Đáp án:

a) $U_{AB}=18V$

b) $I_R=1A$

$I_{Đ_1}=2A$

$I_{Đ_2}=3A$

c) `\mathcalP_{Đ_2}=36W`

d) `\mathcalP_{AB}=54W`

$A=1,8Wh$

Giải thích các bước giải:

Mạch có $(R//Đ_1)\ nt\ Đ_2$

a) $U_{AB}=U_{AC}+U_{CB}=U_{Đ_1}+U_V=6+12=18\ (V)$

b) $I_R=\dfrac{U_R}{R}=\dfrac{U_{Đ_1}}{R}=\dfrac{6}{6}=1\ (A)$ 

`I_{Đ_1}=\mathcalP_{đm}/U_{đm}=12/6=2\ (A)`

$I_{Đ_2}=I_{AC}=\dfrac{U_{AC}}{R_{AC}}=\dfrac{U_{Đ_1}}{\dfrac{R.R_{Đ_1}}{R+R_{Đ_1}}}=\dfrac{U_{Đ_1}}{\dfrac{R.\dfrac{U_{đm}^2}{\mathcal{P}_{đm}}}{R+\dfrac{U_{đm}^2}{\mathcal{P}_{đm}}}}=\dfrac{6}{\dfrac{6.\dfrac{6^2}{12}}{6+\dfrac{6^2}{12}}}=3\ (A)$ 

c) `\mathcalP_{Đ_2}=U_{Đ_2}I_{Đ_2}=U_VI_{Đ_2}=12.3=36\ (W)`

d) `\mathcalP_{AB}=U_{AB}I=U_{AB}I_{Đ_2}=18.3=54\ (W)`

`A=\mathcalP.t=54.{2}/{60}=1,8\ (Wh)`