TL:

$R_{1}=3$ ôm

$R_{2}=2$ ôm

$R_{4}=1$ ôm

$R_{5}=4$ ôm

$U=18V$

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a, $K$ mở: Sđmđ: $R_{1}ntR_{3}ntR_{5}$

$I_{A}=1,8A$

$⇒I=I_{1}=I_{3}=I_{5}=1,8A$

$⇒U_{1}=I.R_{1}=1,8.3=5,4V$

$U_{5}=I.R_{5}=1,8.4=7,2V$

$⇒U_{3}=U-U_{5}-U_{1}=18-5,4-7,2=5,4V$

$⇒R_{3}=\dfrac{U_{3}}{I}=\dfrac{5,4}{1,8}=3$ ôm

b, $K$ đóng: Sđmđ: $[(R_{2}ntR_{4})//(R_{1}ntR_{3})]ntR_{5}$

$R_{24}=R_{2}+R_{4}=2+1=3$ ôm

$R_{13}=R_{1}+R_{3}=3+3=6$ ôm

$R_{2413}=\dfrac{R_{24}.R_{13}}{R_{24}+R_{13}}=\dfrac{3.6}{3+6}=2$ ôm

$R=R_{2413}+R_{5}=2+4=6$ ôm

$⇒I_{5}=I=\dfrac{U}{R}=\dfrac{18}{6}=3A$

$⇒U_{5}=I_{5}.R_{5}=3.4=12V$

$⇒U_{24}=U_{13}=U-U_{5}=18-12=6V$

$⇒I_{2}=I_{4}=\dfrac{U_{24}}{R_{24}}=\dfrac{6}{3}=2A$

$I_{1}=I_{3}=\dfrac{U_{13}}{R_{13}}=\dfrac{6}{6}=1A$

$U_{2}=I_{2}.R_{2}=2.2=4V$

$⇒U_{2}+U_{MC}=U$

$⇒U_{MC}=U-U_{2}=18-4=4V$

$U_{MN}=0V$