\(R=\dfrac{U_1}{I_1}=\dfrac{12}{1,2}=10\left(\Omega\right)\)

\(I_2=I_1+0,3=1,2+0,3=1,5\left(A\right)\)

\(U_2=I_2.R=1,5.10=15\left(V\right)\)

Vậy khi cđdđ qua bóng tăng 0,3 A thì hđt tăng 3 V

Ta có R không đổi:

\(R=\dfrac{U_1}{I_1}=\dfrac{U_2}{I_2}\Leftrightarrow\dfrac{7,2}{I_1}=\dfrac{7,2+10,8}{I_2}\Leftrightarrow7,2.I_2=18.I_1\Rightarrow I_1=0,4.I_2\)

Ta coi R không đổi

\(R=\dfrac{U}{I}\)

=> \(\dfrac{U_1}{I_1}=\dfrac{U_2}{I_2}\Leftrightarrow\dfrac{12}{6.10^{-3}}=\dfrac{U_2}{6.10^{-3}-4.10^{-3}}\Rightarrow U_2=4\left(V\right)\)

\(t_1=\dfrac{s_1}{v_1}=\dfrac{50}{3}\left(s\right)\)

\(v_{tb}=\dfrac{s_1+s_2}{t_1+t_2}=\dfrac{50+80}{\dfrac{50}{3}+20}\approx3,54\left(\dfrac{m}{s}\right)\)