Physics, asked by mohammedbasim4636, 1 year ago

An LR circuit with a battery is connected at i = 0.

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Answered by neeraj5670
0
in given LCR series circuit is connected across a constant voltage source, (the battery) and a switch. Assume that the switch, S is open until it is closed at a time t = 0, and then remains permanently closed producing a “step response” type voltage input. The current, i begins to flow through the circuit but does not rise rapidly to its maximum value of Imax as determined by the ratio of V / R (Ohms Law).

This limiting factor is due to the presence of the self induced emf within the inductor as a result of the growth of magnetic flux, (Lenz’s Law). After a time the voltage source neutralizes the effect of the self induced emf, the current flow becomes constant and the induced current and field are reduced to zero.

We can use Kirchhoff’s Voltage Law, (KVL) to define the individual voltage drops that exist around the circuit and then hopefully use it to give us an expression for the flow of current.

Kirchhoff’s voltage law (KVL) gives us:

The voltage drop across the resistor, Ris I*R (Ohms Law).

The voltage drop across the inductor, Lis by now our familiar expression L(di/dt)

Then the final expression for the individual voltage drops around the LR series circuit..
 

We can see that the voltage drop across the resistor depends upon the current, i, while the voltage drop across the inductor depends upon the rate of change of the current, di/dt. When the current is equal to zero, ( i = 0 ) at time t = 0 the above expression, which is also a first order differential equation..

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