Bài làm:
R45 = R4 + R5 = 10 + R5 (Ω)
R453 = \(\frac{R_{45}.R_3}{R_{45}+R_3}\) = \(\frac{\left(10+R_5\right).10}{10+R_5+10}\) = \(\frac{100+10R_5}{20+R_5}\) (Ω)
R4532 = R2 + R453 = 10 + \(\frac{100+10R_5}{20+R_5}\) = \(\frac{10\left(20+R_5\right)+100+10R_5}{20+R_5}\) = \(\frac{300+20R_5}{20+R_5}\) (Ω)
R45321 = R1 + R4532 = 10 + \(\frac{300+20R_5}{20+R_5}\) = \(\frac{10\left(20+R_5\right)+300+20R_5}{20+R_5}\) = \(\frac{500+30R_5}{20+R_5}\) (Ω)
⇒ I = I1 = I2 = I453 = \(\frac{U}{R}\) = 63 : \(\frac{500+30R_5}{20+R_5}\) = \(\frac{63\left(20+R_5\right)}{500+30R_5}\) = \(\frac{1260+63R_5}{500+30R_5}\) (A)
⇒ U453 = U45 = U3 = I453.R453 = \(\frac{1260+63R_5}{500+30R_5}\).\(\frac{100+10R_5}{20+R_5}\) = \(\frac{63.\left(10+R_5\right)}{50+3R_5}\) (V)
Ta có: U2 = I2.R2 = \(\frac{1260+63R_5}{500+30R_5}\).10 = \(\frac{1260+63R_5}{50+3R_5}\) (V)
Theo đề bài ta có:
U23 = U2 + U3 (R2 nt R3)
⇒ 40,5 = \(\frac{1260+63R_5}{50+3R_5}\) + \(\frac{63.\left(10+R_5\right)}{50+3R_5}\)
⇒ 40,5 = \(\frac{63\left(20+R_5\right)+63\left(10+R_5\right)}{50+3R_5}\)
⇒ 40,5 = \(\frac{63\left(30+2R_5\right)}{50+3R_5}\)
⇔ 40,5.(50 + 3R5) = 63.(30 + 2R5)
⇔ 2025 + 121,5R5 = 1260 + 126R5
⇔ 765 = 4,5R5
⇒ R5 = 170 (Ω).
Vậy R5 = 170Ω.
