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Cấu tạo mạch: \(\left[\left(R_3//R_4\right)ntR_2\right]//R_1\)
\(U_1=U_{234}=U_m=24V\)
\(I_1=\dfrac{U_1}{R_1}=\dfrac{24}{12}=2A\)
\(R_{34}=\dfrac{R_3+R_4}{R_3\cdot R_4}=\dfrac{6+6}{6\cdot6}=\dfrac{1}{3}\Omega\)
\(R_{234}=R_2+R_{34}=9+\dfrac{1}{3}=\dfrac{28}{3}\Omega\)
\(I_2=I_{234}=\dfrac{U_{234}}{R_{234}}=\dfrac{24}{\dfrac{28}{3}}=\dfrac{18}{7}A\)
\(U_2=I_2\cdot R_2=\dfrac{18}{7}\cdot9=\dfrac{162}{7}V\)
\(U_{34}=I_{34}\cdot R_{34}=\dfrac{18}{7}\cdot\dfrac{1}{3}=\dfrac{6}{7}V\)
\(\Rightarrow U_3=U_{34}=\dfrac{6}{7}V\Rightarrow I_3=\dfrac{U_3}{R_3}=\dfrac{1}{7}A\)
\(I_A=I_1+I_3=2+\dfrac{1}{7}=\dfrac{15}{7}A\)
Cho mạch điện như hình vẽ: R1=6 ôm, R2=12 ôm, R3=4 ôm, I1=1A. Tính U1,U2,U3,Uab=?
CTM: \((R_1nt(R_3//R_4))//R_2\)
\(R_{34}=\dfrac{R_3\cdot R_4}{R_3+R_4}=\dfrac{6\cdot2}{6+2}=\dfrac{3}{2}\Omega\)
\(R_{134}=R_1+R_{34}=6+1,5=7,5\Omega\)
\(R_{tđ}=\dfrac{R_{134}\cdot R_2}{R_{134}+R_2}=\dfrac{7,5\cdot6}{7,5+6}=\dfrac{10}{3}\Omega\)
\(I_{34}=I_{134}=\dfrac{U_{134}}{R_{134}}=\dfrac{U_{AB}}{R_{134}}=\dfrac{18}{7,5}=2,4A\)
\(U_3=U_4=U_{34}=I_{34}\cdot R_{34}=2,4\cdot1,5=3,6V\)
\(I_A=I_3=\dfrac{U_3}{R_3}=\dfrac{3,6}{6}=0,6A\)
Có : R1//R2//R3 nên :
\(\Rightarrow\frac{1}{R_{123}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}=\frac{1}{12}+\frac{1}{6}+\frac{1}{4}=\frac{1}{2}\)
\(\Rightarrow R_{123}=2\Omega\)
Có : R123nt R4 nên :
\(\Rightarrow R_{tđ}=R_{123}+R_4=2+4=6\Omega\)
\(\Rightarrow I=\frac{U}{R_{tđ}}=\frac{18}{6}=3A\)
Có : R123nt R4 nên :
\(\Rightarrow I_{123}=I_4=I=3A\)
\(\Rightarrow U_4=I_4.R_4=3.4=12V\)
Có : U123=U-U4=18-12=6V
\(\Rightarrow\)U1=U2=U3=U123=6V
Khi đó : I1=\(\frac{U_1}{R_1}=\frac{6}{12}=0,5A\)
I2=\(\frac{U_2}{R_2}=\frac{6}{6}=1A\)
I3=I-I1-I2=3-1-0,5=1,5A