lois mắt tp

vây à để cho to len 

Bài 2:

a. \(R=R1+\dfrac{R2\cdot R3}{R2+R3}=10+\dfrac{6\cdot3}{6+3}=12\Omega\)

b. \(I=I1=I23=U:R=12:12=1A\left(R1ntR23\right)\)

\(U23=U2=U3=I23\cdot R23=1\cdot\left(\dfrac{6\cdot3}{6+3}\right)=0,5V\left(R2//R3\right)\)

\(\left\{{}\begin{matrix}I2=U2:R2=0,5:6=\dfrac{1}{12}A\\I3=U3:R3=0,5:3=\dfrac{1}{6}A\end{matrix}\right.\)

Giúp mình câu 2c vs bạn 

R1 //{R2  nt(R3//R4)}

\(\Rightarrow Icb=4=I1+I3=\dfrac{Uab}{R1}+I3=\dfrac{Uab}{4}+I3\left(1\right)\)

\(\Rightarrow\dfrac{R3}{R4}=2\Rightarrow R3=2R4\Rightarrow\dfrac{I3}{I4}=0,5\Rightarrow I4=\dfrac{I3}{0,5}\left(A\right)\)

\(\Rightarrow I2=I3+I4\Rightarrow I3+\dfrac{I3}{0,5}=I2\Rightarrow1,5I3=0,5I2\Rightarrow I3=\dfrac{I2}{3}=\dfrac{\dfrac{Uab}{R234}}{3}=\dfrac{\dfrac{Uab}{12}}{3}=\dfrac{Uab}{36}\left(A\right)\left(2\right)\)

\(\left(1\right)\left(2\right)\Rightarrow4=\dfrac{Uab}{4}+\dfrac{Uab}{36}\Rightarrow Uab=14,4V\)

Sao ở dòng đầu bằng U4/R4 vậy

Bài 1:

\(A=P.t=U.I.t=220.2,5.1=550\left(J\right)\)

Bài 2:

\(P=\dfrac{U^2}{R}\Rightarrow R=\dfrac{U^2}{P}=\dfrac{220^2}{1000}=48,4\left(\Omega\right)\)

\(A=P.t=1000.4.60.60=14400000\left(J\right)\)

Nhiệt lượng lò sưởi tỏa ra trong 40 ngày:

\(A=40.14400000=576000000\left(J\right)=160\left(kWh\right)\)

Tiền điện phải trả:

\(160.2100=336000\left(đồng\right)\)

Bài 3:

Điện trở nồi bếp điện:

\(R=\rho\dfrac{l}{S}=1,1.10^{-6}.\dfrac{30}{0,2.10^{-6}}=165\left(\Omega\right)\)

Nhiệt lượng tỏa ra:

\(A=P.t=\dfrac{U^2}{R}.t=\dfrac{220^2}{165}.15.60=264000\left(J\right)\)

Bài 1:

\(R_{tđ}=R_1+R_2+R_3=7+3+2=12\left(\Omega\right)\)

\(I=I_1=I_2=\dfrac{U}{R_{tđ}}=\dfrac{24}{12}=2\left(A\right)\)

\(\left\{{}\begin{matrix}U_1=I_1.R_1=2.7=14\left(V\right)\\U_2=I_2.R_2=2.3=6\left(V\right)\\U_3=I_3.R_3=2.2=4\left(V\right)\end{matrix}\right.\)

Bài 2:

\(R_{tđ}=\dfrac{R_1.R_2}{R_1+R_2}=\dfrac{3.6}{3+6}=2\left(\Omega\right)\)

\(U=U_1=U_2=2,4V\)

\(\left\{{}\begin{matrix}I=\dfrac{U}{R_{tđ}}=\dfrac{2,4}{2}=1,2\left(A\right)\\I_1=\dfrac{U_1}{R_1}=\dfrac{2,4}{3}=0,8\left(A\right)\\I_2=\dfrac{U_2}{R_2}=\dfrac{2,4}{6}=0,4\left(A\right)\end{matrix}\right.\)

Bạn tự vẽ sơ đồ mạch điện nhé!

Bài 1:

\(MCD:R1ntR2ntR3\)

\(=>R=R1+R2+R3=7+3+2=12\Omega\)

\(I=I1=I2=I3=\dfrac{U}{R}=\dfrac{24}{12}=2A=>\left\{{}\begin{matrix}U1=I1\cdot R1=2\cdot7=14V\\U2=I2\cdot R2=2\cdot3=6V\\U3=I3\cdot R3=2\cdot2=4V\end{matrix}\right.\)

cô nhã đã xem

:)) thế cô có nói j k . 

Bài 3:

\(MCD:R1nt\left(R2//R3\right)\)

\(=>R=R1+\dfrac{R2\cdot R3}{R2+R3}=30+\dfrac{15\cdot10}{15+10}=36\Omega\)

\(I=I1=I23=\dfrac{U}{R}=\dfrac{24}{36}=\dfrac{2}{3}A\)

\(U23=U2=U3=I23\cdot R23=\dfrac{2}{3}\cdot\dfrac{15\cdot10}{15+10}=4V=>\left\{{}\begin{matrix}I2=U2:R2=4:15=\dfrac{4}{15}A\\I3=\dfrac{U3}{R3}=4:10=\dfrac{2}{5}A\end{matrix}\right.\)

\(A=UIt=24\cdot\dfrac{2}{3}\cdot5\cdot60=4800J\)

Bài 4:

\(R=\dfrac{U^2}{P}=\dfrac{220^2}{880}=55\Omega\)

\(I=\dfrac{P}{U}=\dfrac{880}{220}=4A\)

\(A=UIt=220\cdot4\cdot5\cdot60=264000J\)