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MCD: R1nt R2 nt R3

 Cường độ dòng điện đi quả R1 R2 R3 là

\(I=I_1=I_2=I_3=\dfrac{U_{AB}}{R_{tđ}}=\dfrac{U_{AB}}{R_1+R_2+R_3}=\dfrac{24}{12+12+24}=\dfrac{1}{24}\left(A\right)\)

a. Ta có:  R₂ = 3R₁

Điện trở R₁ là:

R_tđ = R₁ + R₂

R_tđ = R₁ + 3R₁

24 = 4R₁

=> R₁ = 24/4 = 6(ôm)

b) Vì R₁ nt R₂ nt R₃ => Điện trở tương đương của mạch:

R_tđ = R₁ + R₂ + R₃ = 29 + 15 + 27 = 71 (ôm)

c) Vì R₁ // R₂ // R₃ => Điện trở tương đương của mạch:

 \(\text{\dfrac{1}{Rtđ} = }\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}=\dfrac{1}{250}+\dfrac{1}{50}+\dfrac{1}{750}=\dfrac{19}{750}\)

=> R_tđ = \(\dfrac{750}{19}=39,47\) (ôm)

\(MCD:R1ntR2ntR3\)

\(=>R=R1+R2+R3=8+14+8=30\Omega\)

\(=>I=\dfrac{U}{R}=\dfrac{?}{30}=?A\)

Bổ sung: \(U=40V\)

\(=>I=\dfrac{U}{R}=\dfrac{40}{30}=0,75A\)

\(a,R_{23}=R_2+R_3=30+30=60\left(\Omega\right)\)

\(R_m=\dfrac{R_{23}.R_1}{R_{23}+R_1}=\dfrac{60.15}{60+15}=12\left(\Omega\right)\)

\(b,I_m=\dfrac{U_{AB}}{R_m}=\dfrac{12}{12}=1\left(A\right)\)

\(I_1+I_{23}=1\left(A\right)\)

\(\dfrac{I_1}{I_{23}}=\dfrac{R_{23}}{R_1}=\dfrac{60}{15}=\dfrac{4}{1}\)

\(\rightarrow I_1=0,8\left(A\right);I_{23}=0,2\left(A\right)\)

\(\rightarrow I_2=I_3=0,2\left(A\right)\)

\(R_{12}=\dfrac{15.30}{15+30}=10\left(\Omega\right)\)

\(R_m=R_{12}+R_3=10+30=40\left(\Omega\right)\)

\(I_m=\dfrac{U_{AB}}{R_m}=\dfrac{12}{40}=0,3\left(A\right)\)

\(b,I_{12}=I_3=0,3\left(A\right)\)

\(\dfrac{I_1}{I_2}=\dfrac{R_2}{R_1}=\dfrac{30}{15}=\dfrac{2}{1}\)

\(\rightarrow I_1=0,2\left(A\right);I_2=0,1\left(A\right)\)

a)\(R_{tđ}=R_1+R_2=6+10=16\Omega\)

   \(P=R\cdot I^2=16\cdot0,5^2=4W\)

b)\(R_{tđ}=\dfrac{R_3\cdot R_{12}}{R_3+R_{12}}=\dfrac{8\cdot16}{8+16}=\dfrac{16}{3}\Omega\)

   \(I_m=0,5A\)

   \(\Rightarrow U=I\cdot R=0,5\cdot\dfrac{16}{3}=\dfrac{8}{3}V\)

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

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

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