a, $K$ mở:
Sđmđ: $(R_{1}ntR_{3})//(R_{2}ntR_{4})$
$R_{13}=R_{1}+R_{3}=4+x$
$R_{24}=R_{2}+R_{4}=6+6=12$ ôm
Ta có: $I_{1}=\dfrac{U}{R_{13}}=\dfrac{U}{4+x}$
$I_{2}=\dfrac{U}{R_{24}}=\dfrac{U}{12}$
Ta có: $I_{1}=2I_{2}$
$⇔\dfrac{U}{4+x}=2.\dfrac{U}{12}$
$⇔4+x=6$
$⇔x=2$ ôm
$⇒R_{3}=2$ ôm
$I_{AB}=\dfrac{3U}{12}=\dfrac{U}{4}$
b, $K$ đóng:
Sđmđ: $(R_{1}//R_{2})nt(R_{3}//R_{4})$
$R_{12}=\dfrac{R_{1}.R_{2}}{R_{1}+R_{2}}=\dfrac{4.6}{4+6}=2,4$ ôm
$R_{34}=\dfrac{R_{3}.R_{4}}{R_{3}+R_{4}}=\dfrac{2.6}{2+6}=1,5$ ôm
$⇒R_{tđ}=R_{12}+R_{34}=2,4+1,5=3,9$ ôm
$⇒I_{12}=I_{34}=\dfrac{U}{3,9}$
$⇒U_{12}=I_{12}.R_{12}=\dfrac{8U}{13}V$
$⇒I_{1}=\dfrac{U_{12}}{R_{1}}=\dfrac{8U}{52}$
$⇒I_{2}=I_{12}-I_{1}=\dfrac{U}{3,9}-\dfrac{8U}{52}=\dfrac{4U}{39}$
$U_{34}=U-U_{12}=U-\dfrac{8U}{13}=\dfrac{5U}{13}V$
$⇒I_{3}=\dfrac{U_{34}}{R_{3}}=\dfrac{5U}{26}$
$⇒I_{4}=I_{12}-I_{3}=\dfrac{U}{3,9}-\dfrac{5U}{26}=\dfrac{5U}{78}$
Vì $I_{1}>I_{4}$
$⇒I_{CD}=I_{1}-I_{4}=\dfrac{8U}{52}-\dfrac{5U}{78}=\dfrac{7U}{78}$
