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Ta có R1//R2=>Rtd=\(\dfrac{R1.R2}{R1+R2}=\dfrac{2R2.R2}{2R2+R2}=\dfrac{2R2^2}{3R2}=\dfrac{2R2}{3}=\dfrac{U}{I}=\dfrac{12}{1,5}=8\Omega=>R2=12\Omega=>R1=24\Omega\)
Vì R1//R2=>U1=U2=U=12V
=>\(I1=\dfrac{U1}{R1}=\dfrac{12}{12}=1A\)
=>\(I2=\dfrac{U2}{R2}=\dfrac{12}{24}=0,5A\)
Ta có: I1 = \(\dfrac{U_1}{R_{ }}=\dfrac{18}{R_{ }1}\)
Mà I2 =I1 +3 tương đương với \(\dfrac{18}{R_2}=\dfrac{18}{R_1}+3\)
\(\dfrac{18}{R_2}=\dfrac{18}{2R_2}+3\)
Giải phương trình: R2=3 Ôm, R1=6 Ôm
Từ đó tình được: I1=3A, I2=6A
Đáp số:
\(\Rightarrow\left\{{}\begin{matrix}I1=\dfrac{18}{R1}\\I2=\dfrac{18}{R2}\end{matrix}\right.\)\(\Rightarrow I2=I1+3\Rightarrow\dfrac{18}{R2}=\dfrac{18}{2R2}+3\Rightarrow\left\{{}\begin{matrix}R2=3\Omega\\R1=6\Omega\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}I1=\dfrac{18}{6}=3A\\I2=\dfrac{18}{3}=6A\end{matrix}\right.\)
a)Ta có (R1//R3)nt(R2//R4)=> Rtđ=R13+R24=\(\dfrac{R1.R3}{R1+R3}+\dfrac{R2.R4}{R2+R4}=1+2=3\Omega\)
=> I=\(\dfrac{U}{Rt\text{đ}}=\dfrac{5}{3}A\)
Vì R13ntR24=>I13=I24=I=\(\dfrac{5}{3}A\)
Vì R1//R3=> U1=U3=U13=I13.R13=\(\dfrac{5}{3}.1=\dfrac{5}{3}V\)
=> I1=\(\dfrac{U1}{R1}=\dfrac{5}{3}:2=\dfrac{5}{6}A;I3=\dfrac{U3}{R3}=\dfrac{5}{3}:2=\dfrac{5}{6}A\)
Vì R2//R4=> U2=U4=U24=I24.R24=\(\dfrac{5}{3}.2=\dfrac{10}{3}V\)
=> I2=\(\dfrac{U2}{R2}=\dfrac{10}{3}:3=\dfrac{10}{9}A;I4=\dfrac{U4}{R4}=\dfrac{10}{3}:6=\dfrac{5}{9}A\)
Vì I1<I2=> Chốt dương tại D
=> I1+Ia=I2=> Ia=I2-I1=\(\dfrac{5}{18}A\)
Vậy ampe kế chỉ 5/18 A
a) Ta có: \(\dfrac{I_1}{I_2}=\dfrac{R_2}{R_1}=\dfrac{0,5I_2}{I_2}=0,5\)
\(\Rightarrow\dfrac{R_2}{R_1}=0,5\Rightarrow R_2=0,5R_1\)
\(I_1=0,5I_2\Rightarrow R_2=..0,5..R_1\)
b) Ta có: \(\dfrac{I_1}{I_2}=\dfrac{R_2}{R_1}=\dfrac{I_1}{0,4I_1}=2,5\)
\(\Rightarrow\dfrac{R_2}{R_1}=2,5\Rightarrow R_2=2,5R_1\)
\(I_2=0,4I_1\Rightarrow R_2=...2,5R_1...\)
c) Ta có: \(\dfrac{I_1}{I_2}=\dfrac{R_2}{R_1}=\dfrac{3R_1}{R_1}=3\)
\(\Rightarrow\dfrac{I_1}{I_2}=3\Rightarrow I_1=3I_2\)
\(R_2=3R_1\Rightarrow...I_1=3I_2...\)
d) Ta có: \(\dfrac{I_1}{I_2}=\dfrac{R_2}{R_1}=\dfrac{R_2}{2R_2}=0,5\)
\(\Rightarrow\dfrac{I_1}{I_2}=0,5\Rightarrow I_1=0,5I_2\)
\(R_1=2R_2\Rightarrow I_1=0,5I_2\)
a) Nếu I1 = 0,5I2 => R1 = 2R2
b) I2 = 0,4 I1 => R2 = 2,5 R1
c) R2 = 3R1 => I2 = 1/3 I1
d) R1 = 2R2 => I1 = 0,5I2