Question

What is the response of an inductor for DC voltage?

Answer 1

When a DC voltage is applied to an inductor, the inductor initially resists the change in current. This happens because an inductor opposes any change in current through it by generating a back electromotive force (EMF) according to Lenz's Law. However, once the current has settled, the inductor essentially "behaves like a wire," offering very little resistance to the DC current.

Here’s a breakdown of the response over time:

1. At the moment of application (t = 0):
   

   - When DC voltage is suddenly applied, the inductor resists the change in current due to its inductance.
   - Initially, the current through the inductor starts at zero and increases gradually. This happens because the inductor generates an opposing voltage (back EMF), which resists the rise in current.

1. After a short time (during the transient period):
   

   - The current increases at a rate determined by the inductor's inductance (L) and the applied voltage (V). This is governed by the equation:  
     \[ V = L \frac{dI}{dt} \]  
     where \( dI/dt \) is the rate of change of current.
   - The rate of current increase is not instant, but it gradually approaches a steady value.

1. After a long time (steady state):
   

   - Once the current has had time to build up and reach a steady value, the inductor behaves like a short circuit (a piece of wire with very little resistance) for DC.
   - The current through the inductor reaches its maximum steady-state value, determined by Ohm's Law:  
     \[ I = \frac{V}{R} \]  
     where \( R \) is the resistance of the circuit (if any).

Summary:

1. During the transient phase, the inductor resists changes in current.
   

1. At steady state, the inductor behaves like a wire, with little to no opposition to DC current flow. The only opposition to current is due to any resistance present in the circuit.
   

So, for a long time after a DC voltage is applied, the inductor essentially acts as a short circuit.