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Sơ đồ mạch: R1 // [R3 nt (R2 // R4)]
\(R_{24}=\dfrac{R_2.R_4}{R_2+R_4}=\dfrac{10.5}{10+5}=\dfrac{10}{3}\)(Ω)
\(R_{234}=R_3+R_{24}=5+\dfrac{10}{3}=\dfrac{25}{3}\)(Ω)
\(R_{AB}=\dfrac{R_1.R_{234}}{R_1+R_{234}}=\dfrac{10.\dfrac{25}{3}}{10+\dfrac{25}{3}}=\dfrac{50}{11}\)(Ω)
Sơ đồ mạch: R1 // [(R2 // R3) nt R4]
Điện trở tương đương là :
\(R_{23}=\dfrac{R_2\cdot R_3}{R_2+R_3}=\dfrac{4\cdot4}{4+4}=2\left(\text{Ω }\right)\)
\(R_{234}=R_{23}+R_4=6\left(\text{Ω }\right)\)
\(R_{AB}=\dfrac{R_1\cdot R_{234}}{R_1+R_{234}}=\dfrac{3\cdot6}{3+6}=2\left(\text{Ω }\right)\)
Sơ đồ mạch: R1 // [R2 nt (R3 // R4)]
\(R_{34}=\dfrac{R_3\cdot R_4}{R_3+R_4}=\dfrac{6\cdot6}{6+6}=3\left(\text{Ω}\right)\)
\(R_{234}=R_2+R_{34}=9+3=12\left(\text{Ω}\right)\)
\(R_{tđ}=\dfrac{R_1\cdot R_{234}}{R_1+R_{234}}=\dfrac{12\cdot12}{12+12}=6\left(\text{Ω}\right)\)
theo mạch điện như hình vẽ
\(=>\left(R1ntR3\right)//R2]ntR4\)
do đó \(=>Rtd=R4+\dfrac{\left(R1+R3\right)R2}{R1+R3+R2}\)
\(=6+\dfrac{\left(12+6\right)9}{12+6+9}=12\left(om\right)\)
\(\text{Sơ đồ mạch: [R1 // (R4 nt R5)] nt (R2 // R3) nt R0 }\)
\(R_{4,5}=R_4+R_5=3\left(\text{Ω}\right)\)
\(R_{145}=\dfrac{R_1\cdot R_{45}}{R_1+R_{45}}=\dfrac{1\cdot3}{1+4}=0.75\left(\text{Ω}\right)\)
\(R_{23}=\dfrac{R_2\cdot R_3}{R_2+R_3}=\dfrac{2\cdot6}{2+6}=1.5\left(\text{Ω}\right)\)
Điện trở tương đương của mạch là
\(R_{tđ}=R_0+R_{23}+R_{145}=0.5+1.5+0.75=2.75\left(\text{Ω}\right)\)
Bài 3:
a. Cần mắc vào HĐT 220V để sáng bình thường.
b. \(I=P:U=1100:220=5A\)
c. \(A=Pt=1100.2.30=66000\)Wh = 66kWh = 237 600 000J
d. \(R=p\dfrac{l}{S}\Rightarrow l=\dfrac{R.S}{p}=\dfrac{\left(220:5\right).0,45.10^{-6}}{1,10.10^{-6}}=18\left(m\right)\)
Bài 4:
a. \(Q_{toa}=A=I^2Rt=2,4^2\cdot120\cdot25=17280\left(J\right)\)
b. \(Q_{thu}=mc\Delta t=1.4200.75=315000\left(J\right)\)
\(H=\dfrac{Q_{thu}}{Q_{toa}}100\%=\dfrac{17280}{315000}100\%\approx5,5\%\)
Baì 1:
a. \(R=R1+R2=4+6=10\Omega\)
\(I=I1=I2=U:R=18:10=1,8A\left(R1ntR2\right)\)
b. \(R1nt\left(R2\backslash\backslash\mathbb{R}3\right)\)
\(R'=R1+\left(\dfrac{R2.R3}{R2+R3}\right)=4+\left(\dfrac{6.12}{6+12}\right)=8\Omega\)
\(I'=U:R'=18:8=2,25A\)
Bài 2:
a. \(R=\dfrac{R1.R2}{R1+R2}=\dfrac{15.10}{15+10}=6\Omega\)
b. \(U=U1=U2=18V\left(R1\backslash\backslash\mathbb{R}2\right)\)
\(\Rightarrow\left\{{}\begin{matrix}I1=U1:R1=18:15=1,2A\\I2=U2:R2=18:10=1,8A\end{matrix}\right.\)
R1 nt(R2//R3)(theo ct \(R23=\dfrac{R2R3}{R2+R3}\))
a,\(\Rightarrow Rab=R1+\dfrac{R2R3}{R2+R3}=15+\dfrac{30.30}{30+30}=30\Omega\)
b,\(\Rightarrow I1=I23=\dfrac{U}{Rab}=\dfrac{12}{30}=0,4A\) do R2=R3
\(\Rightarrow U23=I23.\left(\dfrac{R2R3}{R2+R3}\right)=6V=U2=U3\Rightarrow I2=I3=\dfrac{U2}{R2}=0,2A\)
a) \(R_{tđ}=\dfrac{R_{23}.R_1}{R_{23}+R_1}=\dfrac{\left(R_2+R_3\right).R_1}{\left(R_2+R_3\right)+R_1}=\dfrac{\left(6+4\right).2}{\left(6+4\right)+2}=\dfrac{5}{3}\left(\Omega\right)\)
b) \(R_{tđ}=R_1+R_{23}=R_1+\dfrac{R_2.R_3}{R_2+R_3}=2+\dfrac{6.4}{6+4}=\dfrac{22}{5}\left(\Omega\right)\)
Câu a:
\(R_{23}=R_2+R_3=6+4=10\Omega\)
\(R_{tđ}=\dfrac{R_{23}\cdot R_1}{R_{23}+R_1}=\dfrac{10\cdot2}{10+2}=\dfrac{5}{3}\Omega\)
Câu b:
\(R_{23}=\dfrac{R_2\cdot R_3}{R_2+R_3}=\dfrac{6\cdot4}{6+4}=2,4\Omega\)
\(R_{tđ}=R_1+R_{23}=2+2,4=4,4\Omega\)
Sơ đồ mạch: R1 // [R3 nt (R2 // R4)]
\(R_{24}=\dfrac{R_2\cdot R_4}{R_2+R_4}=\dfrac{5\cdot10}{5+10}=\dfrac{10}{3}\left(\text{Ω}\right)\)
\(R_{234}=R_3+R_{24}=5+\dfrac{10}{3}=\dfrac{25}{3}\left(\text{Ω}\right)\)
\(R_{AB}=\dfrac{R_1\cdot R_{234}}{R_1+R_{234}}=\dfrac{10\cdot\dfrac{25}{3}}{10+\dfrac{25}{3}}=\dfrac{50}{11}\left(\text{Ω}\right)\)
Rac vs Rbc đâu ạ