Is Back EMF Always Equal to Battery EMF in Inductive Circuits?

[harsh95]

Homework Statement

A coil of inductance 0.5 H is connected to a 18V battery. Calculate the rate of growth of current?

Homework Equations

E=-L(dI/dt)

The Attempt at a Solution

Actually I have the solution but the problem is that in the book they have assumed that the back emf that would develop due to increasing current would be the same as emf of the battery? i.e its written as dI/dt=E/L=18/0.5= 36 A/s
How is it possible? Or am I thinking wrong

 

[gneill]

Homework Statement

A coil of inductance 0.5 H is connected to a 18V battery. Calculate the rate of growth of current?

Homework Equations

E=-L(dI/dt)

The Attempt at a Solution

Actually I have the solution but the problem is that in the book they have assumed that the back emf that would develop due to increasing current would be the same as emf of the battery? i.e its written as dI/dt=E/L=18/0.5= 36 A/s
How is it possible? Or am I thinking wrong

The potential across the coil cannot be anything else but the same as that of the battery, assuming an ideal coil and battery. An ideal voltage source will produce ANY amount of current required to maintain its constant potential difference.

 

[harsh95]

But I don't understand why the back emf generated by 18V i.e same as the battery.
We say that inductur has no resistance of itself. Then there is no potential drop when the current is steady
And when the current is changing then EMF is induced which is opposes the cause that produces it . And in the formula E=-LdI/dt, the E is the back emf right? And not the battery potential?

 

[gneill]

But I don't understand why the back emf generated by 18V i.e same as the battery.
We say that inductur has no resistance of itself. Then there is no potential drop when the current is steady
And when the current is changing then EMF is induced which is opposes the cause that produces it . And in the formula E=-LdI/dt, the E is the back emf right? And not the battery potential?

The current will definitely NOT be steady. There will be no steady state for this circuit since there's no resistance to limit the current.

Write the KVL loop equation for the circuit. Solve the resulting differential equation for the current. You'll see that the current increases without bound over time, but that dI/dt is a contant. Thus the back-emf is a constant, and it happens to equal the battery voltage.

 

[Nickie]

?I would say that the assumption in the book is not necessarily correct. The back emf in an inductor is proportional to the rate of change of current, as given by the equation E=-L(dI/dt). However, this does not mean that the back emf will always be equal to the emf of the battery.

In this scenario, the back emf will be equal to the battery emf only when the current is increasing at a constant rate. If the current is increasing at a non-constant rate, the back emf will not be equal to the battery emf.

It is important to note that the back emf is a result of the inductor's self-inductance and the rate of change of current, while the battery emf is a constant value provided by the battery. Therefore, it is possible for the back emf to be different from the battery emf.

In conclusion, the assumption in the book may not always hold true and it is important to consider the specific conditions of the circuit when calculating the back emf and the rate of change of current.