Câu 2.
\(\dfrac{1}{R_{tđ}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}=\dfrac{1}{10}+\dfrac{1}{20}+\dfrac{1}{40}=\dfrac{7}{40}\)
\(\Rightarrow R_{tđ}=\dfrac{40}{7}\Omega\)
\(I_m=\dfrac{U}{R}=\dfrac{30}{\dfrac{40}{7}}=5,25A\)
\(I_1=\dfrac{U_1}{R_1}=\dfrac{30}{10}=3A\)
\(I_2=\dfrac{U_2}{R_2}=\dfrac{30}{20}=1,5A\)
\(I_3=I-I_1-I_2=5,25-3-1,5=0,75A\)