Inductor as a battery in a circuit

[Pushoam]

Homework Statement

Can we consider an inductor as a battery of emf ξ = -$L \frac {dI } { dt}$?

For the above circuit,
$V_A - V_B = V_C -V_D \\ -L \frac {dI } { dt} = IR$
Here, the inductor is acting as a battery of emf $-L \frac {dI } { dt}$.

For the above circuit,
$V_A - V_B = V_C -V_D \\ -L \frac {dI } { dt} = \frac{-Q} C$ which is wrong as LHS is positive and RHS is negative.
So, here, the inductor should act as a battery with emf $L \frac {dI } { dt}$. Right?

 

[mfb]

For the above circuit,
$V_A - V_B = V_C -V_D \\ -L \frac {dI } { dt} = \frac{-Q} C$ which is wrong as LHS is positive and RHS is negative.

Why? dI/dt can be positive or Q can be negative.

I don't think it is useful to call an inductance a battery. It can have a voltage, but unlike for a battery this voltage depends on the time-dependence of the current.

 

[berkeman]

Here, the inductor is acting as a battery of emf −LdIdt -L \frac {dI } { dt} .

Not for long...

Quiz Question -- For how long?

 

[cnh1995]

For the above circuit,
VA−VB=VC−VD−LdIdt=−QCVA−VB=VC−VD−LdIdt=−QCV_A - V_B = V_C -V_D \\ -L \frac {dI } { dt} = \frac{-Q} C which is wrong as LHS is positive and RHS is negative.
So, here, the inductor should act as a battery with emf LdIdtLdIdt L \frac {dI } { dt} . Right?

Is there any initial charge on the capacitor?
Is the current increasing or decreasing?
You should consider these things while writing the KVL equation.

In your circuit, it appears to me that the capacitor is discharging through the inductor, so the inductor is storing energy. So it's the capacitor that is acting as a voltage source here. But again, it's not for long.

 

[Pushoam]

Quiz Question -- For how long?

$V_A - V_B = V_C -V_D \\ -L \frac {dI } { dt} = IR$
$I = I_0 e^{- \frac R L t}$
So, it works as a battery( with decreasing potential difference) for always in principle.

From other posts, what I conclude is:
Whether an inductor acts as a battery of emf - $L \frac {dI } { dt}$ depends on the circuit.
I cannot always replace an inductor by a battery of emf - $L \frac {dI } { dt}$ . Right?

?

 

[cnh1995]

$V_A - V_B = V_C -V_D \\ -L \frac {dI } { dt} = IR$
$I = I_0 e^{- \frac R L t}$
So, it works as a battery( with decreasing potential difference) for always in principle.

From other posts, what I conclude is:
Whether an inductor acts as a battery of emf - $L \frac {dI } { dt}$ depends on the circuit.
I cannot always replace an inductor by a battery of emf - $L \frac {dI } { dt}$ . Right?

You can always replace an inductor with a battery of emf Ldi/dt. There is no minus sign.
That sign will depend on the polarity of the battery and it will show up in the KVL equation for that particular loop.
The polarity of that battery will depend on time response of other circuit components.

 

[Pushoam]

You can always replace an inductor with a battery of emf Ldi/dt. There is no minus sign.
That sign will depend on the polarity of the battery and it will show up in the KVL equation for that particular loop.
The polarity of that battery will depend on time response of other circuit components.

Thanks for it.
I was looking for this information.

 

[cnh1995]

Thanks for it.
I was looking for this information.

The general form of KVL for a particular loop is,

Driving voltage (independent)= sum of the voltage drops across all components in that loop.

When you have an RL circuit excited by a dc source of voltage V, the above equation becomes
V=Ldi/dt+iR.

When you have a source-free (i.e. no independent source) circuit like the ones in your OP, driving voltage in the above equation becomes zero.
So, for RL circuit (first one in your OP),
0=Ldi/dt+iR
Or,
-Ldi/dt=iR.
The inductor is now "supplying" energy.

For the second diagram,
0=Ldi/dt+q/C
Or,
-Ldi/dt=q/c.

If you solve this equation, you'll see that the energy oscillates between L and C and the current is sinusoidal.

 

[Pushoam]

In the above circuit,
KVL says,
$V_A - V_B + V_B - V_C + V_C - V_A = 0$
Since, current always flows from higher potential to lower potential, $V _ A -V_B$ has to be positive.
Since, the current is increasing as time passes, $\frac {dI } {dt}$ is positive.
Hence , $V _ A -V_B = L \frac {dI } {dt}$ .
Is the reasoning correct here?

 

[cnh1995]

Is the reasoning correct here?

Yes.

Since, current always flows from higher potential to lower potential, VA−VBVA−VB V _ A -V_B has to be positive.
Since, the current is increasing as time passes, dIdtdIdt\frac {dI } {dt} is positive.

This analysis is for your own intuition. But you can't always analyse like this, especially when you have an RLC circuit excited with a dc source. That transient can be one of four different types. Better write the KVL equation, solve it and see what it gives. You don't have to know anything else.

 

[Pushoam]

Better write the KVL equation, solve it and see what it gives. You don't have to know anything else.

But, KVL eqn. doesn't tell us whether $V_A - V_B = L \frac {dI } {dt} , ~ or~ V_A - V_B = - L \frac {dI } {dt}$.
Without knowing this, how can one solve the eqn.?

 

[cnh1995]

But, KVL eqn. doesn't tell us whether $V_A - V_B = L \frac {dI } {dt} , ~ or~ V_A - V_B = - L \frac {dI } {dt}$.
Without knowing this, how can one solve the eqn.?

Use the general form of KVL.

Driving voltage (independent)= sum of the voltage drops across all components in that loop.

V=Ldi/dt+iR.
You don't have to worry about the polarity. Treat the voltages across all the components in the loop as voltage drops. The math will take care of the rest.

The solution of the DE will be complete once you put the initial condition(or boundary condition) into the solution. Initially, the current is zero. Put that into the solution and you'll get the complete time response.

 

[berkeman]

I=I0e−RLt I = I_0 e^{- \frac R L t}
So, it works as a battery( with decreasing potential difference) for always in principle.

Well, not really forever. As you show, it has a decay with a time constant of L/R. That's the key point I was making. Good that you worked it out!