R1=1
R2=R5=5
R3=R6=3
R4=R7=2
\(U_{AB}=12V\)
a) Tim \(R_{AB}\)
b)Tim \(Í_1,I_2,I_3,I_4,I_5,I_6,I_7\)
c)Tim \(U_1,U_2,U_3,U_4,U_5,U_6,U_7\)
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a:
ĐKXĐ: \(q\notin\left\{0;1;-1\right\}\)
\(HPT\Leftrightarrow\left\{{}\begin{matrix}u1\cdot q^4-u1=15\\u1\cdot q^3-u1\cdot q=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{q^4-1}{q^3-q}=\dfrac{15}{6}=\dfrac{5}{2}\\u1\left(q^4-1\right)=15\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2q^4-2=5q^3-5q\\u1\left(q^4-1\right)=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2q^4-5q^3+5q-2=0\\u1\left(q^4-1\right)=15\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left(q-2\right)\left(q-1\right)\left(q+1\right)\left(2q-1\right)=0\\u1\left(q^4-1\right)=15\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}q=2\\q=\dfrac{1}{2}\end{matrix}\right.\\u1\left(q^4-1\right)=15\end{matrix}\right.\)
TH1: q=2
=>\(u1=\dfrac{15}{2^4-1}=\dfrac{15}{15}=1\)
TH2: q=1/2
=>\(u1=\dfrac{15}{\dfrac{1}{16}-1}=15:\dfrac{-15}{16}=-16\)
b:
\(HPT\Leftrightarrow\left\{{}\begin{matrix}u1-u1\cdot q^2+u1\cdot q^4=65\\u1+u1\cdot q^6=325\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{q^4-q^2+1}{q^6+1}=\dfrac{1}{5}\\u1\left(1+q^6\right)=325\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{1}{q^2+1}=\dfrac{1}{5}\\u1\left(q^6+1\right)=325\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}q^2=4\\u1\left(q^6+1\right)=325\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}q\in\left\{2;-2\right\}\\u1\left(q^6+1\right)=325\end{matrix}\right.\Leftrightarrow u1=\dfrac{325}{65}=5\)
c: \(HPT\Leftrightarrow\left\{{}\begin{matrix}u1\cdot q^3+u1\cdot q^5=-540\\u1\cdot q+u1\cdot q^3=-60\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{q^5+q^3}{q^3+q}=9\\u1\left(q+q^3\right)=-60\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}q^2=9\\u1\left(q+q^3\right)=-60\end{matrix}\right.\)
TH1: q=3
\(u1=-\dfrac{60}{3+3^3}=-\dfrac{60}{30}=-2\)
TH2: q=-3
=>\(u1=-\dfrac{60}{-3-27}=\dfrac{60}{30}=2\)
Hướng dẫn giải
Mạch điện gồm ( R 1 n t R 2 ) / / R 3 / / R 4 n t R 5 / / R 6 n t R 7 .
( R 1 n t R 2 ) nên R 12 = R 1 + R 2 = 4 + 20 = 24 Ω
R 12 / / R 3 nên R 123 = R 12 R 3 R 12 + R 3 = 24.12 24 + 12 = 8 Ω
R 123 / / R 4 nên R 1234 = R 123 . R 4 R 123 + R 4 = 8.8 8 + 8 = 4 Ω
R 1234 n t R 5 nên R 12345 = R 1234 + R 5 = 4 + 20 = 24 Ω
R 12345 / / R 6 nên R 123456 = R 12345 . R 6 R 12345 + R 6 = 24.12 24 + 12 = 8 Ω
R 123456 n t R 7 nên điện trở tương đương của đoạn mạch:
R 1234567 = R 123456 + R 7 = 8 + 8 = 16 Ω
R1=R2=R3=R4=R5=R6=R7=R8=21(om)
a, K1,K2 đều mở R7 nt R8 \(=>Rtd=R1+R2=42\Omega\)
b, K1 mở,K2 đóng (R4 nt R5 nt R6)//(R7 nt R8)
\(=>RTd=\dfrac{\left(R4+R5+R6\right)\left(R7+R8\right)}{R4+R5+R6+R7+R8}=.,....\)
c,K1 đóng , K2 mở R3//(R1 ntR2)//(R7 nt R8)
\(=>\dfrac{1}{Rtd}=\dfrac{1}{R3}+\dfrac{1}{R1+R2}+\dfrac{1}{R7+R8}=>Rtd=.....\)
d, K1,K2 đóng R3 //(R1 nt R2) //(R4 nt R5 nt R6)//(R7 nt R8)
\(=>\dfrac{1}{Rtd}=\dfrac{1}{R3}+\dfrac{1}{R1+R2}+\dfrac{1}{R4+R5+R6}+\dfrac{1}{R7+R8}=>Rtd=....\)
Bài này còn yêu cầu vẽ lại hình để nêu cách mắc dễ hơn ạ. Bạn vẽ hình đc ko ạ
CTM: \(R_7nt\left\{R_6//\left[R_5nt(\left(R_1ntR_2\right)//R_3//R_4)\right]\right\}\)
a)\(R_{12}=R_1+R_2=1+5=6\Omega\)
\(\dfrac{1}{R_{1234}}=\dfrac{1}{R_{12}}+\dfrac{1}{R_3}+\dfrac{1}{R_4}=\dfrac{1}{6}+\dfrac{1}{3}+\dfrac{1}{2}=1\Omega\Rightarrow R_{1234}=1\Omega\)
\(R_{12345}=R_5+R_{1234}=5+1=6\Omega\)
\(R_{123456}=\dfrac{R_6\cdot R_{12345}}{R_6+R_{12345}}=\dfrac{3\cdot6}{3+6}=2\Omega\)
\(R_{AB}=R_7+R_{123456}=2+2=4\Omega\)
b)\(I_7=I_{AB}=\dfrac{U_{AB}}{R_{AB}}=\dfrac{12}{4}=3A=I_{123456}\)
\(\Rightarrow U_6=U_{12345}=U_{123456}=3\cdot2=6V\Rightarrow I_6=\dfrac{U_6}{R_6}=\dfrac{6}{3}=2A\)
\(I_{1234}=I_5=\dfrac{U_{12345}}{R_{12345}}=\dfrac{6}{6}=1A\)
\(U_{12}=U_3=U_4=U_{1234}=1\cdot1=1V\)
\(\left\{{}\begin{matrix}I_3=\dfrac{U_3}{R_3}=\dfrac{1}{3}A\\I_4=\dfrac{U_4}{R_4}=\dfrac{1}{2}A\end{matrix}\right.\)
\(I_1=I_2=\dfrac{U_{12}}{R_{12}}=\dfrac{1}{6}A\)