Đoạn mạch RLC nối tiếp có \(R = 50\Omega ;L = 159mH,C = 31,8\mu F.\)
\(u = 120\cos 100\pi t(V) \Rightarrow {U_0} = 120(V);\omega = 100\pi (rad/s)\)
Ta có :\({Z_L} = L\omega = {159.10^{ - 3}}.100\pi = 50(\Omega )\)
\({Z_C} = \displaystyle{1 \over {C\omega }} = \displaystyle{1 \over {31,{{8.10}^{ - 6}}.100\pi }} = 100(\Omega )\)
\( \Rightarrow \) \(Z = \sqrt {{R^2} + {{({Z_L} - {Z_C})}^2}} = \sqrt {{{50}^2} + {{(50 - 100)}^2}} = 50\sqrt 2 (\Omega )\)
\( \Rightarrow \) \({I_0} = \displaystyle{{{U_0}} \over {{Z_{AB}}}} = {{120} \over {50\sqrt 2 }} = 1,2\sqrt 2 (A)\)
\(\tan \varphi = \displaystyle{{{Z_L} - {Z_C}} \over R} = {{50 - 100} \over {50}} = - 1 \Rightarrow \varphi = {{ - \pi } \over 4}\)
=> u trễ pha hơn i một góc \(\dfrac{\pi}{4}\) hay \(\varphi_u -\varphi_i = -\dfrac{\pi}{4}\)
=> \(\varphi_i=\varphi_u + \dfrac{\pi}{4}=0+\dfrac{\pi}{4}=\dfrac{\pi}{4} rad\)
Vậy : \(i = {I_0}\cos (100\pi t + \varphi_i )\)
\(\Leftrightarrow i = 1,2\sqrt 2 \cos (100\pi t + \displaystyle{\pi \over 4})(A)\)
