Công suất tiêu thụ:$P=\frac{U^{2}}{\sqrt{R^{2}+Z_{C}^{2}}}cos\varphi=\frac{U^{2}}{\sqrt{R^{2}+Z_{C}^{2}}}\times\frac{R}{\sqrt{R^{2}+Z_{C}^{2}}}=U^{2}\frac{1}{R+\frac{Z_{C}^{2}}{R}}$Ứng với $R_{1}$ và $R_{2}$ ta có:$\frac{R_{1}}{R_{1}^{2}+Z_{C}^{2}}=\frac{R_{2}}{R_{2}^{2}+Z_{C}^{2}}$$\Rightarrow R_{1}R_{2}^{2}+Z_{C}^{2}R_{1}=R_{2}R_{1}^{2}+Z_{C}^{2}R_{2}$$\Rightarrow R_{1}R_{2}(R_{2}-R_{1})=Z_{C}^{2}(R_{2}-R_{1})$$\Rightarrow R_{1}R_{2}=Z_{C}^{2}$Bây giờ ta có:$P^{2}=\frac{U^{2}R_{1}}{R_{1}^{2}+Z_{C}^{2}}\times\frac{U^{2}R_{2}}{R_{2}^{2}+Z_{C}^{2}}=U^{4}\frac{Z_{C}^{2}}{Z_{C}^{4}+Z_{C}^{2}(R_{1}^{2}+R_{2}^{2}+Z_{C}^{4})}=U^{4}\frac{Z_{C}^{2}}{Z_{C}^{4}+Z_{C}^{2}\left[ {(R_{1}+R_{2})^{2}-R_{1}R_{2}} \right]+Z_{C}^{4}}=U^{4}\frac{Z_{C}^{2}}{Z_{C}^{4}+Z_{C}^{2}\left[ {200^{2}-Z_{C}^{2}} \right]+Z_{C}^{4}}=\frac{U^{4}}{200^{2}}$Vậy: $P=\frac{U^{2}}{200}=50(W)$b) Ta có:$R_{1}=50$ suy ra $R_{2}=150$$\Rightarrow Z_{C}=\sqrt{R_{1}R_{2}}=86.6(\Omega)$Lại có:$R+\frac{Z_{C}^{2}}{R}\geq 2Z_{C}$Cực tiểu xảy ra khi: $R=\frac{Z_{C}^{2}}{R}$ hay $R=Z_{C}=86.6(\Omega)$Mẫu số cực tiểu thì $P$ cực đại.
Công suất tiêu thụ:$P=\frac{U^{2}}{\sqrt{R^{2}+Z_{C}^{2}}}cos\varphi=\frac{U^{2}}{\sqrt{R^{2}+Z_{C}^{2}}}\times\frac{R}{\sqrt{R^{2}+Z_{C}^{2}}}=U^{2}\frac{1}{R+\frac{Z_{C}^{2}}{R}}$Ứng với $R_{1}$ và $R_{2}$ ta có:$\frac{R_{1}}{R_{1}^{2}+Z_{C}^{2}}=\frac{R_{2}}{R_{2}^{2}+Z_{C}^{2}}$$\Rightarrow R_{1}R_{2}^{2}+Z_{C}^{2}R_{1}=R_{2}R_{1}^{2}+Z_{C}^{2}R_{2}$$\Rightarrow R_{1}R_{2}(R_{2}-R_{1})=Z_{C}^{2}(R_{2}-R_{1})$$\Rightarrow R_{1}R_{2}=Z_{C}^{2}$Bây giờ ta có:$P^{2}=\frac{U^{2}R_{1}}{R_{1}^{2}+Z_{C}^{2}}\times\frac{U^{2}R_{2}}{R_{2}^{2}+Z_{C}^{2}}=U^{4}\frac{Z_{C}^{2}}{Z_{C}^{4}+Z_{C}^{2}(R_{1}^{2}+R_{2}^{2}+Z_{C}^{4})}=U^{4}\frac{Z_{C}^{2}}{Z_{C}^{4}+Z_{C}^{2}\left[ {(R_{1}+R_{2})^{2}-R_{1}R_{2}} \right]+Z_{C}^{4}}=U^{4}\frac{Z_{C}^{2}}{Z_{C}^{4}+Z_{C}^{2}\left[ {200^{2}-Z_{C}^{2}} \right]+Z_{C}^{4}}=\frac{U^{4}}{200^{2}}$Vậy: $P=\frac{U^{2}}{200}=50(W)$b) Ta có:$R_{1}=50$ suy ra $R_{2}=150$$\Rightarrow Z_{C}=\sqrt{R_{1}R_{2}}=86.6(\Omega)$Lại có:$R+\frac{Z_{C}^{2}}{R}\geq 2Z_{C}$Cực tiểu xảy ra khi: $R=\frac{Z_{C}^{2}}{R}$ hay $R=Z_{C}=86.6(\Omega)$
Công suất tiêu thụ:$P=\frac{U^{2}}{\sqrt{R^{2}+Z_{C}^{2}}}cos\varphi=\frac{U^{2}}{\sqrt{R^{2}+Z_{C}^{2}}}\times\frac{R}{\sqrt{R^{2}+Z_{C}^{2}}}=U^{2}\frac{1}{R+\frac{Z_{C}^{2}}{R}}$Ứng với $R_{1}$ và $R_{2}$ ta có:$\frac{R_{1}}{R_{1}^{2}+Z_{C}^{2}}=\frac{R_{2}}{R_{2}^{2}+Z_{C}^{2}}$$\Rightarrow R_{1}R_{2}^{2}+Z_{C}^{2}R_{1}=R_{2}R_{1}^{2}+Z_{C}^{2}R_{2}$$\Rightarrow R_{1}R_{2}(R_{2}-R_{1})=Z_{C}^{2}(R_{2}-R_{1})$$\Rightarrow R_{1}R_{2}=Z_{C}^{2}$Bây giờ ta có:$P^{2}=\frac{U^{2}R_{1}}{R_{1}^{2}+Z_{C}^{2}}\times\frac{U^{2}R_{2}}{R_{2}^{2}+Z_{C}^{2}}=U^{4}\frac{Z_{C}^{2}}{Z_{C}^{4}+Z_{C}^{2}(R_{1}^{2}+R_{2}^{2}+Z_{C}^{4})}=U^{4}\frac{Z_{C}^{2}}{Z_{C}^{4}+Z_{C}^{2}\left[ {(R_{1}+R_{2})^{2}-R_{1}R_{2}} \right]+Z_{C}^{4}}=U^{4}\frac{Z_{C}^{2}}{Z_{C}^{4}+Z_{C}^{2}\left[ {200^{2}-Z_{C}^{2}} \right]+Z_{C}^{4}}=\frac{U^{4}}{200^{2}}$Vậy: $P=\frac{U^{2}}{200}=50(W)$b) Ta có:$R_{1}=50$ suy ra $R_{2}=150$$\Rightarrow Z_{C}=\sqrt{R_{1}R_{2}}=86.6(\Omega)$Lại có:$R+\frac{Z_{C}^{2}}{R}\geq 2Z_{C}$Cực tiểu xảy ra khi: $R=\frac{Z_{C}^{2}}{R}$ hay $R=Z_{C}=86.6(\Omega)$
Mẫu số cực tiểu thì $P$ cực đại.