Câu 3:

<tóm tắt bạn tự làm>

MCD:R1ntR2ntR3

Điện trở R3 là

ta có:\(R_{tđ}=R_1+R_2+R_3\Rightarrow R_3=R_{tđ}-R_1-R_2=55-15-30=10\left(\Omega\right)\)

Bài 1: 

a, \(\)\(\)\(=>R2//\left[R4nt\left(R3//R5\right)\right]\)

\(=>Rtd=\dfrac{R2\left[R4+\dfrac{R3.R5}{R3+R5}\right]}{R2+R4+\dfrac{R3.R5}{R3+R5}}=\dfrac{1.\left[1+\dfrac{1}{1+1}\right]}{1+1+\dfrac{1}{1+1}}=0,6\left(ôm\right)\)

\(=>I=\dfrac{Uab}{Rtd}=\dfrac{10}{0,6}=\dfrac{50}{3}A=I1\)

\(=>Uab=U2345=10V=U2=U345\)

\(=>I2=\dfrac{U2}{R2}=\dfrac{10}{1}=10A\)

\(=>I345=\dfrac{U345}{R345}=\dfrac{10}{1+\dfrac{1.1}{1+1}}=\dfrac{20}{3}A=I4=I35\)

\(=>U35=I35.R35=\dfrac{20}{3}.\dfrac{1.1}{1+1}=\dfrac{10}{3}V=U3=U5\)

\(=>I3=\dfrac{U3}{R3}=\dfrac{\dfrac{10}{3}}{1}=\dfrac{10}{3}A,\)

\(=>I5=\dfrac{U5}{R5}=\dfrac{10}{3}A\)

b, \(I1=0,1A=Im=I2345\)

\(=>Uab=I2345.R2345=0,1.\dfrac{6\left[8+\dfrac{6.12}{6+12}\right]}{6+8+\dfrac{6.12}{6+12}}=0,4V\)

Giúp mình bài 3 đc ko ạ, mình cảm ơn rất nhiều

Bài 1:

a. \(R=R1+R2=20+40=60\Omega\)

b. \(I=I1=I2=\dfrac{U}{R}=\dfrac{24}{60}=0,4A\left(R1ntR2\right)\)

Bài 3:

\(P_2>P_1\left(40>10\right)\Rightarrow\) đèn 2 sáng hơn.

Bài 2:

a. \(U_b=U-U_d=12-9=3V\)

\(I=I_d=I_b=0,5A\left(R_dntR_b\right)\)

\(\Rightarrow R_b=U_b:I_b=3:0,5=6\Omega\)

b. \(R=p\dfrac{l}{S}\Rightarrow l=\dfrac{R.S}{p}=\dfrac{20\cdot0,5\cdot10^{-6}}{1,1\cdot10^{-6}}\approx9,1\left(m\right)\)