Câu hỏi: Đặt điện áp
A.
B.
C.
D.
A.
B.
C.
D.
\begin{aligned}
& \text { Khi } \mathrm{L}=\mathrm{L}_1 \text { thì } \varphi_{\mathrm{i}}=\varphi_{u_L}-\frac{\pi}{2}=\frac{\pi}{6}-\frac{\pi}{2}=-\frac{\pi}{3} \\
& \tan \varphi_1=\frac{Z_{L 1}-Z_C}{R} \Rightarrow \tan \frac{\pi}{3}=\frac{Z_{L 1}-10 \sqrt{3}}{10} \Rightarrow Z_{L 1}=20 \sqrt{3} \Omega \stackrel{L_2=\frac{L_1}{3}}{\longrightarrow} Z_{L 2}=\frac{20 \sqrt{3}}{3} \Omega . \\
& \text { Khi } \mathrm{L}=\mathrm{L}_2 \text { thì } i=\frac{u}{R+\left(Z_{L 2}-Z_C\right) j}=\frac{220 \sqrt{2} \angle 0}{10+\left(\frac{20 \sqrt{3}}{3}-10 \sqrt{3}\right) j}=11 \sqrt{6} \angle \frac{\pi}{6} .
\end{aligned}
& \text { Khi } \mathrm{L}=\mathrm{L}_1 \text { thì } \varphi_{\mathrm{i}}=\varphi_{u_L}-\frac{\pi}{2}=\frac{\pi}{6}-\frac{\pi}{2}=-\frac{\pi}{3} \\
& \tan \varphi_1=\frac{Z_{L 1}-Z_C}{R} \Rightarrow \tan \frac{\pi}{3}=\frac{Z_{L 1}-10 \sqrt{3}}{10} \Rightarrow Z_{L 1}=20 \sqrt{3} \Omega \stackrel{L_2=\frac{L_1}{3}}{\longrightarrow} Z_{L 2}=\frac{20 \sqrt{3}}{3} \Omega . \\
& \text { Khi } \mathrm{L}=\mathrm{L}_2 \text { thì } i=\frac{u}{R+\left(Z_{L 2}-Z_C\right) j}=\frac{220 \sqrt{2} \angle 0}{10+\left(\frac{20 \sqrt{3}}{3}-10 \sqrt{3}\right) j}=11 \sqrt{6} \angle \frac{\pi}{6} .
\end{aligned}
Đáp án D.