Esercizio n. 5

DiAlfredo Di Fiore

soluzione

In questo caso avremo

\cos\varphi_{cc}=\frac{P_{ccn}}{\sqrt{3}\,V_{Icc}\,I_{1n}} \qquad\text{con}\qquad P_j=P_{ccn}=3\,R_{Icc}\,I_{1n}^2

\cos\varphi_{cc}=\frac{350}{\sqrt{3}\cdot 38 \cdot 24}=0.22

R_{Icc}=\frac{P_{ccn}}{3\,I_{1n}^{2}} =\frac{350}{3\cdot 24^{2}} =0.203\ \Omega

Z_{Icc}=\frac{V_{Icc}}{\sqrt{3}\,I_{1n}} =\frac{38}{\sqrt{3}\cdot 24} =0.914\ \Omega

X_{Icc}=\sqrt{Z_{Icc}^{2}-R_{Icc}^{2}} =\sqrt{0.914^{2}-0.203^{2}} =0.891\ \Omega

R_{Icc}=R_{1}+R_{1}'=R_{1}+K_{0}^{2}R_{2} \ \Rightarrow\ 0.203=0.12+2.8^{2}\,R_{2} \ \Rightarrow\ R_{2}=\frac{0.203-0.12}{2.8^{2}}=0.0106\ \Omega

Dato che

Z_{1cc}=\frac{V_{Icc}}{\sqrt{3}\,I_{1n}} \ \Rightarrow\ I_{1n}=\frac{V_{Icc}}{Z_{1cc}\sqrt{3}} =\frac{380}{0.914\sqrt{3}} =240\ \text{A}\quad\text{(abbastanza alta)}

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