a. Bạn tự vẽ sơ đồ nhé!

b. \(R_{td}=R1+R2=10+20=30\Omega\)

c. \(U=IR=0,3.30=9V\)

c. \(R3=p\dfrac{l}{S}=1,1.10^{-6}\dfrac{6}{0,3.10^{-6}}=22\Omega\)

\(R'=R+R3=30+22=52\Omega\)

\(\Rightarrow I=\dfrac{U}{R}=\dfrac{9}{52}=\dfrac{9}{52}\simeq0,17A\)

Vậy CĐDĐ sẽ giảm xuống.

R A B   =   R 1   +   R 23   =   8   +   12   =   20 ω

U 2   =   I A .   R 2   =   1 , 5 .   20   =   30 V

Điện trở toàn mạch của đoạn mạch AB là:

R_tm = \(\dfrac{U}{I}=\dfrac{48}{1,6}=30\left(\Omega\right)\)

Mặt khác vì R1//R2//R3 nên

\(\dfrac{1}{R_{tm}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}\)

mà R2 = \(\dfrac{1}{2}R_1\)

R₃ = \(\dfrac{1}{3}R_1\)

=> \(\dfrac{1}{30}=\dfrac{1}{R_1}+\dfrac{1}{\dfrac{1}{2}R_1}+\dfrac{1}{\dfrac{1}{3}R_1}=\dfrac{1}{R_1}+\dfrac{2}{R_1}+\dfrac{3}{R_1}=\dfrac{6}{R_1}\)

=> R₁ = 6 . 30 = 180\(\Omega\)

=> R₂ = 90\(\Omega\)

=> R₃ = 60 \(\Omega\)

=> I₁ = \(\dfrac{U}{R_1}=\dfrac{48}{180}=\dfrac{4}{15}A\)

I₂=\(\dfrac{U}{R_2}=\dfrac{48}{90}=\dfrac{8}{15}A\)

I₃ = \(\dfrac{U}{R_3}=\dfrac{48}{60}=0,8A\)

\(\dfrac{1}{R}=\dfrac{1}{R1}+\dfrac{1}{R2}+\dfrac{1}{R3}=\dfrac{1}{4}+\dfrac{1}{10}+\dfrac{1}{15}=\dfrac{5}{12}\Rightarrow R=2,4\left(\Omega\right)\)

\(\Rightarrow I=\dfrac{U}{R}=\dfrac{24}{2,4}=10\left(A\right)\)

3) a) a) K mở thì ta có mạch

((R2ntR4)//R1)ntR3

=>Rtđ=\(\dfrac{\left(R2+R4\right).R1}{R2+R4+R1}+R3=3,6\Omega\)

=>\(I=\dfrac{U}{Rt\text{đ}}=\dfrac{7,2}{3,6}=2A\)

Vì R241ntR3=>I241=I3=I=2A

Vì R24//R1=>U24=U1=U241=I241.R241=2.1,6=3,2V

Vì R2ntR4=>I2=I4=I24=\(\dfrac{U24}{R24}=\dfrac{3,2}{8}=0,4A\)

Vì ampe kế nối tiếp R2=>Ia=I2=0,4A

Vậy ampe kế chỉ 0,4A

b) K đóng ta có mạch

((R2//R3)ntR1)//R4

=>R23=1\(\Omega\)

=>R231=3\(\Omega\)

=>Rtđ=2\(\Omega\)

=>\(I=\dfrac{U}{Rt\text{đ}}=\dfrac{7,2}{2}=3,6A\)

Vì R231//R4=>U231=U4=U=7,2V

Vì R23ntR1=>I23=I1=I231=\(\dfrac{U231}{R231}=\dfrac{7,2}{3}=2,4A\)

Vì R2//R3=>U2=U3=U23=I23.R23=2,4.1=2,4V

=>I2=\(\dfrac{U2}{R2}=\dfrac{2,4}{2}=1,2A\)

Vì ampe kế nỗi tiếp R2=>I2=Ia=1,2A

Vậy ampe kế chỉ 1,2A

1) a) Mắc vôn kế thì ta có mạch:

((R2ntR3)//R1)ntR4

=>Rtđ=\(\dfrac{R23.R1}{R23+R1}+R4=4+2=6\Omega\)

=>\(I=\dfrac{U}{rt\text{đ}}=\dfrac{18}{6}=3A\)

vì R241ntR4=>I231=I4=I=3A

=>U4=I4.R4=3.2=6V

Vì R23//R1=>U23=U1=U231=I231.R231=3.4=12V

vì R2ntR3=>I2=I3=I23=\(\dfrac{U23}{R23}=\dfrac{12}{12}=1A\)

=>U3=I3.R3=6V

Uv=Umn=U3+U4=12V

b) Thay bằng ampe kế ta có mạch :

((R3//R4)ntR2)//R1

=>Rtđ= \(\dfrac{\left(\dfrac{R3.R4}{R3+R4}+R2\right).R1}{\dfrac{R3.R4}{R3+R4}+R2+R1}=\dfrac{10}{3}\Omega\)

=>\(I=\dfrac{U}{Rt\text{đ}}=18:\dfrac{10}{3}=5,4A\)

Vì R342//R1=>U342=U1=U=18V

Vì R34ntR2=>I34=I2=I342=\(\dfrac{U342}{R342}=\dfrac{18}{7,5}=2,4\)A

Vì R3//R4=>U3=U4=U34=I34.R34=2,4.1,5=3,6V

=>\(I3=\dfrac{U3}{R3}=\dfrac{3,6}{6}=0,6A\)

Ta có I2=I3+Ia=>2,4=0,6+Ia=>Ia=1,8A

Chiều dòng điện đi từ M-N