Question

[Hoàng Anh Đào]

U_MN = 63V; R₁ = R₂ = R₃ = R₄ = 10Ω. Tìm R₅ để U₂₃ = 40,5V.

Phân tích mạch điện: [(R₄ nt R₅) // R₃] nt R₂ nt R₁

Đang cần gấp, giúp mình với nha! ♥♥

Answer 1

[Netflix]

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.