Question

Đã trả lời bởi chuyên gia

Answer 1

Từ đồ thị: $\frac{U_{L}^{\max}}{U_{L 1}} = \frac{U_{L}^{\max}}{U_{L 2}} = \frac{7}{4},$ và $U_{L}^{\max} = \frac{U}{\sqrt{1 - n^{- 2}}} \overset{\frac{U_{L}^{\max}}{U} = \frac{14}{6} = \frac{7}{3}}{\to} \Rightarrow n = \frac{7 \sqrt{10}}{20} \Rightarrow \cos^{2} φ_{L} = \frac{2}{1 + n} = 0, 95.$

$U_{L} = L ω . \frac{U}{R} \cos φ \Rightarrow \frac{1}{ω^{2}} = {(\frac{U}{U_{L}} . \frac{L}{R} \cos φ)}^{2} \Rightarrow \{\begin{matrix} \frac{1}{ω_{1}^{2}} = {(\frac{U}{U_{L 1}} . \frac{L}{R} \cos φ_{1})}^{2} \\ \frac{1}{ω_{2}^{2}} = {(\frac{U}{U_{L 2}} . \frac{L}{R} \cos φ_{2})}^{2} \\ \frac{1}{ω_{L}^{2}} = {(\frac{U}{U_{L}^{\max}} . \frac{L}{R} \cos φ_{L})}^{2} \end{matrix} .$ (1).

$\begin{array}{l} \frac{2}{ω_{L}^{2}} = \frac{1}{ω_{1}^{2}} + \frac{1}{ω_{2}^{2}} \overset{(1)}{\to} \\ \cos^{2} φ_{1} + \cos^{2} φ_{2} = 2. {(\frac{U_{L}}{U_{L}^{\max}})}^{2} \cos^{2} φ_{L} = 2. {(\frac{4}{7})}^{2} .0, 95 = 0, 62. \end{array}$ (2)

Ta có: $P_{1} = U I_{1} \cos φ_{1} = U \frac{U}{Z_{1}} \cos φ_{1} = \frac{U^{2}}{R} \cos^{2} φ_{1}, P_{2} = U I_{2} \cos φ_{2} = U \frac{U}{Z_{2}} \cos φ_{2} = \frac{U^{2}}{R} \cos^{2} φ_{2} .$

$\Rightarrow P_{1} + P_{2} = \frac{U^{2}}{R} (\cos^{2} φ_{1} + \cos^{2} φ_{2}) \overset{(2)}{\to} P_{1} + P_{2} = P_{C H} .2. {(\frac{U_{L}}{U_{L}^{\max}})}^{2} \cos^{2} φ_{L} .$

Thế số: $P_{1} + P_{2} = P_{C H} .2. {(\frac{U_{L}}{U_{L}^{\max}})}^{2} \cos^{2} φ_{L} = 287.2. {(\frac{4}{7})}^{2} 0, 95 = 178, 1 W$ .

Chọn C.

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Answer 2