Câu hỏi của Nguyễn Văn Tương

Question

nguyễn thị hương giang · Accepted Answer

$R_{23}=\dfrac{R_2\cdot R_3}{R_2+R_3}=\dfrac{24\cdot8}{24+8}=6\Omega$

$R_{tđ}=R_1+R_{23}=12+6=18\Omega$

$I_1=I_m=\dfrac{\xi}{R_{tđ}+r}=\dfrac{12}{18+4}=\dfrac{6}{11}A$

$U_2=U_{23}=I_{23}\cdot R_{23}=\dfrac{6}{11}\cdot6=\dfrac{36}{11}V$

$P_2=\dfrac{U^2_2}{R_2}=\dfrac{\left(\dfrac{36}{11}\right)^2}{24}=0,45W$

$A=P\cdot t=\xi\cdot I\cdot t=12\cdot\dfrac{6}{11}\cdot10\cdot60=3927,27J$

$H=\dfrac{R_{tđ}}{r+R_{tđ}}=\dfrac{18}{4+18}=81,82\%$

Câu trả lời:

$R_{23}=\dfrac{R_2\cdot R_3}{R_2+R_3}=\dfrac{24\cdot8}{24+8}=6\Omega$

$R_{tđ}=R_1+R_{23}=12+6=18\Omega$

$I_1=I_m=\dfrac{\xi}{R_{tđ}+r}=\dfrac{12}{18+4}=\dfrac{6}{11}A$

$U_2=U_{23}=I_{23}\cdot R_{23}=\dfrac{6}{11}\cdot6=\dfrac{36}{11}V$

$P_2=\dfrac{U^2_2}{R_2}=\dfrac{\left(\dfrac{36}{11}\right)^2}{24}=0,45W$

$A=P\cdot t=\xi\cdot I\cdot t=12\cdot\dfrac{6}{11}\cdot10\cdot60=3927,27J$

$H=\dfrac{R_{tđ}}{r+R_{tđ}}=\dfrac{18}{4+18}=81,82\%$