Tóm tắt :
\(U=36V\)
\(R_1//R_2//R_3\)
\(I=4A\)
\(R_1=2R_2=3R_3\)
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R1 = ?
R2 = ?
R3 = ?
GIẢI :
Điện trở tương đương toàn mạch là :
\(R_{tđ}=\dfrac{U}{I}=\dfrac{36}{4}=9\left(\Omega\right)\)
Ta có : \(R_1=2R_2=3R_3\)
=> \(R_1=\dfrac{R_1}{2}=\dfrac{R_1}{3}\)
\(=>R_{tđ}=\dfrac{1}{\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}}=\dfrac{1}{\dfrac{1}{R_1}+\dfrac{1}{\dfrac{R_1}{2}}+\dfrac{1}{\dfrac{R_1}{3}}}\)
\(=>9=\dfrac{1}{\dfrac{1}{R_1}+\dfrac{2}{R_1}+\dfrac{3}{R_1}}\)
\(\Rightarrow9=\dfrac{1}{\dfrac{3}{R_1}}\)
\(\Rightarrow9=\dfrac{R_1}{3}\)
\(\Rightarrow R_1=27\Omega\)
Điện trở R2 là :
\(R_2=\dfrac{R_1}{2}=\dfrac{27}{2}=13,5\left(\Omega\right)\)
Điện trở R3 là:
\(R_3=\dfrac{R_1}{3}=\dfrac{27}{3}=9\left(\Omega\right)\)
Vậy : \(\left\{{}\begin{matrix}R_1=27\Omega\\R_2=13,5\Omega\\R_3=9\Omega\end{matrix}\right.\)
