Bài 6: Bài tập vận dụng định luật Ôm

a, khi K mở \(=>\left(R1ntR2\right)\)\(nt\left(R4//R5\right)\)

A2 chỉ 0,5 A\(=>I4=I\left(A2\right)=0,5A=>U4=U5=I4.R4=0,5.80\)\(=40V\)

\(=>I5=\dfrac{U5}{R5}=\dfrac{40}{20}=2A=>I45=I4+I5=2+0,5=2,5A\)\(=Im=I12\)=>số chỉ ampe kế(A1)=2,5A

\(=>R12=R1+R2=4+4=8\left(Om\right)\)

\(=>U12=I12.R12=8.2,5=20V\)\(=>UAB=U12+U4=60V\)

b,khi đóng K\(=>R1nt\left\{[R2nt\left(R4//R5\right)]//R3\right\}\)

\(=>R245=R2+\dfrac{R4.R5}{R4+R5}=4+\dfrac{80.20}{80+20}=20\left(om\right)\)

\(=>R2345=\dfrac{R3.R245}{R3+R345}=\dfrac{5.20}{5+20}=4\left(om\right)\)

\(=>Rtd=R1+R2345=4+4=8\left(om\right)\)

\(=>Im=\dfrac{UAB}{Rtd}=\dfrac{60}{8}=7,5A=I1=I2345\)

\(=>A1\) chỉ 7,5 A

\(=>U2345=I2345.R2345=7,5.4=30V\)\(=U245=U3\)

\(=>I245=\dfrac{U245}{R245}=\dfrac{30}{20}=1,5A=I45\)

\(=>U45=I45.R45=16.1,5=24V=U4\)

\(=>I4=\dfrac{U4}{R4}=\dfrac{24}{80}=0,3A\)\(=>A2\) chỉ 0,3A

gọi điện trở đó là R1

\(=>R\left(đ\right)//R1\)

\(=>R\left(đ\right)=\dfrac{U\left(đm\right)}{I\left(đm\right)}=\dfrac{12}{0,5}=24\left(om\right)\)

\(=>U1=U\left(đ\right)=I1.R1=36.0,3=10.8V=U\left(đ\right)\)

\(=>U\left(đ\right)< U\left(đm\right)\)

=>đèn sáng yếu hơn bình thường

nếu \(Im>I1,I2\) thì không mắc được 2 bóng đèn có cường độ \(I1,I2\)

nối tiếp nhé

a, \(R1ntR2=>Rtd=R1+R2=10+20=30\left(om\right)\)

b, \(=>Im=\dfrac{U}{Rtd}=\dfrac{12}{30}=0,4A=I1=I2\)

\(=>U1=I1R1=0,4.10=4V\)

\(=>U2=U-U1=12-4=8V\)

c, \(=>R1nt\left(R2//R3\right)\)

\(=>U23=U-U1=12-0,5.10=7V\)

\(=>I1=I23=0,5A\)

\(=>R23=\dfrac{U23}{I23}=\dfrac{7}{0,5}=14\left(om\right)\)

\(=>R23=\dfrac{R2.R3}{R2+R3}=\dfrac{20R3}{20+R3}=14=>R3=47\left(om\right)\)

làm tầm 1-2 bài thôi bn nhé, mấy bài kia làm tương tự thôi b11 cx gần tương tự bài 10

B12 

a, K mở ampe kế chỉ 3A\(=>Im=I23=3A\)

\(=>Rtd=\dfrac{R2.R3}{R2+R3}=\dfrac{Uab}{I12}=\dfrac{R2^2}{2R2}=\dfrac{60}{3}=>R2=R3=40\left(om\right)\)

b, khi K đóng ampe kế A1 chỉ 2A\(=>I1=2A\)

\(=>R1//R2//R3\)

\(=>\dfrac{1}{Rtd}=\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{R1}\)

\(=>\dfrac{1}{\dfrac{U}{Im}}=\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{R1}\)

\(=>\dfrac{Im}{60}=\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{R1}\)

\(=>\dfrac{\left(I1+I2+I3\right)}{60}=\dfrac{1}{40}+\dfrac{1}{40}+\dfrac{1}{R1}\)

\(=>\dfrac{2+\dfrac{60}{40}+\dfrac{60}{40}}{60}=\dfrac{1}{40}+\dfrac{1}{40_{ }}+\dfrac{1}{R1}\)

\(=>R1=30\left(om\right)\)

b13:

a, K đóng =>R1 nt R2

\(=>Rtd=R1+R2=2+4=6\left(om\right)\)

\(=>Im=Ia=\dfrac{Uab}{Rtd}=\dfrac{24}{6}=4A\)=>số chỉ ampe kế là 4A

\(=>Im=I1=I2=4A\)

\(=>U2=I2.R2=4.4=16V=Uv\)

=>số chỉ vôn kế là 16V

b, K mở \(=>R1ntR2ntR3\)

\(=>Rtd=R1+R2+R3=2+4+6=12\left(om\right)\)

\(=>Im=Ia=\dfrac{Uab}{Rtd}=\dfrac{24}{12}=2A\)=>số chỉ ampe kế là 2A

\(=>Im=I2=>U2=I2R2=2.4=8V=Uv\)

=>số chỉ vôn kế là 8V

a, \(R1//R2=>Rtd=\dfrac{10R2}{10+R2}=\dfrac{U}{Ia}=\dfrac{24}{3}=8=>R2=40\left(ôm\right)\)

b, cần mắc \(R3nt\left(R1//R2\right)\)

\(=>Rtd=\dfrac{Uab}{\dfrac{1}{2}.3}=\dfrac{24}{1,5}=16\left(ôm\right)\)

\(=>16=R3+\dfrac{R1.R2}{R1+R2}=R3+\dfrac{10.40}{10+40}=>R3=8\left(ôm\right)\)

a,\(R1nt\left(R2//R3\right)=>Rtd=R1+\dfrac{R2R3}{R2+R3}=4+\dfrac{6.3}{6+3}=6\left(om\right)\)

b,\(=>I1=I23=\dfrac{Uab}{Rtd}=\dfrac{9}{6}=1,5A\)

\(=>U23=I23.R23=1,5.\dfrac{6.3}{6+3}=3V=U2=U3\)

\(=>I2=\dfrac{U2}{R2}=\dfrac{3}{6}=0,5A,=>I3=\dfrac{U3}{R3}=\dfrac{3}{3}=1A\)

c,\(=>Im=Ix=I23=\dfrac{1}{3}.1,5=0,5A\)

\(=>RTd=Rx+\dfrac{R2.R3}{R2+R3}=Rx+\dfrac{6.3}{6+3}=\dfrac{U}{Im}=\dfrac{9}{0,5}=18\)

\(=>Rx=16\left(om\right)\)

\(=>R1nt\left(R2//R3\right)nt\left[R4//\left(R5ntR6\right)\right]\)

\(=>RTd=R1+\dfrac{R2R3}{R2+R3}+\dfrac{R4\left(R5+R6\right)}{R4+R5+R6}\)

\(=12,6+\dfrac{4.6}{4+6}+\dfrac{30\left(15+15\right)}{30+15+15}=30\left(om\right)\)

\(=>Im=\dfrac{U}{Rtd}=\dfrac{30}{30}=1A=I1=I23=I456\)

\(=>U23=I23.R23=1.\dfrac{4.6}{4+6}=2,4V=U2=U3\)

\(=>I2=\dfrac{U2}{R2}=\dfrac{2,4}{4}=0,6A=>I3=I23-I2=0,4A\)

\(=>U456=I456.R456=\dfrac{30\left(15+15\right)}{30+15+15}=15V=U4=U56\)

\(=>I4=\dfrac{U4}{R4}=\dfrac{15}{30}=0,5A\)

\(=>I56=I5=I6=\dfrac{U56}{R56}=\dfrac{15}{15+15}=0,5A\)