Question

Answer 1

\(\left\{{}\begin{matrix}\xi_{hh}=4\xi=4\cdot1,5=6V\\r_{hh}=4r=4\cdot0,5=2\Omega\end{matrix}\right.\)

a)CTM: \(R_1nt\left(R_2//R_3\right)\)

\(R_{23}=\dfrac{R_2\cdot R_3}{R_2+R_3}=\dfrac{4\cdot12}{4+12}=3\Omega\)

\(R_{tđ}=R_1+R_{23}=1+3=4\Omega\)

\(I_A=I=\dfrac{\xi}{r_{hh}+R_{tđ}}=\dfrac{6}{2+4}=1A\)

b)\(t=32phút10s=1930s\)

Khối lượng bạc nhận được ở catot:

\(m=\dfrac{1}{96500}\cdot\dfrac{A}{n}\cdot It=\dfrac{1}{96500}\cdot\dfrac{108}{1}\cdot1\cdot1930=2,16g\)

c)\(U_N=\xi-I\cdot r=6-1\cdot2=4V\)

\(I_N=\dfrac{U_N}{R_{tđ}}=\dfrac{4}{4}=1A=I_{23}\)

\(U_3=U_{23}=I_{23}\cdot R_{23}=1\cdot3=3V\)

\(P_3=\dfrac{U_3^2}{R_3}=\dfrac{3^2}{12}=0,75W\)