A constant voltage is applied to a series R-L circuit at t = 0 by clos

1.	A constant voltage is applied to a series R-L circuit at t = 0 by closing a switch. The voltage across L is 25V at t = 0 and drops to 5 V at t = 0.025 second. If L = 2H, what must be the value of R ?
Answer» At t = 0, i = 0, hence there is no iR drop and the applied voltage must equal the back e.m.f. in the coil. Hence, the voltage across L at t = 0 represents the applied voltage. At t = 0.025 second, voltage across L is 5 V, hence voltage across R = 25 − 5 = 20 V ∴ iR = 20 V − at t = 0.025 second. Now i = I_m(I − e^−t/λ) Here I_m = 25/R ampere, t = 0.025 second ∴ i = 25/R(1 - e^-0.025/λ) R x 25/R(1 - e^-0.0025/λ) = 20 or e^0.025/λ = 5 ∴ 0.025/λ = 2.3log₁₀5 = 1.6077 ∴ λ = 0.025/1.6077 Now λ = L/R = 2/R ∴ 2/R = 0.025/1.6077 ∴ R = 128.56 Ω

A constant voltage is applied to a series R-L circuit at t = 0 by closing a switch. The voltage across L is 25V at t = 0 and drops to 5 V at t = 0.025 second. If L = 2H, what must be the value of R ?

Answer»

At t = 0, i = 0, hence there is no iR drop and the applied voltage must equal the back e.m.f. in the coil.

Hence, the voltage across L at t = 0 represents the applied voltage.

At t = 0.025 second, voltage across L is 5 V, hence voltage across

R = 25 − 5 = 20 V

∴ iR = 20 V − at t = 0.025 second.

Now i = I_m(I − e^−t/λ)

Here I_m = 25/R ampere, t = 0.025 second

∴ i = 25/R(1 - e^-0.025/λ)

R x 25/R(1 - e^-0.0025/λ) = 20 or e^0.025/λ = 5

∴ 0.025/λ = 2.3log₁₀5 = 1.6077

∴ λ = 0.025/1.6077

Now λ = L/R = 2/R

∴ 2/R = 0.025/1.6077

∴ R = 128.56 Ω

Discussion

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