Time constant of an inductor coil

[Jahnavi]

Homework Statement

Homework Equations

The Attempt at a Solution

I think I can solve this problem . But I am bit unsure about how current changes in the circuit and the coil when it is shorted .

Before the coil is shorted , there will be no induced EMF across the terminals of the coil .

When the coil is shorted , is the potential difference across the coil zero (because of the wire placed across it or is it equal to the induced EMF(because of the change in current ) ?

 

[Charles Link]

For this one, you don't need to know any electrical circuit theory. All you need is the definition of time constant $t_c$: For a function $I(t)$ that obeys $I(t)=I_o e^{-t/t_c}$, it is said to have time constant $t_c$.

 

[Jahnavi]

For this one, you don't need to know any electrical circuit theory.

But I am interested in the circuit theory

Could you clear my doubts .

 

[Charles Link]

With no voltage source in the circuit, it obeys $L \frac{dI}{dt}+IR=0$ where there must necessarily be some small resistance $R$. This differential equation has a solution of the form $I(t)=I_o e^{-R t/L }$ so that $t_c=\frac{L}{R}$.

 

[Jahnavi]

With no voltage source in the circuit, it obeys $L \frac{dI}{dt}+IR=0$ where there must necessarily be some small resistance $R$. This differential equation has a solution of the form $I(t)=I_o e^{-R t/L }$ so that $t_c=\frac{L}{R}$.

When the inductor is shorted , what happens to the battery ? I think it remains connected .

 

[Charles Link]

When the inductor is shorted , what happens to the battery ? I think it remains connected .

In a very simple circuit, you have a DC voltage $V_o$, a resistor $R_1$ and an inductor in series. The current in the inductor will be $I_o=\frac{V_o}{R_1}$. You then short circuit the inductor, so that the only resistance in the inductor part of the circuit will be a very small resistance $R$ which is mostly from the wires making up the inductor. The battery and resistor $R_1$ still have current $I_o=\frac{V_o}{R_1}$ and the current for this part of the circuit will also be in the short-circuit wire. (If I could draw a picture, it would be easier).

 

[Jahnavi]

This is the picture I have in my mind .

Assume R to be the internal resistance of the coil . Initially S1 is closed and a constant current E/R is flowing through the coil .

Then coil is short circuited by closing switch S2 while keeping S1 closed . Are you interpreting it similarly or are you opening S1 while closing S2 ?

I suppose the two interpretations of shorting the coil are different .

 

[Charles Link]

I have another (larger) resistor $R_1$ right next to the battery. That way when $S_2$ is closed , there is no need to open $S_1$ at the same time. And to be more exact $I_o=\frac{V_o}{R_1+R }$.$\\$ And what happens with the inductor is it acquires an EMF $\mathcal{E}=L \frac{dI}{dt}$ that tries to maintain the current in the inductor and the EMF attempts to keep the current from changing.

 

[Jahnavi]

OK . Thanks

After S2 is closed (keeping S1 closed) , what will be the current through the large resistor R1 (next to the battery) ?

 

[Charles Link]

The large resistor $R_1$ next to the battery will have current $I=\frac{V_o}{R_1}$. Actually, you could let $R_1$ and $R$ be whatever resistor values you want, and the current in the inductor circuit, after it is short-circuited will have $I(t)=I_o e^{-R t/L}$ where $I_o=\frac{V_o}{R_1+R}$.

 

[Jahnavi]

OK .

So the current through R1 remains constant even after the coil is shorted (after closing S2) ?

And the current through S2( middle branch) will be the sum of the currents through R1 and Coil/R . Right ?

Does that mean the current through the coil decays through the middle wire without affecting the branch containing the battery and R1 ?

Does that also mean that the induced EMF in the coil does not disturb the current through the battery and R1 ?

 

[Charles Link]

See also a couple items I added to the previous post. $\\$ There will be a change in the current through the battery because initially $I_b=\frac{V_o}{R_1+R}$ and after switch $S_2$ is closed, $I_b=\frac{V_o}{R_1}$. $\\$ And $S_2$ is assumed to be a short-circuit, so once $S_2$ is closed, the battery can not tell what the inductor is doing, i.e. it will not be affected by any EMF's that the inductor generates.

 

[Jahnavi]

OK .

After S2 is closed current through S2( middle branch) will be the sum of the currents through R1(constant value) and Coil/R(varying) . Right ?

The current flowing through S2 will be varying ?

Does that mean the current through the coil decays through the middle wire while the current through the battery remains constant with a value Vo/R1 ?

 

[Charles Link]

The current from the inductor will always go from right to left through $S_2$ in the way you drew it, and that current, (which is time varying, $I(t)=I_o e^{-Rt/L}$, and eventually drops to zero) is superimposed on a left to right DC current of $I=\frac{V_o}{R_1}$ from the battery portion of the circuit. Once the inductor current goes to near zero, the current in $S_2$ will be left to right and will be $I=\frac{V_o}{R_1}$. So that $I(t)_{S_2}=\frac{V_o}{R_1}-\frac{V_o}{R_1+R} e^{-Rt/L}$ (with left to right as positive). $\\$ And yes, you are correct=the current in the battery stays constant once the switch $S_2$ is closed. :)

 

[Jahnavi]

Wow !

Back to the original question I asked in the OP .

When the coil is shorted , is the potential difference across the coil zero (because of the wire placed across it or is it equal to the induced EMF(because of the change in current ) ?

I think potential difference across the terminals of the coil will be 0 .The entire induced EMF will drop across the internal resistance of the coil .

Correct ?

 

[Charles Link]

Wow !

Back to the original question I asked in the OP .
I think potential difference across the terminals of the coil will be 0 .The entire induced EMF will drop across the internal resistance of the coil .

Correct ?

If you add a resistor $R$ in the circuit, then you find there is a voltage generated by the inductor. Otherwise, (if $R$ is simply inside the inductor), if you put a voltmeter across $S_2$, (to measure the inductor voltage), you would measure zero, because the resistive voltage drop is internal to the inductor, just as you stated. Yes, you have that correct. :)

 

[Jahnavi]

I have another (larger) resistor $R_1$ right next to the battery. That way when $S_2$ is closed , there is no need to open $S_1$ at the same time.

Why is it that books invariably open S1 just after S2 is closed ? Every book does that .

Is it to protect the battery from getting damaged by a flow of large current?

And since you put a large resistor , this possibility was eliminated because the current was limited ?

 

[Jahnavi]

Thank you so much Charles !

You are quite amazing