thi nang luong tren tu nay mat di nhu vay nang luong dien tu chi con lai la
$W'=W-W_{tC1}=W-\frac{3W}{8}=\frac{5W}{8}$
Khi do ta co $W'=\frac{5W}{8} <=>\frac{1}{2}U_{01}^{2}C_{b1}=\frac{1}{2}.\frac{5}{8}U_{o2}^{2}C_{b2}$
$ <=> U_{o2}=\frac{\sqrt{5}}{2\sqrt{2}}\frac{U_{o1}\sqrt{C_{b1}}}{\sqrt{C_{b2}}}$
$ <=> U_{o2}=\frac{\sqrt{5}}{2\sqrt{2}}\frac{U_{01}\sqrt{2C_{1}}}{\sqrt{C_{1}}}$
$<=>U_{o2} =3\sqrt{5}V$