Tóm tắt:
R1 // R2
\(R1=40\Omega\)
\(R2=60\Omega\)
\(U=12V\)
a. R = ?\(\Omega\)
b. I, I1, I2 = ?A
GIẢI:
a. \(R=\dfrac{R1.R2}{R1+R2}=\dfrac{40.60}{40+60}=24\Omega\)
b. \(U=U1=U2=12V\left(R1\backslash\backslash\mathbb{R}2\right)\)
\(\Rightarrow\left\{{}\begin{matrix}I1=U1:R1=12:40=0,3A\\I2=U2:R2=12:60=0,2A\end{matrix}\right.\)
Tóm tắt:
\(R_1//R_2\)
\(R_1=40\Omega,R_2=60\Omega\)
\(U_{AB}=12V\)
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a) \(R_{tđ}=?\)
b) \(I_1=?,I_2=?\)
\(R_{tđ}=\dfrac{R_1.R_2}{R_1+R_2}=\dfrac{40.60}{40+60}=24\left(\Omega\right)\)
\(U=U_1=U_2=12V\)
\(\left\{{}\begin{matrix}I_1=\dfrac{U_1}{R_1}=\dfrac{12}{40}=0,3\left(A\right)\\I_2=\dfrac{U_2}{R_2}=\dfrac{12}{60}=0,2\left(A\right)\end{matrix}\right.\)
Tóm tắt: \(U_{AB}=12V;R_1=40\Omega;R_2=60\Omega\)
a)\(R_{tđ}=?\)
b)\(I_1=?;I_2=?\)
Bài giải:
a)Điện trở tương đương:\(R_{12}=\dfrac{R_1\cdot R_2}{R_1+R_2}=\dfrac{40\cdot60}{40+60}=24\Omega\)
b)\(U_1=U_2=12V\)
\(I_1=\dfrac{12}{40}=0,3A;I_2=\dfrac{12}{60}=0,2A\)
