Question

Cho mạch điện sau: R1=20ohm, R2=30ohm, R3=10ohm, R4=40ohm.Hiệu điện thế giữa hai đầu mạch chính là 24V.

a)Tính R1234

b)Tính I1,I2,I3,I4,I_MN;U1,U2,U3,U4,U_MB,U_MN.

Answer 1

Ta có (R1//R2)nt(R3//R4)

->R1234=Rtđ=\(\dfrac{20.30}{20+30}+\dfrac{10.40}{10+40}=20\Omega\)

b)I=\(\dfrac{U}{Rt\text{đ}}=\dfrac{24}{20}=1,2A\) Vì R12ntR34->I12=I34=I=Imn=1,2A

Vì R1//R2->U1=U2=Umb=U12=I12.R12=1,2.12=14,4V

->I1=\(\dfrac{U1}{R1}=\dfrac{14,4}{20}=0,72A;I2=\dfrac{U2}{R2}=\dfrac{14,4}{30}=0,48A\)

Vì R3//R4=>U3=U4=U34=I34.R34=1,2.8=9,6V

-> I3=\(\dfrac{U3}{R3}=\dfrac{9,6}{10}=0,96A;I4=\dfrac{U4}{R4}=\dfrac{9,6}{40}=0,24A\)

Umn=U=24V