Bài làm:
a) Vì \(R_1\text{/}\text{/}R_2\) nên: \(U=U_1=U_2=4\left(V\right)\)
Cường độ chạy qua điện trở R1 là:
\(I_1=\dfrac{U_1}{I_1}=\dfrac{4}{30}=\dfrac{2}{15}\approx0,13\left(A\right)\)
Cường độ chạy qua điện trở R2 là:
\(I_2=\dfrac{U_2}{I_2}=\dfrac{4}{15}\approx0,27\left(A\right)\)
Vì \(R_1\text{/}\text{/}R_2\) nên: \(I_{mc}=I_1+I_2=\dfrac{2}{15}+\dfrac{4}{15}=0,4\left(A\right)\)
b) - Sơ đồ mạch điện : \(\left(R_1\text{/}\text{/}R_2\right)ntR_3\)
Từ sơ đồ mạch điện: \(\Rightarrow R_{12}=\dfrac{R_1\cdot R_2}{R_1+R_2}=\dfrac{30\cdot15}{30+15}=10\left(\Omega\right)\)
\(\Rightarrow R_{AB}=R_{12}+R_3=10+6=16\left(\Omega\right)\)
Cường độ chạy qua mạch chính là:
\(I_{mc}=\dfrac{U}{R_{AB}}=\dfrac{4}{16}=0,25\left(A\right)\)
Vì \(R_{12}ntR_3\) nên: \(I_{mc}=I_{12}=I_3=0,25\left(A\right)\)
Vì \(R_1\text{/}\text{/}R_2\) nên: \(U_1=U_2=I_{12}\cdot R_{12}=0,25\cdot10=2,5\left(V\right)\)
\(\Rightarrow\left\{{}\begin{matrix}I_1=\dfrac{U_1}{R_1}=\dfrac{2,5}{30}\approx0,083\left(A\right)\\I_2=\dfrac{U_2}{R_2}=\dfrac{2,5}{15}\approx0,17\left(A\right)\end{matrix}\right.\)
Vậy ................................................
a) R1//R2
\(\Rightarrow I_1=\dfrac{U}{R_1}=\dfrac{4}{30}=\dfrac{2}{15}\left(A\right)\); \(I_2=\dfrac{U}{R_2}=\dfrac{4}{15}\left(A\right)\)
\(I_{mc}=I_1+I_2=\dfrac{2}{15}+\dfrac{4}{15}=0,4\left(A\right)\)
b) (R1//R2) nt R3
\(R_{tđ}=R_{12}+R_3=\left(\dfrac{1}{\dfrac{1}{R_1}+\dfrac{1}{R_2}}\right)+R_3=\left(\dfrac{1}{\dfrac{1}{30}+\dfrac{1}{15}}\right)+6=16\Omega\)
\(I=\dfrac{U}{R_{tđ}}=I_{12}=I_3=\dfrac{4}{16}=0,25\left(A\right)\)
\(U_{12}=U_1=U_2=I_{12}.R_{12}=0,25.10=2,5\left(V\right)\)
\(I_1=\dfrac{U_1}{R_1}=\dfrac{2,5}{30}=\dfrac{1}{12}\left(A\right);I_2=\dfrac{U_2}{R_2}=\dfrac{2,5}{15}=\dfrac{1}{6}\left(A\right)\)
