Question

[Phạm phú hưng]

Giải giúp mình với ạ Gấp lắm nha mn

Answer 1

[nthv_.]

a. \(R=R1+\left(\dfrac{R2.R3}{R2+R3}\right)=15+\left(\dfrac{12.18}{12+18}\right)=\dfrac{545}{36}\left(\Omega\right)\)

b. \(I=I1=I23=\dfrac{U}{R}=\dfrac{24}{\dfrac{545}{36}}=\dfrac{864}{545}\left(A\right)\left(R1ntR23\right)\)

\(U23=U2=U3=I23.R23=\dfrac{864}{545}\left(\dfrac{12.18}{12+18}\right)\simeq11,4V\)(R2//R3)

\(\left\{{}\begin{matrix}I2=U2:R2=11,4:12=0,95A\\I3=U3:R3=11,4:18=\dfrac{19}{30}A\end{matrix}\right.\)

c. \(P=UI=24.\dfrac{864}{545}\simeq38,05\)(W)

d. \(A=Pt=38,05.1,5=57,075\)(Wh)

Answer 2

[nguyễn thị hương giang]

a)\(R_{tđ}=R_1+\dfrac{R_2\cdot R_3}{R_2+R_3}=15+\dfrac{12\cdot18}{12+18}=22,2\Omega\)

b)\(I_1=I_m=\dfrac{U}{R}=\dfrac{24}{22,2}=\dfrac{40}{37}A\)

   \(U_1=I_1\cdot R_1=\dfrac{40}{37}\cdot15=\dfrac{600}{37}V\)\(\Rightarrow U_2=U_3=U_{23}=24-\dfrac{600}{37}=\dfrac{288}{37}V\)

 \(\Rightarrow I_2=\dfrac{U_2}{R_2}=\dfrac{24}{37}A;I_3=\dfrac{U_3}{R_3}=\dfrac{16}{37}A\)

c)\(P=U_m\cdot I_m=24\cdot\dfrac{40}{37}=\dfrac{960}{37}\approx25,945W\)

d)\(A=UIt=24\cdot\dfrac{40}{37}\cdot1,5\cdot3600=140108,1081J\approx0,04kWh\)