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Bài 1:
a. \(R=R1+\left(\dfrac{R2.R3}{R2+R3}\right)=30+\left(\dfrac{15\cdot10}{15+10}\right)=36\Omega\)
b. \(I=I1=I23=\dfrac{U}{R}=\dfrac{24}{36}=\dfrac{2}{3}A\left(R1ntR23\right)\)
\(U23=U2=U3=I23\cdot R23=\dfrac{2}{3}\cdot\left(\dfrac{15.10}{15+10}\right)=4V\left(R2\backslash\backslash R3\right)\)
\(\rightarrow\left\{{}\begin{matrix}I2=U2:R2=4:15=\dfrac{4}{15}A\\I3=U3:R3=4:10=0,4A\end{matrix}\right.\)
Bài 2:
a. \(R=\dfrac{R1.\left(R2+R3\right)}{R1+R2+R3}=\dfrac{6\cdot\left(2+4\right)}{6+2+4}=3\Omega\)
b. \(U=IR=2.3=6V\)
Điện trở tương đương: \(R=\dfrac{\left(R1+R2\right)R3}{R1+R2+R3}=\dfrac{\left(15+25\right)10}{15+25+10}=8\Omega\)
\(U=U12=U3=12V\)(R12//R3)
\(I=U:R=12:8=1,5A\)
\(I3=U3:R3=12:10=1,2A\)
\(R1ntR2\Rightarrow I12=I1=I2\)
Mà: \(I12=I-I3=1,5-1,2=0,3A\)
\(\Rightarrow I12=I1=I2=0,3A\)
\(R_{78}=R_7+R_8=1+1=2\left(\text{Ω}\right)\)
\(R_{678}=\dfrac{R_6\cdot R_{78}}{R_6+R_{78}}=\dfrac{1\cdot2}{1+2}=\dfrac{2}{3}\left(\text{Ω}\right)\)
\(R_{5678}=R_5+R_{678}=1+\dfrac{2}{3}=\dfrac{5}{3}\left(\text{Ω}\right)\)
\(R_{45678}=\dfrac{R_4\cdot R_{5678}}{R_4+R_{5678}}=\dfrac{1\cdot\dfrac{5}{3}}{1+\dfrac{5}{3}}=\dfrac{5}{8}\left(\text{Ω}\right)\)
\(R_{345678}=R_3+R_{45678}=1+\dfrac{5}{8}=\dfrac{13}{8}\left(\text{Ω}\right)\)
\(R_{2345678}=\dfrac{R_2\cdot R_{345678}}{R_2+R_{345678}}=\dfrac{1\cdot\dfrac{13}{8}}{1+\dfrac{13}{8}}=\dfrac{13}{21}\left(\text{Ω}\right)\)
\(R_{AB}=R_1+R_{2345678}=1+\dfrac{13}{21}=\dfrac{34}{21}\left(\text{Ω}\right)\)