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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:
\(A_1=U_1.I_1.t=I_1^2.R_1.t=2^2\cdot3\cdot300=3600J\)
a) \(R_Đ=\dfrac{U^2}{P}=\dfrac{10^2}{10}=10\left(ÔM\right)\)
\(R_{TĐ}=\dfrac{R_1.R_2}{R_1+R_2}+R_Đ=\dfrac{8.7}{8+7}+10=13,73\left(ÔM\right)\)
b)
\(U_{12}=U_1=U_2=R_{12}.I=3,73.1,8=6,71\left(V\right)\)
\(\Rightarrow P_1=\dfrac{U^2_1}{R_1}=\dfrac{6,71^2}{8}=5,63\left(W\right)\)
\(\Rightarrow A_1=P_1.t=5,63.600=3378\left(J\right)\) ( Đổi \(10P=600s\))
c)
\(P_2=\dfrac{U_2^2}{R_2}=\dfrac{6,71^2}{7}=6,43\left(W\right)\)
\(\Rightarrow\) \(P_1< P_2\)
\(R_Đ=\dfrac{U^2_Đ}{P_Đ}=\dfrac{10^2}{10}=10\Omega;I_{Đđm}=\dfrac{P_Đ}{U_Đ}=\dfrac{10}{10}=1A\)
a)\(R_{12}=R_1+R_2=8+7=15\Omega\)
\(R_{tđ}=\dfrac{R_{12}\cdot R_Đ}{R_{12}+R_Đ}=\dfrac{15\cdot10}{15+10}=6\Omega\)
b)\(U_m=R_{tđ}\cdot I_m=6\cdot1,8=10,8V\)
\(I_1=I_2=I_{12}=\dfrac{U_{12}}{R_{12}}=\dfrac{U}{R_{12}}=\dfrac{10,8}{15}=0,72A\)
\(A_1=U_1I_1t=R_1I_1^2t=8\cdot0,72^2\cdot10\cdot60=2488,32J\)
c)Công suất: \(P=U\cdot I=R\cdot I^2\)
Mặt khác: \(R_1ntR_2\Rightarrow I_1=I_2=I\) mà \(R_1>R_2\)
Nên \(P_1>P_2\)
\(R_{12}=\dfrac{15.30}{15+30}=10\left(\Omega\right)\)
\(R_m=R_{12}+R_3=10+30=40\left(\Omega\right)\)
\(I_m=\dfrac{U_{AB}}{R_m}=\dfrac{12}{40}=0,3\left(A\right)\)
\(b,I_{12}=I_3=0,3\left(A\right)\)
\(\dfrac{I_1}{I_2}=\dfrac{R_2}{R_1}=\dfrac{30}{15}=\dfrac{2}{1}\)
\(\rightarrow I_1=0,2\left(A\right);I_2=0,1\left(A\right)\)
a, Rtd=10+30=40\(\Omega\) \(I=\dfrac{U}{R_{tđ}}=\dfrac{12}{40}=0,3\left(A\right)\)
b, \(P_2=I^2R_2=0,9\left(W\right)\)
c, \(Q=I^2.R_{tđ}.18.60=3888\left(J\right)\)
d,\(I_2=\dfrac{3}{5}.0,3=0,18\left(A\right)\) \(\Rightarrow I_3=0,3-I_2=0,12\left(A\right)\)
\(U_{23}=12-0,3.30=3\left(V\right)\)\(\Rightarrow R_3=\dfrac{3}{0,12}=25\left(\Omega\right)\)
\(a,R_{23}=R_2+R_3=30+30=60\left(\Omega\right)\)
\(R_m=\dfrac{R_{23}.R_1}{R_{23}+R_1}=\dfrac{60.15}{60+15}=12\left(\Omega\right)\)
\(b,I_m=\dfrac{U_{AB}}{R_m}=\dfrac{12}{12}=1\left(A\right)\)
\(I_1+I_{23}=1\left(A\right)\)
\(\dfrac{I_1}{I_{23}}=\dfrac{R_{23}}{R_1}=\dfrac{60}{15}=\dfrac{4}{1}\)
\(\rightarrow I_1=0,8\left(A\right);I_{23}=0,2\left(A\right)\)
\(\rightarrow I_2=I_3=0,2\left(A\right)\)
Mạch điện như nào vậy bạn