Bài 1:
\(R_{tđ}=R_1+R_2+R_3=7+3+2=12\left(\Omega\right)\)
\(I=I_1=I_2=\dfrac{U}{R_{tđ}}=\dfrac{24}{12}=2\left(A\right)\)
\(\left\{{}\begin{matrix}U_1=I_1.R_1=2.7=14\left(V\right)\\U_2=I_2.R_2=2.3=6\left(V\right)\\U_3=I_3.R_3=2.2=4\left(V\right)\end{matrix}\right.\)
Bài 2:
\(R_{tđ}=\dfrac{R_1.R_2}{R_1+R_2}=\dfrac{3.6}{3+6}=2\left(\Omega\right)\)
\(U=U_1=U_2=2,4V\)
\(\left\{{}\begin{matrix}I=\dfrac{U}{R_{tđ}}=\dfrac{2,4}{2}=1,2\left(A\right)\\I_1=\dfrac{U_1}{R_1}=\dfrac{2,4}{3}=0,8\left(A\right)\\I_2=\dfrac{U_2}{R_2}=\dfrac{2,4}{6}=0,4\left(A\right)\end{matrix}\right.\)
Bạn tự vẽ sơ đồ mạch điện nhé!
Bài 1:
\(MCD:R1ntR2ntR3\)
\(=>R=R1+R2+R3=7+3+2=12\Omega\)
\(I=I1=I2=I3=\dfrac{U}{R}=\dfrac{24}{12}=2A=>\left\{{}\begin{matrix}U1=I1\cdot R1=2\cdot7=14V\\U2=I2\cdot R2=2\cdot3=6V\\U3=I3\cdot R3=2\cdot2=4V\end{matrix}\right.\)
Bài 2:
*SỬA ĐỀ: hai điện trở mắc song song chứ không phải ba nhé!*
\(MCD:R1//R2???\)
\(=>R=\dfrac{R1\cdot R2}{R1+R2}=\dfrac{3\cdot6}{3+6}=2\Omega\)
\(U=U1=U2=2,4V=>\left\{{}\begin{matrix}I=\dfrac{U}{R}=\dfrac{2,4}{2}=1,2A\\I1=\dfrac{U1}{R1}=\dfrac{2,4}{3}=0,8A\\I2=\dfrac{U2}{R2}=\dfrac{2,4}{6}=0,4A\end{matrix}\right.\)
