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a)Nối M và B bằng một vôn kế rất lớn.
Khi đó CTM là: \(\left(R_1//\left(R_2ntR_3\right)\right)ntR_4\)
Ta có: \(U_V=U_3+U_4\)
\(R_{23}=R_2+R_3=6+6=12\Omega\)
\(R_{123}=\dfrac{R_1\cdot R_{23}}{R_1+R_{23}}=\dfrac{6\cdot12}{6+12}=4\Omega\)
\(R_{tđ}=R_{123}+R_4=4+2=6\Omega\)
\(I_4=I_{123}=I=\dfrac{U}{R_{tđ}}=\dfrac{18}{6}=3A\Rightarrow U_4=I_4\cdot R_4=3\cdot2=6V\)
\(U_{23}=U_{123}=I_{123}\cdot R_{123}=3\cdot4=12V\)
\(I_2=I_3=I_{23}=\dfrac{U_{23}}{R_{23}}=\dfrac{12}{12}=1A\Rightarrow U_3=I_3\cdot R_3=1\cdot6=6V\)
Vậy \(U_V=U_3+U_4=6+6=12V\)
b)Nối M với B bằng một ampe kế lớn.
Khi đó CTM là \(\left(R_1nt\left(R_3//R_4\right)\right)//R_2\)
Ta có: \(I_A=I_2+I_3\)
\(R_{34}=\dfrac{R_3\cdot R_4}{R_3+R_4}=\dfrac{6\cdot2}{6+2}=1,5\Omega\)
\(R_{134}=R_1+R_{34}=6+1,5=7,5\Omega\)
\(R_{tđ}=\dfrac{R_{134}\cdot R_2}{R_{134}+R_2}=\dfrac{7,5\cdot6}{7,5+6}=\dfrac{10}{3}\Omega\)
\(U_2=U_{134}=U=18V\)
\(I_2=\dfrac{U_2}{R_2}=\dfrac{18}{6}=3A\)
\(I_{34}=I_{134}=\dfrac{U_{134}}{R_{134}}=\dfrac{U}{R_{134}}=\dfrac{18}{7,5}=2,4A\)
\(U_3=U_4=U_{34}=I_{34}\cdot R_{34}=2,4\cdot1,5=3,6V\)
\(I_3=\dfrac{U_3}{R_3}=\dfrac{3,6}{6}=0,6A\)
Vậy \(I_A=I_2+I_3=3+0,6=3,6A\)
a, (R1//R2)nt(R3//R4)
\(R_1\)//\(R_2\Rightarrow R_{12}=\dfrac{R_1.R_2}{R_1+R_2}=\dfrac{4.10}{4+10}=\dfrac{40}{14}=\dfrac{20}{7}\Omega\)
\(R_3\)//\(R_4\Rightarrow R_{34}=\dfrac{R_3.R_4}{R_3+R_4}=\dfrac{12.15}{12+15}=\dfrac{180}{27}=\dfrac{20}{3}\Omega\)
R12 nt R 34 => \(R_{tđ}=R_{12}+R_{34}=\dfrac{20}{7}+\dfrac{20}{3}=\dfrac{200}{21}\approx9,5\left(\Omega\right)\)
b, R12 nt R 34 \(\Rightarrow I_{12}=I_{34}=I=\dfrac{U_m}{R_{tđ}}=\dfrac{15}{9,5}\approx1,58\left(\Omega\right)\)
\(R_1\)//R2\(\Rightarrow U_1=U_2=U_{12}=I_{12}.R_{12}=1,58.\dfrac{20}{7}\approx4,5\Omega\)
\(\Rightarrow I_1=\dfrac{U_{12}}{R_1}=\dfrac{4,5}{4}=1,125\left(A\right)\); \(I_2=\dfrac{U_{12}}{R_2}=\dfrac{4,5}{10}=0,45\left(A\right)\)
R3//R4\(\Rightarrow U_3=U_4=U_{34}=I_{34}.R_{34}=1,58.\dfrac{20}{3}\approx10,5\Omega\)
\(\Rightarrow I_3=\dfrac{U_{34}}{R_3}=\dfrac{10,5}{12}=0,875\left(A\right)\)\(\Rightarrow I_4=\dfrac{U_{34}}{R_4}=\dfrac{10,5}{15}=0,7\left(A\right)\)
c, Ia = 0
Khi: \(\dfrac{R_1}{R_2}=\dfrac{R_3}{R_5}\Leftrightarrow\dfrac{4}{10}=\dfrac{12}{R_5}\Rightarrow R_5=30\left(\Omega\right)\)