Tóm tắt:
\(R1=40\Omega\)
\(R2=60\Omega\)
\(U=24V\)
a. R = ?\(\Omega\)
b. I = ?A
U2 = ?V
c. P = ?W
d. R3 // R2
I' = 0,3A
R3 = ?\(\Omega\)
GIẢI:
a. \(R=R1+R2=40+60=100\Omega\)
b. \(I=I1=I2=\dfrac{U}{R}=\dfrac{24}{100}=0,24A\left(R2ntR2\right)\)
\(U2=I2.R2=0,24.60=14,4V\)
c. \(P=UI=24.0,24=5,76\)W
d. \(I'=I_1'=I_{23}=0,3A\left(R1ntR23\right)\)
\(U23=U2=U3=U-U1=24-\left(0,3.40\right)=12V\left(R1\backslash\backslash\mathbb{R}2\right)\)
\(R23=U23:I23=12:0,3=40\Omega\)
\(\dfrac{1}{R3}=\dfrac{1}{R23}-\dfrac{1}{R2}=\dfrac{1}{40}-\dfrac{1}{60}=\dfrac{1}{120}\Rightarrow R3=120\Omega\)