Bài 1:
$R_{2}=\dfrac{1}{2}.R_{1}$
$R_{3}=\dfrac{1}{3}.R_{1}$
Ta có: $\dfrac{1}{R}=\dfrac{1}{R_{1}}+\dfrac{1}{R_{2}}+\dfrac{1}{R_{3}}$
$=\dfrac{1}{R_{1}}+\dfrac{2}{R_{1}}+\dfrac{3}{R_{1}}=\dfrac{6}{R_{1}}$
$⇒R=\dfrac{R_{1}}{6}$
Ta có: $R=\dfrac{U}{I}$
$⇔\dfrac{R_{1}}{6}=\dfrac{48}{1,6}$
$⇒R_{1}=180$ ôm
$⇒R_{2}=90$ ôm
$R_{3}=60$ ôm
Bài 2:
a, $I=\dfrac{U}{R}=\dfrac{36}{100}=0,36A$
b,
$R'=\dfrac{U}{I'}=\dfrac{36}{1,5}=24$ ôm
Th1: Phần bị cắt bỏ có điện trở $100-24=76$ ôm
Th2: $R'=\dfrac{R_{1}.R_{2}}{R}=\dfrac{(R-R_{2}).R_{2}}{R}$
$⇔24=\dfrac{(100-R_{2}).R_{2}}{100}$
$⇔R_{2}=60$ ôm $⇒R_{1}=40$ ôm
hoặc $R_{2}=40$ ôm $⇒R_{1}=60$ ôm
Vì $R_{1}>R_{2}$
$⇒R_{1}=60$ ôm; $R_{2}=40$ ôm