A circuit of resistance 10 Ω and inductance 0.1 H in series has a direct voltage of 200 V suddenly applied to it. Find the voltage drop across the inductance at the instant of switching on and at 0.01 second. Find also the flux-linkages at these instants.
(i) Switching instant
At the instant of switching on, i = 0, so that iR = 0 hence all applied voltage must drop across the inductance only. Therefore, voltage drop across inductance = 200 V. Since at this instant i = 0, there are no flux-linkages of the coil.
(ii) When t = 0.01 second
As time passes, current grows so that the applied voltage is partly dropped across the resistance and partly across the coil. Let us first find iR drop for which purpose, we need the value of i at t = 0.01 second.
Now, time period of the circuit is λ = L/R = 0.1/10 = 0.01 second. Since the given time happens to be equal to time constant,
∴ i = (200/10) × 0.632 = 12.64 A; iR = 152.64 × 10 = 126.4 V
Drop across inductance = √(2002 – 126.42) = 155V
Now, L = NΦ/i or NΦ = Li
∴ Flux-linkages Li = 0.1 × 12.64 = 1.264 Wb-turns.