A circuit of resistance R ohms and inductance L henries has a direct v

1.	A circuit of resistance R ohms and inductance L henries has a direct voltage of 230 V applied to it. 0.3 second second after switching on, the current in the circuit was found to be 5 A. After the current had reached its final steady value, the circuit was suddenly short-circuited. The current was again found to be 5 A at 0.3 second second after short-circuiting the coil. Find the value of R and L.
Answer» For growth ; 5 = I_m (1 −e ^−0.3/λ) ...(i) For decay; 5 = I_m e ^−0.3/λ Equating the two, we get, I_me ^−0.3/λ = (1 −e^{− 0.3/λ})I_m or 2e^−0.3/λ = 1 ∴ e ^−0.3/λ = 0.5 or λ = 0.4328 Putting this value in (i), we get, 5 = I_m = e ^0.3/0.4328or I_m = 5 e ^{+ 0.3/0.4328} = 5 × 2 = 10 A. Now, I_m = V/R ∴ 10 = 230/R or R = 230/10 = 23 Ω (approx.) As λ = L/R = 0.4328 ; L = 0.4328 × 23 = 9.95 H

A circuit of resistance R ohms and inductance L henries has a direct voltage of 230 V applied to it. 0.3 second second after switching on, the current in the circuit was found to be 5 A. After the current had reached its final steady value, the circuit was suddenly short-circuited. The current was again found to be 5 A at 0.3 second second after short-circuiting the coil. Find the value of R and L.

Answer»

For growth ; 5 = I_m (1 −e ^−0.3/λ) ...(i)

For decay; 5 = I_m e ^−0.3/λ

Equating the two, we get, I_me ^−0.3/λ = (1 −e^{− 0.3/λ})I_m

or 2e^−0.3/λ = 1

∴ e ^−0.3/λ = 0.5 or λ = 0.4328

Putting this value in (i), we get,

5 = I_m = e ^0.3/0.4328or I_m = 5 e ^{+ 0.3/0.4328} = 5 × 2 = 10 A.

Now, I_m = V/R

∴ 10 = 230/R or R = 230/10 = 23 Ω (approx.)

As λ = L/R = 0.4328 ; L = 0.4328 × 23 = 9.95 H

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