Vì (R3 // (R1 nt R2 )) → R_td =\(\frac{R3.\left(R1+R2\right)}{R3+R2+R1}\)=\(\frac{10.\left(2+8\right)}{10+2+8}\)=5Ω

Hiệu điện thế giữa hai đầu đoạn mạch là:

từ P=\(\frac{U^2}{R_{td}}\)→U=\(\sqrt{P.R_{td}}\)=\(\sqrt{3,6.5}\)=\(\sqrt{18}\) (V)

Mà R3//(R1 nt R2)→U=U₃=U₁₂→U₃=\(\sqrt{18}\) V

Công suất tiêu thụ của điện trở R3 là;

P=\(\frac{U^2_3}{R_3}\) = \(\frac{\left(\sqrt{18}\right)^2}{10}\)=\(\frac{18}{10}\)=1,8 (W)

Vậy P₃=1,8 W

1,8W

ta có : R_td = \(\frac{R3.\left(R2+R1\right)}{R3+R2+R1}\)=5 ôm

=> I_{tm = \(\sqrt{\frac{P}{R}}\)=\(\frac{3\sqrt{2}}{5}\)}

=> P₃ = I² . R₃ =7,2 W

mọi người trả lời hộ với ?

a)\(R_{tđ}=R_1+R_2=6+10=16\Omega\)

   \(P=R\cdot I^2=16\cdot0,5^2=4W\)

b)\(R_{tđ}=\dfrac{R_3\cdot R_{12}}{R_3+R_{12}}=\dfrac{8\cdot16}{8+16}=\dfrac{16}{3}\Omega\)

   \(I_m=0,5A\)

   \(\Rightarrow U=I\cdot R=0,5\cdot\dfrac{16}{3}=\dfrac{8}{3}V\)

\(R_{12}=R_1+R_2=2+8=10\Omega\)

\(R_{tđ}=\dfrac{R_{12}\cdot R_3}{R_{12}+R_3}=\dfrac{10\cdot10}{10+10}=5\Omega\)

\(U_m=\sqrt{P\cdot R_{tđ}}=\sqrt{3,6\cdot5}=3\sqrt{2}V\)

\(\Rightarrow U_{12}=U_m=3\sqrt{2}V\)

\(I_2=I_{12}=\dfrac{U_{12}}{R_{12}}=\dfrac{3\sqrt{2}}{10}A\)

\(P_2=I_2\cdot U_2=I^2_2\cdot R_2=\left(\dfrac{3\sqrt{2}}{10}\right)^2\cdot8=1,44W\)

CTM: \(R_1nt\left(R_2//R_3\right)\)

a)\(R_{23}=\dfrac{R_2\cdot R_3}{R_2+R_3}=\dfrac{2\cdot6}{2+6}=1,5\Omega\)

\(R_{tđ}=R_1+R_{23}=3+1,5=4,5\Omega\)

\(I=\dfrac{U}{R_{tđ}}=\dfrac{9}{4,5}=2A\)

b)Công đoạn mạch sản ra trong thời gian \(t=5min=300s\) là:

\(A=UIt=9\cdot2\cdot300=5400J\)

c)\(I_1=I=2A\)

Điện năng tiêu thụ trên điện trở \(R_1\) trong thời gian \(t=5min=300s\) là:

\(A_1=U_1.I_1.t=I_1^2.R_1.t=2^2\cdot3\cdot300=3600J\)

\(R_{tđ}=\dfrac{\left(R_1+R_2\right)\cdot R_3}{R_1+R_2+R_3}=4\Omega\)

\(I_1=I_2=0,3A\)\(\Rightarrow U_1=0,3\cdot8=2,4V;U_2=0,3\cdot12=3,6V\)

\(U_m=U_1+U_2=2,4+3,6=6V\)

\(P=\dfrac{U^2}{R}=\dfrac{6^2}{4}=9W\)

\(R_{12}=R_1+R_2=8+12=20\left(\Omega\right)\)

\(R_{tđ}=\dfrac{R_{12}.R_3}{R_{12}+R_3}=\dfrac{20.5}{20+5}=4\left(\Omega\right)\)

\(I_{12}=I_1=I_2=0,3A\)

\(U=U_{12}=U_3=I_{12}.R_{12}=0,3.20=6\left(V\right)\)

\(P=\dfrac{U^2}{R}=\dfrac{6^2}{4}=9\left(W\right)\)

\(R_{tđ}=\dfrac{R_1.R_2}{R_1+R_2}=\dfrac{5.10}{5+10}=\dfrac{10}{3}\left(\Omega\right)\)

\(U=U_1=U_2=15V\)

\(\left\{{}\begin{matrix}I=\dfrac{U}{R_{tđ}}=\dfrac{15}{\dfrac{10}{3}}=4,5\left(A\right)\\I_1=\dfrac{U_1}{R_1}=\dfrac{15}{5}=3\left(A\right)\\I_2=\dfrac{U_2}{R_2}=\dfrac{15}{10}=1,5\left(A\right)\end{matrix}\right.\)

\(P=U.I=15.4,5=67,5\left(W\right)\)

a. \(R=\dfrac{R1.R2}{R1+R2}=\dfrac{5.10}{5+10}=\dfrac{10}{3}\left(\Omega\right)\)

b. \(U=U1=U2=15V\)(R1//R2)

\(\left\{{}\begin{matrix}I1=U1:R1=15:5=3A\\I2=U2:R2=15:10=1,5A\end{matrix}\right.\)

c. \(\left\{{}\begin{matrix}P_m=U_mI_m=15.\left(3+1,5\right)=67,5\\P1=U1.I1=15.3=45\\P2=U2.I2=15.1,5=22,5\end{matrix}\right.\)(W)