Dạng 1: Liên hệ giữa các giá trị hiệu dụng

Xét mạch: 
+ Cảm kháng: $Z_L=L.\omega =L.2.\pi.f$
+ Dung kháng: $Z_C=\frac{1}{C.\omega} =\frac{1}{C.2.\pi.f}$
* Tổng trở:
$Z_{AB}=\sqrt{(R_1+R_2)^2+(Z_L-Z_C)^2}$
$Z_{AP}=\sqrt{R_1^2+(Z_L-Z_C)^2}$
$Z_{AN}=\sqrt{R_1^2+Z_L^2}$
$Z_{AM}=\sqrt{R_1^2}=R_1$
$Z_{MP}=\sqrt{(Z_L-Z_C)^2}=\left | Z_L-Z_C \right |$
$Z_{NB}=\sqrt{R^2_2+Z_C^2}$
* Điện áp: $U_{AB}=\sqrt{(U_{R_1}+U_{R_2})^2+(U_L-U_C)^2}$
$U_{AP}=\sqrt{(U^2_{R_1}+(U_L-U_C)^2}$
$U_{AN}=\sqrt{(U^2_{R_1}+U_L^2}$
$U_{AM}=U_{R_1}$
$U_{AM}=\left | U_L-U_C \right |$
$U_{NB}=\sqrt{U^2_{R_2}+U^2_C}$
* Cường độ dòng điện:
$I=\frac{U_R}{R}=\frac{U_L}{Z_L}=\frac{U_C}{Z_C}=\frac{U_{RC}}{Z_RC}= \frac{U_{MN}}{Z_{MN}}=...=\frac{U _{dau}}{Z_{do}}$
* Độ lệch pha:
$tan\varphi _{MN}=\frac{Z_L-Z_C}{R}$
$tan\varphi _{RC}=\frac{-Z_C}{R}$
$tan\varphi _{RL}=\frac{Z_L}{R}$
$tan\varphi _{LC}=\frac{Z_L-Z_C}{O}=\pm \infty$
Nhớ: 
+ Thuộc kỹ cung góc và các giá trị tan, cos của hàm lượng giác.
+ $tan\varphi=\frac{a}{O}=\left\{\begin{matrix} +\infty \ neu \ a>0\\ -\infty \ neu \ a<0 \end{matrix}\right.$
+ $\varphi _{MN}=\varphi_{u _{MN}}-\varphi_ i$
VD1: Cho mạch điện: $R=100\Omega , C=15,9\mu F, f=50Hz, U_{MN}=U_{AB}=200V$
a. Tìm L?
b. Tìm $\varphi _{AM}, \varphi _{AB}$?
Giải
a.
$Z_C=\frac{1}{C.2.\pi.f}=\frac{1}{15,9.10^{-6}.2\pi.50}=200\Omega$
$U_{AM}=U_{AB}\Rightarrow Z_{ZM}=Z_{AB}\Rightarrow Z_{ZM}^2=Z_{AB}^2$
$\Rightarrow R^2+Z_L^2=R^2+(Z_L-Z_C)^2$
$\Rightarrow Z_L^2=(Z_L-Z_C)^2\Rightarrow \left | Z_L-Z_C \right |=Z_L$
$\Rightarrow \bigg \lbrack \begin{matrix} Z_L-Z_C=Z_L\Rightarrow Z_C=0 \ \ \ (loai)\\ Z_L-Z_C=-Z_L\Rightarrow Z_L=\frac{Z_C}{2}=\frac{200}{2}=100\Omega \end{matrix}$
$Z_L=L.2\pi.f\Rightarrow L=\frac{Z_L}{2\pi.f}=\frac{100}{2\pi.50}=\frac{1}{\pi}(H)$
b.
$tan_{\varphi _{AM}}=\frac{Z_L-O}{R}=\frac{100}{100}=1\Rightarrow \varphi _{AM}=\frac{\pi}{4}$
$tan_{\varphi _{AB}}=\frac{Z_L-Z_C}{R}=\frac{100-200}{100}=-1 \Rightarrow \varphi _{AB}=-\frac{\pi}{4}$
VD2: Đặt điện áp $u=U\sqrt{2}cos2.\pi.f(v)$ vào 2 đầu mạch RLC ghép nối tiếp thì $U_R=60V, U_L=120V, U_C=40V$
a. Tìm U?
b. Thay tụ C bằng tụ C' thì U'_R = 80V, Tìm U'_L và U'_C?
Giải
a.
$U=\sqrt{U^2_R+(U_L-U_C)^2}=\sqrt{60^2+(120-40)^2}=100V$
b. Ta có:
$U^2=U'^2_R+(U'_L-U'_C)^2$
$\Rightarrow 100^2=80^2+(U'_L-U'_C)^2$
$\Rightarrow \left | U'_L-U'_C \right |=\sqrt{100^2-80^2}=60$
Tỉ số $\frac{U'_L}{U'_R}=\frac{U_L}{U_R}=\frac{Z_L}{R}=\frac{120}{60}=2$

$\Rightarrow U'_L=2U'_R=2.80=160V$
$\Rightarrow \left | 160-U'_C \right |=60$
$\Rightarrow \bigg \lbrack \begin{matrix} 160-U'_C=60\Rightarrow U'_C=100V\\ 160-U'_C=-60\Rightarrow U'_C=220V \end{matrix}$