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a) Điện trở dây tóc bóng đèn:
\(R=\dfrac{U_{ĐM}^2}{P_{ĐM}}=\dfrac{220^2}{60}=\dfrac{2420}{3}\)(Ω)
3)Công suất bóng đèn lúc này là
\(P_1=\dfrac{U^2_{NG}}{R}=\dfrac{110^2}{\dfrac{2420}{3}}=15\left(W\right)\)
Công suất lúc này đã giảm:
\(P_{ĐM}:P_1=60:15=4\)(lần)
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a, k mở vẽ lại mạch ((R1ntR4)//R2)ntR3
\(I_1=I_4=2\left(A\right)\)
\(I_2+2=I_3\)
\(U_{AB}=\left(I_2+2\right)R_3+I_2R_2\Rightarrow I_2=\dfrac{5}{3}\left(A\right)\)
\(U_4=U_2-U_1=I_2R_2-I_1R_1=80\left(V\right)\Rightarrow R_4=\dfrac{80}{2}=40\left(\Omega\right)\)
b, k đóng mạch vẽ lại R1//[R2nt(R3//R4)]
\(\Rightarrow I_{234}=\dfrac{420}{R_{234}}=\dfrac{420}{144}=\dfrac{35}{12}\left(A\right)\)
\(\Rightarrow U_4=U_{34}=I_{234}.R_{34}=\dfrac{35}{12}.24=70\left(V\right)\)
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\(Q_{thu}=mc\Delta t=2\cdot4200\cdot75=630000\left(J\right)\)
\(H=\dfrac{Q_{thu}}{Q_{toa}}100\%\Rightarrow Q_{toa}=\dfrac{Q_{thu}}{H}100\%=\dfrac{630000}{80}100\%=787500\left(J\right)\)
\(A=Pt=1100\cdot2\cdot30=66000\)Wh = 66kWh
\(\Rightarrow T=A\cdot1000=66\cdot1000=66000\left(dong\right)\)
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Bài 19:
\(\left\{{}\begin{matrix}U1=I1\cdot R1=2\cdot15=30V\\U2=I2\cdot R2=1\cdot10=10V\end{matrix}\right.\)
\(=>U=U1+U2=30+10=40V\)
Chọn A
Bài 20:
\(\left\{{}\begin{matrix}U1=I1\cdot R1=5\cdot6=30V\\U2=I2\cdot R2=2\cdot9=18V\\U3=I3\cdot R3=3\cdot15=45V\end{matrix}\right.\)
\(=>U=U1+U2+U3=30+18+45=93V\)
Chọn C
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\(R_Đ=\dfrac{U^2_Đ}{P_Đ}=\dfrac{12^2}{6}=24\Omega\)
\(R_{12}=\dfrac{R_1\cdot R_2}{R_1+R_2}=\dfrac{20\cdot20}{20+20}=10\Omega\)
\(R_{tđ}=R_Đ+R_{12}=24+10=34\Omega\)
\(I_A=I_m=\dfrac{U_m}{R_{tđ}}=\dfrac{15}{34}=0,44A\)
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A. nha bn
R1=p.l1/S1=p.2l2/S1
R2=p.l2S2=p.2l2/S1=>R1/R2=p.2.l2/S1:p.2.l2/S1=1=>R1=R2