\(R_1=6\Omega\) nhé bạn
a, Ta có : \(R_1//R_2//R_3\)
\(\Rightarrow\frac{1}{R_{tđ}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\)
\(=\frac{1}{6}+\frac{1}{12}+\frac{1}{16}\)
\(=\frac{5}{16}\)
\(\Rightarrow R_{tđ}=3.2\Omega\)
Cường độ dòng điện qua mạch chính:
\(\Rightarrow I=\frac{U}{R_{td}}=\frac{2,4}{3,2}=0.75A\)
Do \(R_1//R_2//R_3\Rightarrow U_1=U_2=U_3=U\)
\(\Rightarrow I_1=\frac{U_1}{R_1}=\frac{2,4}{6}=0.4A\)
\(I_2=\frac{U_2}{R_2}=\frac{2,4}{12}=0,2A\)
\(I_3=\frac{U_3}{R_3}=\frac{2,4}{16}=0,15A\)
Bạn xem lại cho mình R1 =6 hay 60 ôm nhé
Khi R\(_1\) \(=60\Omega\)
a, \(\Rightarrow\frac{1}{R_{tđ}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\)
\(=\frac{1}{60}+\frac{1}{12}+\frac{1}{16}\)
\(=\frac{13}{80}\)
\(\Rightarrow R_{tđ}=\frac{80}{13}\Omega\approx6.15\Omega\)
b, Do \(R_1//R_2//R_3\Rightarrow U_1=U_2=U_3=U=2,4V\)
\(\Rightarrow I_1=\frac{U_1}{R_1}=\frac{2,4}{60}=0,04\left(A\right)\)
\(I_2=\frac{U_2}{R_2}=\frac{2,4}{12}=0,2\left(A\right)\)
\(I_3=\frac{U_3}{R_3}=\frac{2,4}{16}=0,15\left(A\right)\)
