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\(R=R1+\left(\dfrac{R2.R3}{R2+R3}\right)=10+\left(\dfrac{6.3}{6+3}\right)=12\Omega\)
\(I=U:R=12:12=1A\)
\(\Rightarrow I=I1=I23=1A\left(R1ntR23\right)\)
\(U23=U-U1=12-\left(10,1\right)=2V\)
\(\Rightarrow U23=U2=U3=2V\)(R2//R3)
\(\left\{{}\begin{matrix}I2=U2:R2=2:6=\dfrac{1}{3}A\\I3=U3:R3=2:3=\dfrac{2}{3}A\end{matrix}\right.\)
a) \(R_{23}=\dfrac{6.3}{6+3}=2\left(\Omega\right)\)
\(R_{tđ}=R_1+R_{23}=10+2=12\left(\Omega\right)\)
b) \(I=I_1=I_{23}=\dfrac{U}{R_{tđ}}=\dfrac{12}{12}=1\left(A\right)\)
\(U_1=U-U_{23}=12-2=10\left(V\right)\)
\(U_{23}=U_2=U_3=I_{23}.R_{23}=1.2=2\left(V\right)\)
\(\left\{{}\begin{matrix}I_2=\dfrac{U_2}{R_2}=\dfrac{2}{10}=0,2\left(A\right)\\I_3=\dfrac{U_3}{R_3}=\dfrac{2}{6}=\dfrac{1}{3}\left(A\right)\end{matrix}\right.\)
\(\Rightarrow U=IR=24.0,5=12V\)
vay phai dat vao 2 dau bong den 1 HDT=12V
\(R_{tđ}=\dfrac{R_1\cdot R_2}{R_1+R_2}=\dfrac{24\cdot12}{24+12}=8\Omega\)
\(I=\dfrac{U}{R}=\dfrac{12}{8}=1,5A\)
\(P=\dfrac{U^2}{R}=\dfrac{12^2}{8}=18W\)
\(Q_{tỏa1}=A_1=U_1\cdot I_1\cdot t=12\cdot\dfrac{12}{24}\cdot1\cdot3600=21600J\)
\(Q_{tỏa2}=A_2=U_2\cdot I_2\cdot t=12\cdot\dfrac{12}{12}\cdot1\cdot3600=43200J\)