Question

Cho R1=6 ôm,R2=5ôm,vôn kế chỉ 4V,Uab=12V.Tính R3

Answer 1

Bài 5.

\(U_2=U_V=4V\Rightarrow I_2=\dfrac{U_2}{R_2}=\dfrac{4}{5}=0,8A\)

\(I_1=I_2=I_3=I_{mạch}=0,8A\)

\(U_1=I_1\cdot R_1=0,8\cdot6=4,8V\)

\(U_3=U_{AB}-U_1-U_2=12-4,8-4=3,2V\)

\(R_3=\dfrac{U_3}{I_3}=\dfrac{3,2}{0,8}=4\Omega\)

Answer 2

$R_{tđ} = R_1 + R_2 + R_3 = 6 + 5 + R_3 = 11 + R_3(Ω)$

$I = \dfrac{U}{R_{tđ}} = \dfrac{12}{11 + R_3}(A)$

mà:  $I_2 = \dfrac{U_2}{R_2} = \dfrac{4}{5} = 0,8(A)$
$\Rightarrow I = I_2$
$⇔ \dfrac{12}{11 + R_3} = 0,8$
$⇔ R_3 = 4 Ω$