Xét mạch RLC nối tiếp 
Nếu \(i=I_0cos(\omega t+\varphi _i)\)

\(\Rightarrow \left\{\begin{matrix} u_R=U_{OR}cos(\omega t+\varphi _i)\\ u_L=U_{OL}cos(\omega t+\varphi _i+\frac{\pi}{2})\\ u_C=U_{OC}cos(\omega t+\varphi _i-\frac{\pi}{2}) \end{matrix}\right.\)
u_R cùng pha \(i\Rightarrow u_R=iR\Leftrightarrow i=\frac{u_R}{R}\)
u_L vuông pha \(i\Rightarrow \left ( \frac{u_L}{U_{0L}} \right )+\left ( \frac{i}{I_0} \right )^2=1\)
u_C vuông pha \(i\Rightarrow \left ( \frac{u_C}{U_{0C}} \right )+\left ( \frac{i}{I_0} \right )^2=1\)
u_R vuông pha \(u_L \Rightarrow \left ( \frac{u_R}{U_{OR}} \right )^2+ \left ( \frac{u_L}{U_{OL}} \right )^2=1\)
u_R vuông pha \(u_C \Rightarrow \left ( \frac{u_R}{U_{OR}} \right )^2+ \left ( \frac{u_C}{U_{OC}} \right )^2=1\)

u_R ngược pha \(u_C \Rightarrow \left\{\begin{matrix} u_L=U_{OL}cos(\omega t+\varphi )\\ u_C=U_{OCL}cos(\omega t+\varphi ) \end{matrix}\right.\Rightarrow \frac{u_L}{u_C} =\frac{U_{OL}}{U_{OC}} =\frac{Z_L}{Z_C}\)
* Chú ý:

\(U_{OL}=U_L\sqrt{2}, \ \ I_O=I \sqrt{2}\)
\(\Rightarrow \left ( \frac{u_L}{U_{OL}} \right )+\left ( \frac{i}{I_O} \right )^2=1 \Rightarrow \left ( \frac{u_L}{U_{L}} \right )+\left ( \frac{i}{I} \right )^2=2\)
* Viết biểu thức:
+ Nếu đề cho \(u_{MN}=U_{OMN}.cos(\omega t+\varphi_{u_{MN}} )\)
\(\rightarrow\) Yêu cầu viết biểu thức:
Ta có: \(i=I_O cos(\omega t+\varphi _i)\)
+ Tìm \(Z_{MN}\Rightarrow I_O=\frac{U_{OMN}}{Z_{MN}}\)
+ Tìm \(tan\varphi _{MN}\Rightarrow \varphi _{MN}\Rightarrow \varphi _i=\varphi _{u_{MN}}-\varphi _{MN}\)
Vậy \(i=I_O cos(\omega t+\varphi _i)\)
+ Nếu đề cho \(i=I_O cos(\omega t+\varphi _i)\)
\(\rightarrow\) Yêu cầu viết biểu thức \(u_{MN}\)?
+ Tìm \(Z_{MN}\Rightarrow U_{OMN}=I_O.Z_{MN}\)
+ Tìm \(tan\varphi _{MN}\Rightarrow \varphi _{MN}\Rightarrow \varphi _{u_{MN}} =\varphi _i+\varphi _{MN}\)
Vậy \(u_{MN}=U_{OMN} cos(\omega t + \varphi _{u_{MN}})\)
+ Nếu đề cho u₁ \(\rightarrow\) Yêu cầu viết u₂?
Từ u₁ \(\rightarrow\) Viết biểu thức i \(\rightarrow\) Viết u₂

VD1: Đặt điện áp \(u=200\sqrt{2}cos(100\pi t + \frac{\pi }{6}) (v)\) vào 2 đầu tụ \(C=\frac{10^{-4}}{\pi}F\)
a. Viết biểu thức i?
b. Tại thời điểm \(u=10\sqrt{6}(V)\) thì cường độ dòng điện i bằng bao nhiêu?
Giải
a.
\(Z_C=\frac{1}{C.\omega }=\frac{1}{\frac{10^{-4}}{\pi}.100\pi}=100\Omega\)
\(I_O=\frac{U_{OC}}{Z_C}=\frac{200\sqrt{2}}{100}=2\sqrt{2}A\)
\(\varphi _{u_{C}}=\varphi _i-\frac{\pi}{2}\Rightarrow \varphi _i=\varphi _{u_{C}}+ \frac{\pi}{2}=\frac{\pi}{6}+\frac{\pi}{2}=\frac{2\pi}{3}\)
Vậy \(i=2\sqrt{2}.cos(100\pi t + \frac{2 \pi}{3}) (A)\)
b.\(u_C\perp i\Rightarrow \left ( \frac{u_C}{U_{OC}} \right )^2+\left ( \frac{i}{I_O } \right )^2=1\)
\(\Rightarrow i=\pm I_O\sqrt{1-\left ( \frac{u_C}{U_{O}} \right )^2}=\pm 2\sqrt{2} \sqrt{1-\left ( \frac{100\sqrt{6}}{200\sqrt{2}} \right )^2}\)
\(\Rightarrow i=\pm 2\sqrt{2}.\frac{1}{2}=\pm \sqrt{2}(A)\)
VD2: Đặt điện áp \(u=200cos(100\pi t-\frac{\pi}{4})(V)\) vào 2 đầu mạch RLC ghép nối tiếp có \(R = 50\Omega, L=318mH, C=63,6\mu F\). Viết biểu thức \(i, u_R, u_C, u_{RC}, u_{LC}\).
Giải
\(Z_L=L.\omega =318.10^{-3}.100 \pi=100\Omega\)
\(Z_C=\frac{1}{C.\omega} =\frac{1}{63,6.10^{-6}.100 \pi}=50\Omega\)
\(Z=\sqrt{R^2+(Z_L-Z_C)^2}=\sqrt{50^2+(100-50)^2}=50\sqrt{2}\Omega\)
\(\Rightarrow I_O=\frac{U_O}{Z}=\frac{200}{50\sqrt{2}}=2\sqrt{2}A\)
\(tan\varphi =\frac{Z_L-Z_C}{R}=\frac{100-50}{50}=1\Rightarrow \varphi =\frac{\pi }{4}\)
\(\Rightarrow \varphi _i=\varphi _u-\varphi =-\frac{\pi}{4}-\frac{\pi}{4}=-\frac{\pi}{2}\)
Vậy \(i=2\sqrt{2}.cos(100\pi t-\frac{\pi}{2})(A)\)
\(u_R=100\sqrt{2}.cos(100\pi t-\frac{\pi}{2})(v)\)
\(u_L=200\sqrt{2}.cos(100\pi t-\frac{\pi}{2}+\frac{\pi}{2})(v)\)
\(u_C=100\sqrt{2}.cos(100\pi t-\frac{\pi}{2}-\frac{\pi}{2})(v)\)
\(u_{RC}=U_{ORC}.cos(\omega t+\varphi _{u_{RC}})\)
\(Z_{RC}=\sqrt{R^2+Z_L^2}=\sqrt{50^2+50^2}=50\sqrt{2}\Omega\)
\(\Rightarrow U_{ORC}=I_{RC}=2\sqrt{2}.50\sqrt{2}=200V\)
\(tan\varphi _{RC}=\frac{-Z_C}{R}=\frac{-50}{50}=-1\Rightarrow \varphi _{RC}=-\frac{\pi }{4}\)
\(\Rightarrow \varphi _{u_{RC}}=\varphi _i+\varphi _{RC}=-\frac{\pi}{2}+(-\frac{\pi}{4} )=-\frac{3\pi}{4}\)
Vậy \(u_{RC}=200.cos(100\pi t-\frac{3 \pi}{4})(V)\)
Viết \(u_{LC}=U_{OLC}.cos (\omega t+\varphi _{u_{LC}})\)
\(U_{OLC}=I_O.Z_{LC}\left | Z_L-Z_C \right |=2\sqrt{2}.\left | 100-50 \right |=100\sqrt{2}\)
\(tan\varphi _{LC}=\frac{Z_L-Z_C}{O}=\frac{50}{0}=\infty \Rightarrow \varphi _{LC}=\frac{\pi}{2}\)
\(\Rightarrow \varphi _{u_{LC}}=\varphi _i+\varphi _{LC}=-\frac{\pi}{2}+\frac{\pi}{2}=0\)
Vậy \(u_{LC}=100\sqrt{2}.cos.100\pi t (V)\)
Cài đặt: 
\(shift\rightarrow mode\rightarrow 4\rightarrow R\)
\(mode\rightarrow 2\rightarrow CMPLX\)
\(u=U_O cos(\omega t + \varphi _u)\rightarrow U_O< \varphi _u\)
\(i=I_O cos(\omega t + \varphi _i)\rightarrow I_O<\varphi _i\)
\(Z=\sqrt{R^2+(Z_L-Z_C)^2}\rightarrow R+(Z_L-Z_C)i\)
Viết i
\(i=\frac{u}{Z}=\frac{U_O<\varphi _u}{R+(Z_L-Z_C)i}=\frac{U_{ORC}< \varphi _{u_{RC}}}{R-Z_Ci }=\frac{u_{ORL}<\varphi _{u_{RL}}}{R+Z_L.i}\)
\(=\frac{U_{OL}<\varphi _{u_{L}}}{Z_L.i}=\frac{U_{OC}<\varphi _{u_{C}}}{-Z_C.i}=...\)
Viết u:
\((I_O<\varphi _i).[R+(Z_L-Z_C)i]\)
\(\rightarrow (I_O<\varphi _i).(R-Z_C.i)=U_{ORC}<\varphi _{i_{RC}}\)Tìm Z:
\(\frac{U_O<\varphi _u}{I_O<\varphi _i}=R+(Z_L-Z_C)i\)
Chú ý:
\(shift\rightarrow 2\rightarrow 3=r<\) Đ
\(shift\rightarrow 2\rightarrow 4=a+bi\)