
#1
May313, 08:51 AM

P: 11

I fully understand mathematically using differential equations and also using conservation of energy why current should decay in an RL circuit.
However , I cannot comprehend how this phenomenon can be explained purely based on induced voltages and currents. An intuitive understanding basically. Can anyone help me out? 



#2
May313, 06:56 PM

Mentor
P: 11,989

I'll try to provide a conceptual explanation, based on induced voltages.
If there is a nonzero current, then by Ohm's Law there is a nonzero voltage (due to the resistor). Since there is a nonzero voltage, then by Faraday's law of induction the magnetic flux through the inductor must be changing. The magnetic flux through the inductor is produced by the current. So a changing flux goes handinhand with a changing current. To figure out whether the current change causes an increase or decrease, use Lenz's Law and one of the righthandrules  specifically, the one that tells you the direction of B for a given direction of current through the inductor loops. Hope that helps. 



#3
May413, 12:52 AM

P: 102





#4
May413, 01:16 PM

P: 114

Why does current decay in an RL circuit?
We start with a certain amount of magnetic energy stored in the L. This is associated with a certain current flowing.
So long as there is no resistance in the circuit then the current will flow forever because no energy is discharged. This is a superconductor situation. Put a low resistance into the circuit and with the given current as a starting value then energy will discharge slowly into the resistance and we will see the current decay as the energy is transferred from the magnetic field of the L to heat in the resistor. Put a high resistance into the circuit and with the given starting current there will be higher loss rates in the resistor. There is still the same amount of energy in the L to start with so because the higher resistive load consumes it faster the current decays faster. 



#5
May413, 03:56 PM

HW Helper
P: 3,339

an intuitive explanation: There is some initial current. What the system wants to do, is to dissipate all the energy straight away in the resistor (bringing the current swiftly to zero). But the inductor doesn't like a sudden change in the current. So the inductor only lets the current go down gradually (as an exponential).
And the typical timescale for the decay is given by the inductance, divided by resistance. So in the limit of very small inductance, we get what we expect  the current goes to zero very rapidly, so that we would hardly see the curve, it would look almost like a step function. (Well, the exponential tail never really disappears, but you know what I mean, it looks more like a step function, for systems with smaller inductance). Edit: well, the shape of the decay function is always the same. But the timescale depends on the inductance, so if you had a circuit with inductance much smaller than resistance, then you would need to have a better 'time resolution' to be able to see the curve before it gets very close to zero. 


Register to reply 
Related Discussions  
Question about current decay in RL circuit  General Physics  6  
Current Decay in an RL Circuit  Introductory Physics Homework  1  
Solving a Simple Circuit involving Current Source by the Loop Current Method..  Introductory Physics Homework  2  
current decay in closed circuit with only 1 resistor  Introductory Physics Homework  2  
finding time constant of current decay in closed circuit  Advanced Physics Homework  1 