\(A.R1=\dfrac{4}{3}R2\)

\(P=UI=220\left(\dfrac{220}{48,5}\right)\simeq997,9\)W

\(P=\dfrac{U^2}{R}=\dfrac{220^2}{48,5}=997,9W\)

\(R=\dfrac{U^2}{P}=\dfrac{220^2}{200}=242\Omega\)

\(R=\dfrac{U}{I}=\dfrac{220}{2}=110\left(\Omega\right)\)

\(R=\rho\dfrac{l}{S}\Rightarrow S=\dfrac{\rho.l}{R}=\dfrac{0,4.10^{-6}.5}{110}\approx1,81.10^{-8}\left(m^2\right)\)

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MẠCH SONG SONG

A = 2,5 số = 2,5kWh = 2500Wh

\(\Rightarrow P=\dfrac{A}{t}=\dfrac{2500}{3}=833,\left(3\right)\)W

\(\Rightarrow I=\dfrac{P}{U}=\dfrac{833,\left(3\right)}{220}=3,\left(78\right)A\)

a. \(\left\{{}\begin{matrix}Q_{thu}=mc\left(t-t_1\right)=1,5.4200\left(100-25\right)=472500\left(J\right)\\Q_{toa}=A=Pt=1200.8.60=576000\left(J\right)\end{matrix}\right.\)

b. \(H=\dfrac{Q_{thu}}{Q_{toa}}100\%=\dfrac{472500}{576000}100\%\simeq82\%\)

c. \(Q'=3.Q_{toa}=3.576000=1728000\left(J\right)=0,48\)kWh

\(\Rightarrow T=Q'.30.2000=0,48.30.2000=28800\left(dong\right)\)

\(R=U:I=12:0,1=120\Omega\)

\(R=p\dfrac{l}{S}\Rightarrow S=\dfrac{p.l}{R}=\dfrac{0,40.10^{-6}.6}{120}=2.10^{-8}\left(m^2\right)\)

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