Tóm tắt :
\(U_1=5V\)
\(I_1=100mA\)
\(U_2=U_1+20\%U_1\)
\(I_2-I_1=?\)
GIẢI :
Ta có : \(U_2=U_1+20\%U_1=5+\dfrac{20.5}{100}=6\left(mA\right)\)
\(\dfrac{U_1}{I_1}=\dfrac{U_2}{I_2}\Rightarrow I_2=\dfrac{U_2.I_1}{U_1}=\dfrac{6.100}{5}=120\left(mA\right)\)
Vậy : \(I_2-I_1=120-100=20\left(mA\right)\)
Bài giải :
Ta có : \(U_2=U_1+20\%U_1=5+\dfrac{20,5}{100}=6\) mA
\(\dfrac{U_1}{I_1}=\dfrac{U_2}{I_2}\Rightarrow I_2=\dfrac{U_2.I_1}{U_1}=\dfrac{6.100}{5}=120\) mA
Vậy \(I_2-I_1=120-100=20\)mA