Question

Cho mạch điện như hình vẽ: R1=6 ôm, R2=12 ôm, R3=4 ôm, I2=1A. Tính U1, U2, U3, Uab=?

Answer 1

\(R_{23}=\dfrac{R_2\cdot R_3}{R_2+R_3}=\dfrac{12\cdot4}{12+4}=3\Omega\)

\(R_{TĐ}=R_1+R_{23}=6+3=9\Omega\)

\(U_2=I_2\cdot R_2=1\cdot12=12V\Rightarrow U_{23}=U_3=12V\left(R_1//R_2\right)\)

\(I_3=\dfrac{U_3}{R_3}=\dfrac{12}{4}=3A\)

\(I_{23}=I_2+I_3=1+3=4A\)

\(U_1=I_1\cdot R_1=I_{23}\cdot R_1=4\cdot6=24V\)

\(U_{AB}=U_1+U_{23}=24+12=36V\)

Answer 2

có hình vẽ ko bn

Answer 3

Ta có: \(U_2=I_2R_2=1.12=12\left(V\right)\)

\(\Rightarrow U_3=U_2=12\left(V\right)\)

\(\Rightarrow I_3=\dfrac{U_3}{R_3}=3\left(A\right)\)

\(\Rightarrow I_1=I_2+I_3=4\left(A\right)\)

\(\Rightarrow U_1=I_1R_1=24\left(V\right)\)

\(\Rightarrow U_{AB}=U_1+U_2=24+12=36\left(V\right)\)