R1//R2//R3

a,\(\Rightarrow\dfrac{1}{RTt}=\dfrac{1}{R1}+\dfrac{1}{R2}+\dfrac{1}{R3}\Rightarrow Rtd=12,5\Omega\)

b,\(\Rightarrow\left\{{}\begin{matrix}I1=\dfrac{37,5}{25}=1,5A\\I2=\dfrac{37,5}{50}=0,75A\\I3=\dfrac{37,5}{50}=0,75A\end{matrix}\right.\)\(\Rightarrow Im=\dfrac{37,5}{Rtd}=3A\)

Điện trở mắc song song nên 

Cường độ dòng điện: 

→ Đáp án D

B. R = 9\(\Omega\), I = 0,6A

a. \(R=\dfrac{\left(R1+R2\right)R3}{R1+R2+R3}=\dfrac{\left(80+40\right)60}{80+40+60}=40\Omega\)

b. \(U=U12=U3=IR=40.0,15=6V\)(R12//R3)

\(\left\{{}\begin{matrix}I3=U3:R3=6:60=0,1A\\I12=I1=I2=U12:R12=6:\left(80+40\right)=0,05A\left(R1ntR2\right)\end{matrix}\right.\)

a) R_TĐ = 40Ω

b)I₃ = 0.1A

   I₂ = I₁ = 0.05A

   U₃ = 266.67V

   U₁ = 177.78V

    U₂ = 88.89V

a) Điện trở tương đương là: 

 \(R_{tđ}=\dfrac{1}{\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}}=\dfrac{1}{\dfrac{1}{12}+\dfrac{1}{6}+\dfrac{1}{4}}=2\left(\Omega\right)\)

b) Hiệu điện thế U:

 \(U=I.R=3.2=6\left(V\right)\)

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\(\left\{{}\begin{matrix}R=\dfrac{R1.R2}{R1+R2}=\dfrac{60.30}{60+30}=20\Omega\\I=\dfrac{U}{R}=\dfrac{6}{20}=0,3A\end{matrix}\right.\)

Chọn C

\(R1//R2\Rightarrow Rtd=\dfrac{R1R2}{R1+R2}=24\Omega\Rightarrow Im=\dfrac{U}{Rtd}=\dfrac{12}{24}=0,5A\)

\(\Rightarrow R2//\left(R1ntR3\right)\Rightarrow Im=\dfrac{U}{\dfrac{R2\left(R1+R3\right)}{R2+R1+R3}}=0,4A\)

a,R1//R2 \(=>Rtd=\dfrac{R1.R2}{R1+R2}=8\left(ôm\right)\)

b,\(=>Im=\dfrac{U}{Rtd}=\dfrac{12}{8}=1,5A\)

\(=>I1=\dfrac{U}{R1}=1A,=>I2=\dfrac{U}{R2}=0,5A\)

c,\(=>U1\left(max\right)=I1\left(max\right).R1=24V\)

\(=>U2\left(max\right)=I2\left(max\right)R2=36V>U1\left(max\right)\)

=> phải chọn U1=24V để làm HĐT cho mạch R1//R2 trên

a) \(\dfrac{1}{R_{td}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}\\ =\dfrac{1}{12}+\dfrac{1}{6}+\dfrac{1}{4}\\ =\dfrac{1}{2}\\ =>R_{td}=2\Omega\)

b) \(U=I.R_{td}=3.2=6\left(V\right)\)

\(a,Rtd=\dfrac{R1R2}{R1+R2}=\dfrac{12.24}{12+24}=8\left(om\right)\)

b,\(=>Im=\dfrac{U}{Rtd}=\dfrac{6}{8}=\dfrac{3}{4}A\)