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Đoạn dây được cắt thành hai đoạn bằng nhau nên \(R_1=R_2=\dfrac{R}{2}\)
Điện trở của đoạn mạch:
\(R_{tđ}=\dfrac{R_1\cdot R_2}{R_1+R_2}=\dfrac{\dfrac{R}{2}\cdot\dfrac{R}{2}}{\dfrac{R}{2}+\dfrac{R}{2}}=\dfrac{R^2}{2}:\dfrac{2R}{2}=\dfrac{R}{2}\)
Ta có \(l=\dfrac{RS}{\rho}\Leftrightarrow R=\dfrac{l\rho}{S}\)
Cắt l thành n mảnh \(\Leftrightarrow l'=\dfrac{l}{4}\) \(\Leftrightarrow R'=\dfrac{\dfrac{l}{n}\rho}{S}=\dfrac{R}{n}=\dfrac{216}{n}\\ \Leftrightarrow R_{td}=\dfrac{R'.R'^n}{R'+R'^n}\Leftrightarrow6=\dfrac{\dfrac{216}{n}.\left(\dfrac{216}{n}\right)^n}{\dfrac{216}{n}+\left(\dfrac{216}{n}\right)^n}\Leftrightarrow n\approx0,27\)
\(a,\Rightarrow R=\dfrac{pl}{S}=\dfrac{0,4.10^{-6}.37,5}{0,5.10^{-6}}=30\Omega\)
b,\(\Rightarrow Rtd=\dfrac{x\left(30-x\right)}{x+30-x}=\dfrac{x\left(30-x\right)}{30}\)
\(\Rightarrow30Rtd=x\left(30-x\right)\Rightarrow-x^2+30x-30Rtd=0\)
\(\Rightarrow\Delta\ge0\Rightarrow30^2-4\left(-30Rtd\right).\left(-1\right)\ge0\)
\(\Rightarrow900-120Rtd\ge0\Leftrightarrow-120Rtd\ge-900\Leftrightarrow Rtd\le7,5\Omega\)
\(\Rightarrow Max\left(Rtd\right)=7,5\Leftrightarrow x=15\left(\Omega\right)\)
\(\Rightarrow\)gia tri 2 phan lan luot la \(\left\{{}\begin{matrix}R1=15\Omega\\R2=30-15=15\Omega\end{matrix}\right.\)
Câu 41.
Điện trở dây:
\(R=\rho\cdot\dfrac{l}{S}=\rho\cdot\dfrac{l}{\pi\cdot\dfrac{d^2}{4}}=2,8\cdot10^{-8}\cdot\dfrac{3,14}{\pi\cdot\left(\dfrac{2\cdot10^{-3}}{2}\right)^2}=0,028\Omega\)
a. \(\left\{{}\begin{matrix}I=\dfrac{P}{U}=\dfrac{15}{12}=1,25A\\R=\dfrac{U}{I}=\dfrac{12}{1,25}=9,6\Omega\end{matrix}\right.\)
b. \(R_{td}=R+R_{bd}=10+9,6=19,6\Omega\)
c. \(R_{ss}=\dfrac{U}{I}=\dfrac{12}{5}=2,4\Omega\)
Ta có: \(\dfrac{1}{R_{ss}}=\dfrac{1}{R'}+\dfrac{1}{R''}=\dfrac{2}{R'}\Rightarrow R'=R''=2R_{ss}=2\cdot2,4=4,8\Omega\)
Bài 13:
a. \(I=I1=I2=600mA=0,6A\left(R1ntR2\right)\)
\(\rightarrow U2=P2:I2=5,4:0,6=9V\)
\(\rightarrow U1=U-U2=15-9=6V\)
\(\Rightarrow\left\{{}\begin{matrix}R1=U1:I1=6:0,6=10\Omega\\R2=U2:I2=9:0,6=15\Omega\end{matrix}\right.\)
b. \(5min20s=320s\)
\(\Rightarrow\left\{{}\begin{matrix}Q=UIt=15\cdot0,6\cdot320=2880\left(J\right)\\Q1=U1\cdot I1\cdot t=9\cdot0,6\cdot320=1728\left(J\right)\\Q2=U2\cdot I2\cdot t=6\cdot0,6\cdot320=1152\left(J\right)\end{matrix}\right.\)
Bài 14:
a. \(I=U:R=15:\left(\dfrac{30\cdot15}{30+15}\right)=1,5A\)
b. \(U=U1=U2=15V\left(R1\backslash\backslash R2\right)\)
\(\rightarrow\left\{{}\begin{matrix}I1=U1:R1=15:30=0,5A\\I2=U2:R2=15:15=1A\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}P=UI=15\cdot1,5=22,5\\P1=U1\cdot I1=15\cdot0,5=7,5\\P2=U2\cdot I2=15\cdot1=15\end{matrix}\right.\)(W)
c. \(\Rightarrow\left\{{}\begin{matrix}Q=UIt=15\cdot1,5\cdot12\cdot60=16200\left(J\right)\\Q1=U1\cdot I1\cdot t=15\cdot0,5\cdot12\cdot60=5400\left(J\right)\\Q2=U2\cdot I2\cdot t=15\cdot1\cdot12\cdot60=10800\left(J\right)\end{matrix}\right.\)
Khi cắt chúng thành hai đoạn dây băng nhau ta có: \(R_1=R_2=\dfrac{R}{2}\)
Mắc chúng song song ta có điện trở bộ dây:
\(R_{tđ}=\dfrac{R_1\cdot R_2}{R_1+R_2}=\dfrac{\dfrac{R}{2}\cdot\dfrac{R}{2}}{\dfrac{R}{2}+\dfrac{R}{2}}=\dfrac{\dfrac{R^2}{4}}{R}=\dfrac{R}{4}\)