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Sagot :
Answer: hi your question is incomplete attached below is the complete question
1) attached below
2) a) 31 Ω, 302.9 Ω
b) 17.3 Ω , 26.4 Ω
Explanation:
1) Deriving Eqn 5-6 from Eqn 5-5
attached below
2) using τ’s calculated and measured resistance
use given data
Tm1 = Rm1 * Cm1
= 329.3 * 333 * 10^-9 = 109.65 μs
Tm2 = Rm2 * Cm2
= 329.3 * 200 * 10^-9 = 658.6 μs
a) Capacitance of capacitors
For Cm1
t₁₂ = 72 μs = R₁' Cm1 log²
∴ R₁' = ( 72 * 10^-6 ) / ( 333 * 10^-9 log² )
= 31 Ω
For Cm2
t₁₂ = 40 μs , ∴ R₂' = 302.9 Ω
b) comparing calculated and measured values via percent error
errors
Rm1 - R₁' = 17.3 Ω
Rm2 - R₂' = 26.4 Ω


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