Bài 6: Bài tập vận dụng định luật Ôm

mỗi bài 4 thôi à bạn?

4 \(=>\left(R3//R4\right)nt\left(R1//R2\right)ntR5\)

\(=>\)\(I34=\dfrac{Uab}{Rtd}=\dfrac{6}{R5+\dfrac{R3.R4}{R3+R4}+\dfrac{R1.R2}{R1\left|R2\right|}}\)

\(=\dfrac{6}{0,5+\dfrac{3.1}{3+1}.2}=3A=I3+I4=Ia\)

Hình ?

TH1: K mở =>R0 nt R2

\(=>U1=I0.R0\left(V\right)\)

\(=>Ubd=I0.Rtd=\dfrac{U1}{R0}\left(R0+R2\right)=>Ubd=U1+\dfrac{U1.R2}{R0}\)

\(=>\dfrac{U1.R2}{R0}=Ubd-U1=>R0=\dfrac{U1.R2}{Ubd-U1}\)

Th2: R0 nt (R1//R2)

\(=>U0=U2\)

\(=>Ubd=U2+I0.R12=U2+\dfrac{U2}{R0}.\dfrac{R1.R2}{R1+R2}\)

\(=>Ubd=U2+\dfrac{U2}{R0}.\dfrac{\dfrac{R2}{4}.R2}{\dfrac{R2}{4}+R2}=U2+\dfrac{U2}{R0}.\dfrac{\dfrac{R2^2}{4}}{\dfrac{5R2}{4}}\)

\(=U2+\dfrac{U2}{R0}.\dfrac{R2}{5}=>Ubd=U2+\dfrac{U2.R2}{5R0}\)

\(=>R0=\dfrac{U2.R2}{5\left(Ubd-U2\right)}\)

\(=>\dfrac{U1.R2}{Ubd-U1}=\dfrac{U2.R2}{5\left(Ubd-U2\right)}\)

\(=>Ubd=\dfrac{4U1U2}{5U1-U2}\)

R1 nt(R2 //R3)

\(=>Rtd=R1+\dfrac{R2R3}{R2+R3}=6+\dfrac{9.15}{9+15}=11,625\Omega\)

\(=>Umax=I1.Rtd=5.11,625=58,125V\)

R8 nt {R7//{R5 nt R6nt{(R1 ntR2)//(R3 nt R4)}}

\(=>Rtd=R8+\dfrac{R7\left\{R5+R6+\dfrac{\left(R1+R2\right)\left(R3+R4\right)}{R1+R2+R3+R4}\right\}}{R7+R5+R6+\dfrac{\left(R1+R2\right)\left(R3+R4\right)}{R1+R2+R3+R4}}=16\Omega\)

\(=>I1=I12=1A=>U12=U1234=I12.R12=\left(R1+R2\right)=12V\)

\(=>I1234=\dfrac{U1234}{R5+R6+\dfrac{\left(R1+R2\right)\left(R3+R4\right)}{R1+R2+R3+R4}}=1A=I5=I6=I123456\)

\(=>U1234567=U7=1.R1234567=Rtd-R8=6V=>I7=\dfrac{6}{12}=0,5A=>Im=0,5+1=1,5A=>Uab=Im.Rtd=16.1,5=24V\)

(bấm máy tính lại ktra cho chắc nhé)

R1 nt (R2//R3)

\(=>Rdd1=\dfrac{U\left(đm1\right)^2}{Pđm1}=\dfrac{12^2}{U\left(đm1\right).I\left(đm1\right)}=\dfrac{12^2}{12.2}=6\Omega\)

\(=>R\left(đ2\right)=\dfrac{U\left(đm2\right)^2}{Pđm2}=\dfrac{6^2}{6.1,5}=4\Omega\)

\(=>R\left(đ3\right)=\dfrac{U\left(đm3\right)^2}{P\left(đm3\right)}=\dfrac{9^2}{9.1,5}=6\Omega\)

\(=>I1=I23=\dfrac{U}{R\left(đ1\right)+\dfrac{R\left(đ2\right)R\left(đ3\right)}{R\left(đ2\right)+R\left(đ3\right)}}=2A=I\left(đm1\right)\)

\(=>U23=I23R23=2.\left(\dfrac{4.6}{4+6}\right)=4,8V=U2=U3=>I2=\dfrac{U2}{R\left(đ2\right)}=1,2A< I\left(Đm2\right)=>I3=\dfrac{U3}{R\left(đ3\right)}=0,8A< I\left(đm3\right)\)

Vậy đèn 1 hoạt động bình thường sáng bth

-đèn 2 và đèn 3 sáng yếu hơn bình thường

R1 nt R2 nt R3

\(=>I1=I2=I3=\dfrac{U}{R1+R2+R3}=\dfrac{U}{3R}\left(A\right)\)

R1//R2//R3

\(=>U1=U2=U3=U\) mà các điện trở R1=R2=R3=R

\(=>\dfrac{1}{Rtd}=\dfrac{1}{R}+\dfrac{1}{R}+\dfrac{1}{R}=>\dfrac{1}{Rtd}=\dfrac{3}{R}=>Rtd=\dfrac{R}{3}\Omega\)

\(=>I'=I1=I2=I3=\dfrac{U}{Rtd}=\dfrac{3U}{R}A\)

R2//(R1nt[R5//(R3ntR4))

\(=>R1345=R1+\dfrac{R5\left(R3+R4\right)}{R5+R3+R4}=7\Omega=>Rtd=\dfrac{R2.R1345}{R2+R1345}=14\Omega\)

\(=>I3=I4=I34=>U5=U34=I34.R34=0,5.\left(R3+R4\right)=3V=>I5=\dfrac{U5}{R5}=1A=>I1=I5+I34=1,5A=>U1345=U2=1,5.R1345=10,5V=U2=Um=>I2=\dfrac{U2}{R2}=1,5A\)

a)Hiệu điện thế của nguồn điện là:

  Ta có: \(R=\dfrac{U}{I}\Leftrightarrow U=R.I=50.1,8=90\left(V\right)\)

b) Ta có: \(R'=\dfrac{U}{\dfrac{I}{3}}=\dfrac{90}{\dfrac{1,8}{3}}=150\left(\Omega\right)\)

c) Ta có: \(R''=\dfrac{U}{I"}\Leftrightarrow I"=\dfrac{U}{R"}=\dfrac{90}{15}=6\left(A\right)\)

Bài 1 : 

Tóm tắt : 

U_AB = 35V

R₁ = 15Ω

R₂ = 3Ω

R₃ = 7Ω

R₄ = 10Ω

a) R_AB = ?

b) I₁ , I₂ , I₃ , I₄ = ?

     a)                      \(R_{23}=R_2+R_3\)

                                     = 3 + 7

                                     = 10(Ω)

                     \(R_{234}=\dfrac{R_{23}.R_4}{R_{23}+R_4}=\dfrac{10.10}{10+10}=5\left(\Omega\right)\)

                             \(R_{AB}=R_1+R_{234}\)

                                      = 15 + 5

                                      = 20(Ω)

b)                    \(I_{AB}=\dfrac{U_{AB}}{R_{AB}}=\dfrac{35}{20}=1,75\left(A\right)\)

                      Có : \(I_{AB}=I_1=I_{234}=1,75\left(A_{ }\right)\)

                     \(U_{234}=I_{234}.R_{234}=1,75.5=8,75\left(V\right)\)

                       ⇒ \(U_{23}=U_4=U_{234}=8,75\left(V\right)\)

                           \(I_4=\dfrac{U_4}{R_4}=\dfrac{8,75}{10}=0,875\left(A\right)\)

                          \(I_{23}=\dfrac{U_{23}}{R_{23}}=\dfrac{8,75}{10}=0,875\left(A\right)\)

                            ⇒ \(I_2=I_3=I_{23}=0,875\left(A\right)\)

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