a, Sđmđ: $(R_{1}ntR_{3})//(R_{2}ntR_{4})$
$R_{13}=R_{1}+R_{3}=6+12=18$ ôm
$R_{24}=R_{2}+R_{4}=6+6=12$ ôm
$⇒I_{1}=I_{3}=\dfrac{U}{R_{13}}=\dfrac{12}{18}=\dfrac{2}{3}A$
$I_{2}=I_{4}=\dfrac{U}{R_{24}}=\dfrac{12}{12}=1A$
$U_{1}=I_{1}.R_{1}=\dfrac{2}{3}.6=4V$
$⇒U_{3}=U-U_{1}=12-4=8V$
$U_{2}=I_{2}.R_{2}=1.6=6V$
$⇒U_{4}=12-6=6V$
b, Ta thấy $U_{1}+U_{4}=4+6=10<U$
$⇒$ Mắc cực dương của vôn kế vào $M$
$U_{V}=U-10=12-10=2V$
c, $R_{A}=0$
$⇒U_{MN}=0$
Sđmđ: $(R_{1}//R_{2})nt(R_{3}//R_{4})$
$R_{12}=\dfrac{R_{1}.R_{2}}{R_{1}+R_{2}}=\dfrac{6.6}{6+6}=3$ ôm
$R_{34}=\dfrac{R_{3}.R_{4}}{R_{3}+R_{4}}=\dfrac{12.6}{12+6}=4$ ôm
$⇒R_{tđ}=R_{12}+R_{34}=3+4=7$ ôm
$⇒I_{12}=I_{34}=\dfrac{U}{R_{tđ}}=\dfrac{12}{7}A$
$⇒U_{12}=I_{12}.R_{12}=\dfrac{12}{7}.3=\dfrac{36}{7}V$
$⇒I_{1}=\dfrac{U_{12}}{R_{1}}=\dfrac{6}{7}A$
$U_{34}=U-U_{12}=12-\dfrac{36}{7}=\dfrac{48}{7}V$
$⇒I_{3}=\dfrac{U_{34}}{R_{3}}=\dfrac{4}{7}A$
Ta có: $I_{1}>I_{3}$
$⇒I_{A}$ chiều từ $M→N$
$⇒I_{A}=I_{1}-I_{3}=\dfrac{6}{7}-\dfrac{4}{7}=\dfrac{2}{7}A$
