Ta có: \(U=150-50=100V\)

\(E=\dfrac{U}{d}=\dfrac{100}{0,2}=500\)

cHỌN C

C

Điện trở tương đương của mạch ngoài:

\(R_N=\dfrac{R_1R_2}{R_1+R_2}=\dfrac{18R_1}{R_1+18}\)

Cường độ dòng điện qua mạch:

\(I=\dfrac{\text{ξ}}{R_N+r}=6:\left(\dfrac{18R_1}{R_1+18}+4\right)\)

\(\Rightarrow I=6:\dfrac{22R_1+72}{R_1+18}\)

\(\Rightarrow I=\dfrac{3\left(R_1+18\right)}{11R_1+36}\)

Hiệu điện thế hai đầu mạch:

\(U_N=I.R_N=\dfrac{3\left(R_1+18\right)}{11R_1+36}.\dfrac{18R_1}{R_1+18}\)

\(\Rightarrow U_N=\dfrac{54R_1}{11R_1+36}\)

Cường độ dòng điện qua \(R_1\)

\(I_1=I_A=\dfrac{U_N}{R_1}=\dfrac{54}{11R_1+36}\)

\(\Rightarrow\dfrac{54}{11R_1+36}=0,4\)

\(\Rightarrow4,4R_1=39,6\)

\(\Rightarrow R_1=9\Omega\)

Vậy chọn A.

Đề không cho hình thì mình tự ra đề vậy: \(R_1ntR_2\)

\(R_N=R_1+R_2=5+14=19\Omega\)

\(I=\dfrac{\xi}{r+R_N}=\dfrac{48}{1+19}=2,4A\)

\(t=16'5s=965s\)

Cu có \(A=64đvC\) và có hóa trị ll nên \(n=2\)

\(m_{Cu}=\dfrac{1}{F}\cdot\dfrac{A}{n}\cdot It=\dfrac{1}{96494}\cdot\dfrac{64}{2}\cdot2,4\cdot965=0,768kg=768g\)

Mắc hai nguồn nối tiếp: 

\(\xi_b=n\cdot\xi=2\xi\)

\(r_b=n\cdot r=2r\)

\(\Rightarrow I_1=\dfrac{\xi_{b1}}{r_{b1}}=\dfrac{2\xi}{2r}=\dfrac{\xi}{r}\)

Mắc song song:

\(\xi_b=\xi\)

\(r_b=\dfrac{r}{n}=\dfrac{r}{2}\)

\(\Rightarrow I_2=\dfrac{\xi_{b2}}{r_{b2}}=\dfrac{\xi}{\dfrac{r}{2}}=2\cdot\dfrac{\xi}{r}=2I_1\)

\(\Rightarrow\dfrac{I_1}{I_2}=\dfrac{1}{2}\)

Mắc hai nguồn nối tiếp:

\(\xi_b=n\cdot\xi=2\xi\)

\(r_b=n\cdot r=2r\)

\(\Rightarrow I_1=\dfrac{\xi_b}{r_{b1}+r}=\dfrac{2\xi}{3r}\)

Mắc hai nguồn song song:

\(\xi_b=\xi\)

\(r_b=\dfrac{r}{n}=\dfrac{r}{2}\)

\(\Rightarrow I_2=\dfrac{\xi_{b2}}{r_{b2}+r}=\dfrac{\xi}{\dfrac{r}{2}+r}=\dfrac{\xi}{\dfrac{3}{2}r}=\dfrac{2}{3}\cdot\dfrac{\xi}{r}\)

Vậy \(\dfrac{I_1}{I_2}=1\)

Chọn C

a, \(\xi_b=\xi_1+\xi_2=7,5+7,5=15\left(V\right)\)

\(r_b=r_1+r_2=1+1=2\left(\Omega\right)\)

b,\(R=\dfrac{R_2\cdot\left(R_1+R_3\right)}{R_2+R_1+R_3}=\dfrac{40\cdot\left(40+20\right)}{40+40+20}=24\left(\Omega\right)\)

\(I=\dfrac{\xi_b}{R+r_b}=\dfrac{15}{24+2}=\dfrac{15}{26}\left(A\right)\)

c,\(U_{AB}=I\cdot R=\dfrac{15}{26}\cdot24=\dfrac{180}{13}\left(V\right)\)

\(I_{AD}=I_{DB}=I_{AB}=\dfrac{U_{AB}}{R_1+R_3}=\dfrac{\dfrac{180}{13}}{40+20}=\dfrac{3}{13}\left(A\right)\)

\(U_{AD}=R_1I_{AD}=40\cdot\dfrac{3}{13}=\dfrac{120}{13}\left(V\right)\)

\(U_{DB}=R_1I_{DB}=20\cdot\dfrac{3}{13}=\dfrac{60}{13}\left(V\right)\)

d,\(P_1=U_{AD}\cdot I_{AD}=\dfrac{120}{13}\cdot\dfrac{3}{13}=\dfrac{360}{169}\left(W\right)\)

\(P_2=\dfrac{U_{AB}^2}{R_2}=\dfrac{\left(\dfrac{180}{13}\right)^2}{40}=\dfrac{810}{169}\left(W\right)\)

\(P_3=U_{DB}\cdot I_{DB}=\dfrac{60}{13}\cdot\dfrac{3}{13}=\dfrac{180}{169}\left(W\right)\)

e,\(A_1=P_1t=\dfrac{360}{169}\cdot300\approx639,05\left(J\right)\)

\(P_b=I^2R=\left(\dfrac{15}{26}\right)^2\cdot24=\dfrac{1350}{169}\left(W\right)\)