câu 4

TH1: Uab=10V,Ucd=4V

=> R1 //(R2 nt R3)

\(=>R2=\dfrac{4}{I2}=4\Omega\),\(=>R3=\dfrac{10-4}{I3}=6\Omega\)

TH2: Ucd=6V,Uab=1,5V=>R2//(R1 nt R3)

\(=>R1=\dfrac{1,5}{I1}=\dfrac{1,5}{\dfrac{6-1,5}{6}}=\dfrac{1,5}{\dfrac{3}{4}}=2\Omega\)

câu 5

{R3 nt(R1//R2)}//R4

\(=>Rtd=\dfrac{R4\left\{R3+\dfrac{R1R2}{R1+R2}\right\}}{R3+R4+\dfrac{R1R2}{R2+R1}}=15\Omega\)

\(=>Ia=I2+I4\)

ta đi tính I2 và I4 

\(=>U4=U123=U=30V=>I4=\dfrac{30}{30}=1A\)

\(=>I123=\dfrac{30}{R3+\dfrac{R1R2}{R1+R2}}=1A=I12=>U12=R12=\dfrac{4}{5}V=U2=>I2=\dfrac{U2}{R2}=\dfrac{1}{5}A\)

\(=>Ia=\dfrac{1}{5}+1=\dfrac{6}{5}A\)

R1 nt (R2//R3)

\(,=>Rtd=R1+\dfrac{R2R3}{R2+R3}=60+\dfrac{30.20}{30+20}=72\Omega\)

,\(=>I1=I23=\dfrac{Uab}{Rtd}=\dfrac{12}{72}=\dfrac{1}{6}A\)

\(=>U1=I1R1=\dfrac{1}{6}.60=10V=>U23=12-10=2V=U2=U3\)

\(=>I2=\dfrac{U2}{R2}=\dfrac{2}{30}=\dfrac{1}{15}A=>I3=\dfrac{U3}{R3}=\dfrac{2}{20}=\dfrac{1}{10}A\)

\(=>P1=U\left(đm1\right).I\left(đm1\right)=220.0,8=176W=>R1=\dfrac{U\left(đm1\right)^2}{P1}=275\Omega\)

\(=>P2=U\left(đm2\right)I\left(đm2\right)=220.0,5=110W=>R2=\dfrac{U\left(đm2\right)^2}{P2}=440\Omega\)

R1 nt R2

\(=>I1=I2=\dfrac{440}{R1+R2}=\dfrac{440}{440+275}=\dfrac{8}{13}A\)

\(=>I1=\dfrac{8}{13}A< I\left(đm1\right),I2=\dfrac{8}{13}A>I\left(đm2\right)\)

=>đèn 1 sáng yếu hơn bth , đèn 2 sáng hơn bth(có thể bị cháy)

do đèn 2 có thể cháy nên ko mắc nối tiếp