Question

The two circuits of Fig. have the same time constant of 0.005 second. With the same d.c. voltage applied to the two circuits, it is found that the steady state current of circuit (a) is 2000 times the initial current of circuit (b). Find R₁, L₁ and C.

Answer 1

The time constant of circuit  is 

λ = L₁/R₁ second, and that of circuit is 

λ = CR₂ second. 

∴ L₁/R₁ = 0.005 C× 2 × 10⁶ = 0.005, C = 0.0025 × 10⁻⁶ = 0.0025μF

Steady-state current of circuit  is = V/R₁ = 10/R₁ amperes. 

Initial current of circuit = V/R₂ = 10/2 × 10⁶ = 5 × 10⁻⁶ A* 

Now 10/R₁ = 2000 × 5 × 10⁻⁶ 

∴ R₁ = 1000 Ω. 

Also L₁/R₁ = 0.005 

∴ L₁ = 1000 × 0.005 = 5H