$R_{1}=R_{3}=2$ ôm
$R_{2}=3$ ôm
$R_{4}=6$ ôm
$U_{AB}=5V$
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Sđmđ: $(R_{1}//R_{3})nt(R_{2}//R_{4})$
$R_{13}=\dfrac{R_{1}.R_{3}}{R_{1}+R_{3}}=\dfrac{2.2}{2+2}=1$ ôm
$R_{24}=\dfrac{R_{2}.R_{4}}{R_{2}+R_{4}}=\dfrac{3.6}{3+6}=2$ ôm
$⇒R_{tđ}=R_{13}+R_{24}=1+2=3$ ôm
$⇒I=\dfrac{U_{AB}}{R_{tđ}}=\dfrac{5}{3}A$
$⇒U_{1}=U_{3}=I.R_{13}=\dfrac{5}{3}.1=\dfrac{5}{3}V$
$U_{2}=U_{4}=I.R_{24}=\dfrac{5}{3}.2=\dfrac{10}{3}V$
Ta có: $I_{1}=\dfrac{U_{1}}{R_{1}}=\dfrac{\dfrac{5}{3}}{2}=\dfrac{5}{6}A$
$I_{3}=\dfrac{U_{3}}{R_{3}}=\dfrac{\dfrac{5}{3}}{2}=\dfrac{5}{6}A$
$I_{2}=\dfrac{U_{2}}{R_{2}}=\dfrac{\dfrac{10}{3}}{3}=\dfrac{10}{9}A$
$I_{4}=\dfrac{U_{4}}{R_{4}}=\dfrac{\dfrac{10}{3}}{6}=\dfrac{5}{9}A$
Vì $I_{2}>I_{1}$
$⇒I_{a}=I_{2}-I_{1}=\dfrac{10}{9}-\dfrac{5}{6}=\dfrac{5}{18}A$
