Đáp án:

 a. \({E_b} = 20V,{r_b} = 2\Omega \)

b.\(U = \frac{{180}}{{11}}V,I = \frac{{20}}{{11}}A\)

c. \({A_{nguon}} = 21818,18J,{A_{mach}} = 17851,24J\)

d. m=0,581g

Giải thích các bước giải:

\(\begin{array}{l}
a.\\
{E_b} = 4E = 4.5 = 20V\\
{r_b} = 4r = 4.0,5 = 2\Omega \\
b.\\
{R_{12}} = {R_1} + {R_2} = 3 + 3 = 6\Omega \\
{R_{123}} = \frac{{{R_{12}}{R_3}}}{{{R_{12}} + {R_3}}} = \frac{{6.6}}{{6 + 6}} = 3\Omega \\
R = {R_{123}} + {R_4} = 3 + 6 = 9\Omega \\
I = \frac{{{E_b}}}{{R + {r_b}}} = \frac{{20}}{{9 + 2}} = \frac{{20}}{{11}}A\\
U = IR = \frac{{20}}{{11}}.9 = \frac{{180}}{{11}}V\\
c.\\
{A_{nguon}} = EIt = 20.\frac{{20}}{{11}}.10.60 = 21818,18J\\
{A_{mach}} = UIt = \frac{{180}}{{11}}.\frac{{20}}{{11}}.10.60 = 17851,24J\\
d.\\
m = \frac{{AIt}}{{96500.n}} = \frac{{64.\frac{{20}}{{11}}.965}}{{96500.2}} = 0,581g
\end{array}\)