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a)Điện trở tương đương:
\(\dfrac{1}{R_{tđ}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{16}=\dfrac{5}{16}\)
\(\Rightarrow R_{tđ}=\dfrac{16}{5}\Omega=3,2\Omega\)
b)\(R_1//R_2//R_3\Rightarrow U_1=U_2=U_3=U=2,4V\)
\(I_m=\dfrac{U}{R_{tđ}}=\dfrac{2,4}{3,2}=0,75A\)
\(I_1=\dfrac{U_1}{R_1}=\dfrac{2,4}{6}=0,4A\)
\(I_2=\dfrac{U_2}{R_2}=\dfrac{2,4}{12}=0,2A\)
\(I_3=I_m-I_1-I_2=0,15A\)
a)\(\dfrac{1}{R_{tđ}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{16}=\dfrac{5}{16}\)
\(\Rightarrow R_{tđ}=\dfrac{16}{5}\Omega=3,2\Omega\)
b)\(I_m=\dfrac{U}{R_{tđ}}=\dfrac{2,4}{3,2}=0,75A\)
\(R_1//R_2//R_3\Rightarrow U_1=U_2=U_3=U=2,4V\)
\(I_1=\dfrac{U_1}{R_1}=\dfrac{2,4}{6}=0,4A\)
\(I_2=\dfrac{U_2}{R_2}=\dfrac{2,4}{12}=0,2A\)
\(I_3=I-I_1-I_2=0,75-0,4-0,2=0,15A\)
CTM: \(R_1//\left(R_2ntR_3\right)\)
a)\(R_{23}=R_2+R_3=12+12=24\Omega\)
\(R_{tđ}=\dfrac{R_1\cdot R_{23}}{R_1+R_{23}}=\dfrac{6\cdot24}{6+24}=4,8\Omega\)
b)\(U_1=U_{23}=U=12V\)
\(I_m=\dfrac{U}{R_{tđ}}=\dfrac{12}{4,8}=2,5A\)
\(I_1=\dfrac{U_1}{R_1}=\dfrac{12}{6}=2A\)
\(I_2=I_3=I_{23}=I_m-I_1=2,5-2=0,5A\)
\(\Rightarrow\left\{{}\begin{matrix}a,R1//\left(R2ntR3\right)\Rightarrow Rtd=\dfrac{R1\left(R2+R3\right)}{R1+R2+R3}=6\Omega\\b,\Rightarrow\left\{{}\begin{matrix}U=U1=U23=24V\Rightarrow I1=\dfrac{U1}{R1}=\dfrac{8}{3}A\\I2=I3=\dfrac{U23}{R2+R3}=\dfrac{4}{3}A\\U2=I2.R2=8V\\U3=U-U2=16V\end{matrix}\right.\\c,R1//\left(R2ntRx\right)\Rightarrow Im=1,5.\dfrac{24}{6}=6A\\\Rightarrow Rtd=\dfrac{R1\left(R2+Rx\right)}{R1+R2+Rx}=\dfrac{9\left(6+Rx\right)}{15+Rx}=\dfrac{24}{Im}=4\left(\Omega\right)\Rightarrow Rx=1,2\Omega\end{matrix}\right.\)
a)\(R_{tđ}=R_1+R_2=6+10=16\Omega\)
\(P=R\cdot I^2=16\cdot0,5^2=4W\)
b)\(R_{tđ}=\dfrac{R_3\cdot R_{12}}{R_3+R_{12}}=\dfrac{8\cdot16}{8+16}=\dfrac{16}{3}\Omega\)
\(I_m=0,5A\)
\(\Rightarrow U=I\cdot R=0,5\cdot\dfrac{16}{3}=\dfrac{8}{3}V\)
a) Điện trở tương đương của mạch là:
\(R_{td}=R_1+R_2+R_3=4+2+6=12\Omega\)
b) \(I=\dfrac{U}{R_{td}}=\dfrac{6}{12}=0,5A\)
Do \(R_1ntR_2ntR_3\Rightarrow I_1=I_2=I_3=I\)
\(\Rightarrow I_3=\dfrac{U_3}{R_3}\Rightarrow U_3=I_3R_3=0,5\cdot6=3V\)
a)Ba điện trở mắc nối tiếp: \(R_{tđ}=R_1+R_2+R_3=4+2+6=12\Omega\)
b)\(I_m=\dfrac{U}{R_{tđ}}=\dfrac{6}{12}=0,5A\)
\(R_1ntR_2ntR_3\Rightarrow I_1=I_2=I_3=I=0,5A\)
\(U_3=I_3\cdot R_3=0,5\cdot6=3V\)
\(R_{tđ}=R_1+R_2+R_3=3+5+7=15\Omega\)
\(I_1=I_2=I_3=I_m=2A\)
\(U_1=I_1\cdot R_1=2\cdot3=6V\)
\(U_2=I_2\cdot R_2=2\cdot5=10V\)
\(U_3=I_3\cdot R_3=2\cdot7=14V\)
a) Điện trở tương đương của mạch là:
\(R_{td}=R_1+R_2+R_3=6+12+16=34\Omega\)
b) Hiệu điện thế của mạch là:
\(I=\dfrac{U}{R_{td}}\Rightarrow U=IR_{td}=0,5\cdot34=17V\)