Đáp án:
Giải:
a) Sđmđ: [(R1 nt R3) // R2] nt R4
`R_{13}=45+15=60 \ (\Omega)`
`U_{AD}=60.0,9=54 \ (V)`
`I_2=\frac{54}{90}=0,6 \ (A)`
`I_4=I_a+I_2=0,9+0,6=1,5 \ (A)`
`U_4=1,5.24=36 \ (V)`
`U_{AB}=54+36=90 \ (V)`
b) K mở:
`R_{AD}=\frac{60.90}{60+90}=36 \ (\Omega)`
`R_{td}=36+R_4`
`I=\frac{90}{36+R_4}`
`U_{AD}=IR_{AD}=\frac{3240}{36+R_4}`
`I_a=\frac{U_{AD}}{R_{13}}=\frac{54}{36+R_4}`
K đóng:
Sđmđ:
R1 // [R2 nt (R3 // R4)]
`R_{34}=\frac{15R_4}{15+R_4}`
`R_{234}=90+\frac{15R_4}{15+R_4}=\frac{1350+105R_4}{15+R_4}`
`I_{234}=\frac{90.(15+R_4)}{1350+105R_4}=\frac{1350+90R_4}{1350+105R_4}`
`U_{34}=\frac{20250R_4+1350R_4^2}{20250+2925R_4+105R_4^2}`
`I_a=\frac{1350R_4+90R_4^2}{20250+2925R_4+105R_4^2}`
→ `\frac{54}{36+R_4}=\frac{1350R_4+90R_4^2}{20250+2925R_4+105R_4^2}`
→ `1093500+157950R_4+5670R_4^2=48600R_4+4590R_4^2+90R_4^3`
→ `R_4=45 \ (\Omega)`
c) `I_a=\frac{54}{36+R_4}=0,(6) \ (A)`
`R_{234}=\frac{1350+105R_4}{15+R_4}=101,25 \ (\Omega)`
`R_{td}=\frac{R_1R_{234}}{R_1+R_{234}}=\frac{405}{13} \ (\Omega)`
$I=\dfrac{90}{\dfrac{405}{13}}=2,(8) \ (A)$
`U_{34}=\frac{20250R_4+1350R_4^2}{20250+2925R_4+105R_4^2}=10 \ (V)`
→ `I_4=\frac{10}{45}=0,(2) \ (A)`
Cường độ dòng điện qua khóa K:
`I-I_4=2,(8)-0,(2)=2,(6) \ (A)`
