Bài 1:

a. \(R=R1+R2+R3=6+18+16=40\left(\Omega\right)\)

b. \(I=I1=I2=I3=\dfrac{U}{R}=\dfrac{20}{40}=0,5A\left(R1ntR2ntR3\right)\)

Bài 2:

a. \(R=p\dfrac{l}{S}=0,40.10^{-6}\dfrac{2,5}{0,1.10^{-6}}=10\left(\Omega\right)\)

Bài 3:

a. \(R=\dfrac{U^2}{P}=\dfrac{220^2}{1000}=48,4\left(\Omega\right)\)

b. \(A=Pt=1000.1=1000\)Wh = 1kWh

\(\left\{{}\begin{matrix}v'=s':t'=60:15=4\left(\dfrac{m}{s}\right)\\v''=s'':t''=50:20=2,5\left(\dfrac{m}{s}\right)\\v=\dfrac{s'+s''}{t'+t''}=\dfrac{60+50}{15+20}=\dfrac{22}{7}\left(\dfrac{m}{s}\right)\end{matrix}\right.\)

\(\left\{{}\begin{matrix}v_1=\dfrac{S_1}{t_1}=\dfrac{60}{15}=4\left(\dfrac{m}{s}\right)\\v_2=\dfrac{S_2}{t_2}=\dfrac{50}{20}=2,5\left(\dfrac{m}{s}\right)\end{matrix}\right.\)

\(v_{tb}=\dfrac{S_1+S_2}{t_1+t_2}=\dfrac{60+50}{15+20}=\dfrac{22}{7}\left(\dfrac{m}{s}\right)\)

Lỗi rồi

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\(U=U2=U1=I1.R1=\left(I-I2\right)R1=\left(1,2-0,4\right)6=4,8V\left(R1\backslash\backslash\mathbb{R}2\right)\)

\(\Rightarrow R2=\dfrac{U2}{I2}=\dfrac{4,8}{0,4}=12\left(\Omega\right)\)

Vì R₁ mắc song song với R₂ nên I=I₁+I₂⇒ I₁= I-I₂= 1,2-0,4=0,8A

Hiệu điện thế U₁ 

R₁=U₁/I₁ ⇒ U₁= R₁.I₁ = 6.0,8=4,8 V

Vì U=U₁=U₂ nên có có R₂ là

R₂=U₂/I₂ = 4,8/0,4= 12Ω

Mặt phẳng chứa tia tới và pháp tuyến của gương ở điểm tới. (chắc zậy)

\(R=p\dfrac{l}{S}\Rightarrow l=\dfrac{R.S}{p}=\dfrac{33.0,5.10^{-6}}{1,1.10^{-6}}=15\left(m\right)\)

\(C=\pi d=3,14.0,02=0,0628\left(m\right)\)

\(\Rightarrow N=\dfrac{l}{C}=\dfrac{15}{0,0628}\simeq238,8\left(vong\right)\)

\(R_{23}=\dfrac{R_2.R_3}{R_2+R_3}=\dfrac{15.10}{15+10}=6\left(\Omega\right)\)

\(R_{tđ}=R_1+R_{23}=9+6=15\left(\Omega\right)\)

\(U=I.R_{tđ}=0,3.15=4,5\left(V\right)\)

\(R=R1+\left(\dfrac{R2.R3}{R2+R3}\right)=9+\left(\dfrac{15.10}{15+10}\right)=15\left(\Omega\right)\)

\(\Rightarrow U=IR=0,3.15=4,5V\)

\(\Rightarrow\left\{{}\begin{matrix}i=100^o:2=50^o\\i=i'\Leftrightarrow i'=50^o\end{matrix}\right.\)