Đáp án:
`a) \ n=2`
`b) \ U_{DE}=5,5V`
`c) \ U_v=3V`
Giải:
a) `U_{CE}=U_{AE}-U_{AC}=16,5-5,5=11 \ (V)`
→ `I=\frac{U_{CE}}{R_{CE}}=\frac{11}{2R}`
Ta có:
`\frac{U_{AC}}{R_v}+\frac{U_{AC}}{2R}=I`
→ `\frac{5,5}{nR}+\frac{5,5}{2R}=\frac{11}{2R}`
→ `\frac{5,5.2R+5,5nR}{nR.2R}=\frac{11}{2R}`
→ `\frac{11+5,5n}{2nR}=\frac{11}{2R}`
→ `\frac{11+5,5n}{n}=11`
→ `11+5,5n=11n`
→ `5,5n=11`
→ `n=2`
b) `U_{DE}=\frac{U_{CE}}{2}=\frac{11}{2}=5,5 \ (V)`
c) `R_{DE}=\frac{R_v.R}{R_v+R}=\frac{2R.R}{2R+R}=\frac{2R}{3}`
→ `I_{DE}=\frac{3U_{DE}}{2R}`
Ta có:
`U_{DE}=U_{AE}-U_{AD}`
`U_{DE}=16,5-I_{DE}.R_{AD}`
`U_{DE}=16,5-\frac{3U_{DE}}{2R}.3R`
`U_{DE}=16,5-\frac{9U_{DE}}{2}`
→ `2U_{DE}=33-9U_{DE}`
→ `U_{DE}=3 \ (V)`
→ `U_v=3V`
