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\(=>P1=U\left(đm1\right).I\left(đm1\right)=220.0,8=176W=>R1=\dfrac{U\left(đm1\right)^2}{P1}=275\Omega\)
\(=>P2=U\left(đm2\right)I\left(đm2\right)=220.0,5=110W=>R2=\dfrac{U\left(đm2\right)^2}{P2}=440\Omega\)
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
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a, cường độ dđ mạch
\(I=\dfrac{U}{R_{td}}=\dfrac{12}{10+5}=0,8\left(A\right)\)
\(\Rightarrow U_1=I.R_1=8\left(V\right)\)
\(\Rightarrow U_2=I.R_2=5.0,8=4\left(V\right)\)
b, \(\Rightarrow U_1=\dfrac{4}{2}=2\left(V\right)\)
\(I=I_2=\dfrac{4}{5}=0,8\left(A\right)\)
\(I_1=\dfrac{2}{10}=0,2\left(A\right)\)
\(I_3=I_2-I_1=0,6\left(A\right)\)
\(\Rightarrow R_3=\dfrac{U_1}{I_3}=\dfrac{2}{0,6}=\dfrac{10}{3}\left(\Omega\right)\)
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a) Do mắc song song nên:
\(U_{23}=U_3=U_2=I_2.R_2=0,5.6=3\left(V\right)\)
Cường độ dòng điện I3:
\(I_3=\dfrac{U_3}{R_3}=\dfrac{3}{9}=\dfrac{1}{3}\left(A\right)\)
Do mắc nối tiếp nên:
\(I=I_1=I_{23}=I_2+I_3=0,5+\dfrac{1}{3}=\dfrac{5}{6}\left(A\right)\)
b) \(R_{23}=\dfrac{R_2.R_3}{R_2+R_3}=\dfrac{6.9}{6+9}=3,6\left(\Omega\right)\)
\(R_{AB}=R_{23}+R_1=12+3,6=15,6\left(\Omega\right)\)
Hiệu điện thế U giữa 2 đầu đoạn mạch:
\(U=I.R_{tđ}=\dfrac{5}{6}.15,6=13\left(V\right)\)
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Câu 14:
\(R_{tđ}=R_1+R_2+R_3=12+15+13=40\left(\Omega\right)\)
\(I=I_1=I_2=\dfrac{U}{R}=\dfrac{12}{40}=0,3\left(A\right)\)
Câu 15:
\(R=\rho\dfrac{l}{S}=0,40.10^{-6}.\dfrac{10}{0,1.10^{-6}}=40\left(\Omega\right)\)
\(I=\dfrac{U}{R}=\dfrac{12}{40}=0,3\left(A\right)\)
hơi nhiều nên làm vài bài thoi còn lại bạn tự làm
16. R1 //(R2 nt R3)
\(=>U123=U1=U23=Im.Rtd=2.\left\{\dfrac{R1\left(R2+R3\right)}{R1+R2+R3}\right\}=15\left(V\right)\)
\(=>I1=\dfrac{U1}{R1}=\dfrac{15}{12}=1,25A\)
\(=>I23=I2=I3=\dfrac{U23}{R2+R3}=\dfrac{15}{15+5}=0,75A\)
17. R1 nt(R2 //R3)
a,\(=>Rtd=R1+\dfrac{R2R3}{R2+R3}=4+\dfrac{6.12}{6+12}=8\left(om\right)\)
b,\(=>I1=I23=\dfrac{U}{Rtd}=\dfrac{24}{8}=3A=>U23=U2=U3=I23.R23=3.4=12V=>I2=\dfrac{U2}{R2}=2A,I3=\dfrac{U3}{R3}=1A\)
19. R3 nt(R1//R2)ntRa
\(=>U12=\dfrac{R1R2}{R1+R2}=\dfrac{24R2}{24+R2}=U-Ua-U3=24-1.0,2-1.3,8=>R2=120\left(om\right)\)
20. R3 nt Ra nt(R1//R2)(làm tương tự)
18. (R1 nt R2)//R3
\(a,=>Rtd=\dfrac{R3\left(R1+R2\right)}{R3+R1+R2}=4\left(om\right)\)
\(b,=>U3=U12=12V=>I3=\dfrac{U3}{R3}=1A,=>I1=I2=\dfrac{U12}{R1+R2}=\dfrac{12}{2+4}=2A\)
c,\(=>U1=I1R1=4V=>U2=I2R2=8V\)
(mấy bài còn lại lm tương tự)