Xét mạch:
Độ lệch pha:
\(R=100\Omega , C=15,9\mu F, f=50Hz, U_{MN}=U_{AB}=200V\)
a. Tìm L?
b. Tìm \(\varphi _{AM}, \varphi _{AB}\)?
\(Z_C=\frac{1}{C.2.\pi.f}=\frac{1}{15,9.10^{-6}.2\pi.50}=200\Omega\)
\(U_{AM}=U_{AB}\Rightarrow Z_{ZM}=Z_{AB}\Rightarrow Z_{ZM}^2=Z_{AB}^2\)
\(\Rightarrow R^2+Z_L^2=R^2+(Z_L-Z_C)^2\)
\(\Rightarrow Z_L^2=(Z_L-Z_C)^2\Rightarrow \left | Z_L-Z_C \right |=Z_L\)
\(\Rightarrow \bigg \lbrack \begin{matrix} Z_L-Z_C=Z_L\Rightarrow Z_C=0 \ \ \ (loai)\\ Z_L-Z_C=-Z_L\Rightarrow Z_L=\frac{Z_C}{2}=\frac{200}{2}=100\Omega \end{matrix}\)
\(Z_L=L.2\pi.f\Rightarrow L=\frac{Z_L}{2\pi.f}=\frac{100}{2\pi.50}=\frac{1}{\pi}(H)\)
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'$_{C?}$
\(U=\sqrt{U^2_R+(U_L-U_C)^2}=\sqrt{60^2+(120-40)^2}=100V\)
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 \(\frac{U'_L}{U'_R}=\frac{U_L}{U_R}=\frac{Z_L}{R}=\frac{120}{60}=2\)
\(\Rightarrow U'_L=2U'_R=2.80=160V\)
\(\Rightarrow \left | 160-U'_C \right |=60\)
\(\Rightarrow \bigg \lbrack \begin{matrix} 160-U'_C=60\Rightarrow U'_C=100V\\ 160-U'_C=-60\Rightarrow U'_C=220V \end{matrix}\)
- Cảm kháng: \(Z_L=L.\omega =L.2.\pi.f\)
- Dung kháng: \(Z_C=\frac{1}{C.\omega} =\frac{1}{C.2.\pi.f}\)
- \(Z_{AB}=\sqrt{(R_1+R_2)^2+(Z_L-Z_C)^2}\)
- \(Z_{AP}=\sqrt{R_1^2+(Z_L-Z_C)^2}\)
- \(Z_{AN}=\sqrt{R_1^2+Z_L^2}\)
- \(Z_{AM}=\sqrt{R_1^2}=R_1\)
- \(Z_{MP}=\sqrt{(Z_L-Z_C)^2}=\left | Z_L-Z_C \right |\)
- \(Z_{NB}=\sqrt{R^2_2+Z_C^2}\)
- \(U_{AB}=\sqrt{(U_{R_1}+U_{R_2})^2+(U_L-U_C)^2}\)
- \(U_{AP}=\sqrt{(U^2_{R_1}+(U_L-U_C)^2}\)
- \(U_{AN}=\sqrt{(U^2_{R_1}+U_L^2}\)
- \(U_{AM}=U_{R_1}\)
- \(U_{AM}=\left | U_L-U_C \right |\)
- \(U_{NB}=\sqrt{U^2_{R_2}+U^2_C}\)
Độ lệch pha:
- \(tan\varphi _{MN}=\frac{Z_L-Z_C}{R}\)
- \(tan\varphi _{RC}=\frac{-Z_C}{R}\)
- \(tan\varphi _{RL}=\frac{Z_L}{R}\)
- \(tan\varphi _{LC}=\frac{Z_L-Z_C}{O}=\pm \infty\)
- Thuộc kỹ cung góc và các giá trị tan, cos của hàm lượng giác.
- \(tan\varphi=\frac{a}{O}=\left\{\begin{matrix} +\infty \ neu \ a>0\\ -\infty \ neu \ a<0 \end{matrix}\right.\)
- \(\varphi _{MN}=\varphi_{u _{MN}}-\varphi_ i\)
\(R=100\Omega , C=15,9\mu F, f=50Hz, U_{MN}=U_{AB}=200V\)
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 \(\frac{U'_L}{U'_R}=\frac{U_L}{U_R}=\frac{Z_L}{R}=\frac{120}{60}=2\)
\(\Rightarrow U'_L=2U'_R=2.80=160V\)
\(\Rightarrow \left | 160-U'_C \right |=60\)
\(\Rightarrow \bigg \lbrack \begin{matrix} 160-U'_C=60\Rightarrow U'_C=100V\\ 160-U'_C=-60\Rightarrow U'_C=220V \end{matrix}\)