\(R_{23}=\dfrac{R_2\cdot R_3}{R_2+R_3}=\dfrac{10\cdot10}{10+10}=5\Omega\)

\(R_N=R_1+R_{23}=6+5=11\Omega\)

\(I_m=\dfrac{\xi}{r+R_N}=\dfrac{12}{1+11}=1A\)

\(U_1=I_1\cdot R_1=I\cdot R_1=1\cdot6=6V\)

\(U_2=U_3=U_{23}=I_{23}\cdot R_{23}=1\cdot5=5V\)

\(P_N=\xi\cdot I=12\cdot1=12W\)

\(A_N=P\cdot t=12\cdot10\cdot60=7200J\)

cÂU 9.

\(I=\dfrac{\xi}{r+R_N}=\dfrac{20}{5+10}=\dfrac{4}{3}A\)

\(P_R=I^2_R\cdot R=\dfrac{4}{3}\cdot10=\dfrac{40}{3}W\)

\(P=\xi\cdot I=20\cdot\dfrac{4}{3}=\dfrac{80}{3}W\)

\(H=\dfrac{R}{r+R}=\dfrac{10}{5+10}=\dfrac{2}{3}\approx66,67\%\)

\(t=16\) phút \(5\) giây \(=965s\)

a. Có:

\(R_{12}=R_1+R_2=2+4=6\Omega\)

\(R_N=\dfrac{R_{12}R_3}{R_{12}+R_3}=\dfrac{6.12}{6+12}=4\Omega\)

\(I=\dfrac{E}{R_N+r}=\dfrac{13,5}{4+2}=2,25A\)

\(I_2=\dfrac{R_3}{R_3+R_{12}}.I=\dfrac{12}{12+6}.2,25=1,5A\)

Khối lượng bạc:

\(m=\dfrac{AIt}{nF}=\dfrac{108.1,5.965}{96500.1}=1,62g\)

b. Có:

\(I_{dm}=I_2=1,5A\)

\(U_{dm}=I_{dm}.R=1,5.4=6V\)

\(P_{dm}=I_{dm}.U_{dm}=1,5.6=9W\)

B

Khi biến trở là R:

\(I_1=\dfrac{E}{R+r}\Rightarrow0,5=\dfrac{6}{2+R}\Rightarrow R=10\Omega\)

Khi biến trở là \(R\text{/3}\)

\(I_2=\dfrac{E}{\dfrac{R}{3}+r}=\dfrac{6}{\dfrac{10}{3}+2}=1,125A\)

Vậy ta chọn đáp án B.