Question

Answer 1

bài 3:

a) Điện trở tương đương:

\(R_{Td}=\dfrac{R_1R_2}{R_1+R_2}=\dfrac{4.12}{4+12}=3\Omega\)

b) U=\(R_{td}.I=3.2=6\left(V\right)\)

c) ta có: U=U₁=U₂=6V

=> I₁=\(\dfrac{U_1}{R_1}=\dfrac{6}{4}=1,5\left(A\right)\)

=> I=I₁+I₂

=> I₂= I- I₁=\(2-1,5=0,5\left(A\right)\)

Answer 2

Bài 4:

\(R_{td}=\dfrac{R_1R_2}{R_1+R_2}=\dfrac{8.12}{8+12}=4,8\Omega\)
b) \(U=I.R_{td}=1,5.4,8=7,2\left(V\right)\)
c) U=U₁=U₂=7,2(V)

=> \(I_1=\dfrac{U_1}{R_1}=\dfrac{7,2}{8}=0,9\left(A\right)\)

I= I₁ +I₂ => I₂= I - I₁= 1,5-0,9=0,6(A)