Question

Answer 1

a, Khi mở máy thì

\(U=E=7,5W\) 

Ta có

\(I=\dfrac{E}{R+r}\Leftrightarrow1=\dfrac{7,5}{3+2+r}\Rightarrow r=2,5\) 

b, 

\(I=\dfrac{U}{R}=\dfrac{5}{3+2}=1=I_đ=I_R\) 

mà I_đm \(=\dfrac{P}{U}=\dfrac{3}{3}=1\) 

\(\Rightarrow\) đèn sáng bình thường

c, Khi K mở

\(q_1=0\\ q_2=UC_2=7,5.0,2.10^{-6}=1,5.10^{-6}\\ \Rightarrow\Delta q=\left|q_2-q_1\right|=1,5.10^{-6}\) 

Khi K đóng

\(q_1=UC_1=5.0,3.10^{-6}=1,5.10^{-6}\\ q_2=UC_2=IRC_2=1,2.0,2.10^{-6}=0,4.10^{-6}\\ \Rightarrow\Delta q=1,1.10^{-6}\\ n=\dfrac{\left|\Delta q'-\Delta q\right|}{\left|e\right|}=2,5.10^{12}\)

Answer 2

B/sung ý a)