Bài 6: Bài tập vận dụng định luật Ôm

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 //R3)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

nối tiếp nhé

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)