Tóm tắt:

R₁ = 4\(\Omega\)

R₂ = 16\(\Omega\)

U₁ = 3,2V

I = 1A

a. R_tđ = ?\(\Omega\)

b. I₁ = ?A

c. I₂ = ?A

GIẢI:

a. Điện trở tương đương của đoạn mạch: (R₁.R₂) : (R₁ + R₂) = (4.16) : (4 + 16) = 3,2 (\(\Omega\))

b. Cường độ dòng điện qua R₁ :

I₁ = U₁ : R₁ = 3,2 : 4 = 0,8 (A)

Do mạch mắc song song nên U = U₁ = U₂ = 3,2 (V)

c. Cường độ dòng điện qua R₂:

I₂ = U₂ : R₂ = 3,2 : 16 = 0,2 (A)

\(1.R1//R2\Rightarrow\left\{{}\begin{matrix}Rtd=\dfrac{R1R2}{R1+R2}=3,2A\\I1=\dfrac{U1}{R1}=0,8A\\I2=I-I1=1-0,8=0,2A\end{matrix}\right.\)

\(2.\Rightarrow\left\{{}\begin{matrix}R=\dfrac{pl}{S}=\dfrac{0,5.10^{-6}.100}{10^{-6}}=50\Omega\\I=\dfrac{U}{R}=\dfrac{5,8}{50}=0,116A\end{matrix}\right.\)

\(3.\Rightarrow l=\dfrac{RS}{p}=\dfrac{330.2,5.10^{-6}}{2,8.10^{-8}}\approx29464m\)

\(4.\Rightarrow p=\dfrac{RS}{l}=\dfrac{1,7.10^{-6}}{100}=1,7.10^{-8}\Rightarrow l:Cu\)

\(5.\Rightarrow\left\{{}\begin{matrix}S=\dfrac{pl}{R}=1.10^{-7}\left(m^2\right)\\U=IR=5V\\n=\dfrac{L}{c}=\dfrac{2,5}{d.\pi}=\dfrac{2,5}{\dfrac{2}{100}.\pi}=39\left(vong\right)\end{matrix}\right.\)

a,\(\Rightarrow I1=Im-I2=1,2-0,4=0,8A\)

b,\(\Rightarrow U1=I1R1=4,8V\)

c,\(\Rightarrow R2=\dfrac{U1}{I2}=12\Omega\)

d,\(\Rightarrow Rtd=\dfrac{R1R2}{R1+R2}=4\Omega\)

a, CĐDĐ qua R₁ :

Ta có: \(I=I_1+I_2\Leftrightarrow I_1=I-I_2=1,2-0,4=0,8\left(A\right)\)

b, HĐT giữa 2 đầu R₁:

Ta có: \(I_1=\dfrac{U_1}{R_1}\Leftrightarrow U_1=I_1.R_1=0,8.6=4,8\left(V\right)\)

c, Điện trở R₂:

   \(R_2=\dfrac{U_2}{I_2}=\dfrac{U_1}{I_2}=\dfrac{4,8}{0,4}=12\left(\Omega\right)\)

d, Điện trở tđ của mạch:

    \(R_{tđ}=\dfrac{R_1R_2}{R_1+R_2}=\dfrac{6.12}{6+12}=4\left(\Omega\right)\)

a.ta có R1//R2 ⇒I=I1+I2⇒I1=I-I2=1,2-0.4=0,8A
b.U1=I1.R1=0,8.6=4,8V
c.Ta có U=U1=U2=4,8V
R2=\(\dfrac{U2}{I2}=\dfrac{4,8}{0,4}=12\)Ω
d.R tương đương=\(\dfrac{R1.R2}{R1+R2}=\dfrac{6.12}{6+12}=4\)Ω

a) \(R_{tđ}=\dfrac{U}{I}=\dfrac{30}{1,2}=25\left(\Omega\right)\)

b) \(\dfrac{1}{R_{tđ}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}\)

\(\Rightarrow R_2=\dfrac{1}{\dfrac{1}{R_{tđ}}-\dfrac{1}{R_1}}=\dfrac{1}{\dfrac{1}{25}-\dfrac{1}{60}}=\dfrac{300}{7}\left(\Omega\right)\)

c) Do mắc song song nên \(U=U_1=U_2=30V\)

\(I_1=\dfrac{U_1}{R_1}=\dfrac{30}{60}=0,5\left(A\right)\)

\(I_2=\dfrac{U_2}{R_2}=\dfrac{30}{\dfrac{300}{7}}=0,7\left(A\right)\)

a/ $R_{td}=\dfrac{1}{\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}}=\dfrac{1}{\dfrac{1}{30}+\dfrac{1}{15}+\dfrac{1}{25}}=\dfrac{50}{7}\Omega$

b/ $U_3=I_3R_3=0,75.25=18,75V$

$U=U_3=18,75V$

$I=\dfrac{U}{R_{td}}=\dfrac{18,75}{\dfrac{50}{7}}=2,625A$

c/ $U_2=U_3=18,75V$

$I_2=\dfrac{U_2}{R_2}=\dfrac{18,75}{15}=1,25A$