đề ko yc Ampe kế và vôn kế nên mạch dfieenj chỉ có điện trở thôi bn nhé

Ta có mạch (((R5ntR6)//R4)nt(R2//R3)ntR1

R56=30\(\Omega\)=>R564=\(\dfrac{30.30}{30+30}=15\Omega\)

R23=\(\dfrac{4.6}{4+6}=2,4\Omega\)=>Rtđ=R1+R23+R456=30\(\Omega\)

=>I=I1=I23=I456=\(\dfrac{U}{Rtđ}=1A\)

Vì R2//R3=>U2=U3=U23=I23.R23=2,4V=>I2=\(\dfrac{U2}{R2}=0,6A;I3=\dfrac{U3}{R3}=0,4A\)

Vì R4//R56=>U4=U56=U456=I456.R456=15V

=>\(I4=\dfrac{U4}{R4}=0,5A\)

Vì R5ntR6=>I5=I6=I56=\(\dfrac{U56}{R56}=0,5A\)

Vậy................

ý là thế này hả bn?

(R1ntR2)//(R3ntR4)

a,\(=>Rtd=\dfrac{\left(R1+R2\right)\left(R3+R4\right)}{R1+R2+R3+R4}=\dfrac{\left(10+15\right)\left(10+25\right)}{10+15+10+25}=\dfrac{175}{12}\left(om\right)\)

b,\(=>U12=U34=36V\)

\(=>I12=I1=I2=\dfrac{U12}{R12}=\dfrac{36}{10+15}=1,44A\)

\(=>I34=I3=I4=\dfrac{U34}{R34}=\dfrac{36}{10+25}=\dfrac{36}{35}A\)

b)R12=R1+R2=4+4=8\(\Omega\)

R123=\(\frac{R12.R3}{R12+R3}\)=\(\frac{8.6}{8+6}=\frac{27}{7}\)

R=R123+R4=\(\frac{24}{7}+9=\frac{87}{7}\)

c)I=I4=I123=U/R=60:\(\frac{87}{7}\)=\(\frac{140}{29}\)I

U4=R4.I4=9.140/29=1260/29V

U3=U12=U-U4=60-1260/29=480/29V

I3=U3/R3=\(\frac{480}{29}:6=\frac{80}{29}\)A

I1=I2=I12=U12/R12=480/29:8=60/29A

U1=U2=U12/2=240/29 V

R1 ở đâu vậy bn

\(R_{tđ}=\dfrac{\left(R_1+R_2\right)R_3}{R_1+R_2+R_3}=\dfrac{\left(R+R\right)R}{R+R+R}=\dfrac{2R^2}{3R}=\dfrac{2}{3}R\)

(R1ntR2)//R3 hả bạn ?

Đúng rồi bn giúp mk với

Bài 1:

\(R_{12}=\dfrac{R_1.R_2}{R_1+R_2}=\dfrac{100.100}{100+100}=50\left(\Omega\right)\)

Điện trở toàn mạch là: 

\(R_{tđ}=R_{23}+R_3=50+50=100\left(\Omega\right)\)

Bài 2:

Ta có: \(R_{tđ}=R_1+R_2+R_3=150\left(\Omega\right)\)

Mà \(R_1=R_2=R_3\)

\(\Rightarrow R_1=R_2=R_3=150:3=50\left(\Omega\right)\)

Bài 3:

Điện trở dây dẫn là:

\(R=\dfrac{U}{I}=\dfrac{18}{2,5}=7,2\left(\Omega\right)\)

\(R_{tm}=\frac{r_1r_2}{r_1+r_2}+r_3\)