Calculating low frequency cutoff for this R-C microphone circuit

[songoku]

Homework Statement

Please see below

Relevant Equations

Q = CV

time constant = RC

I want to ask about question (c). My idea is to compare the period and time constant. The period is 0.05 s and time constant is 0.005 s.

Time constant is the time needed for capacitor to discharge until the charge stored in it becomes 37% of initial charge. But I don't know how to relate the period and time constant to answer the question.

Thanks

 

[kuruman]

But I don't know how to relate the period and time constant to answer the question.

You have already related the two. You know that the period is 10 times the time constant. Look at the drawing. It shows sound wave signal in - electrical signal out. Then consider what a microphone is designed to do. Part (b) should guide your thinking.

 

[songoku]

You have already related the two. You know that the period is 10 times the time constant. Look at the drawing. It shows sound wave signal in - electrical signal out. Then consider what a microphone is designed to do. Part (b) should guide your thinking.

Microphone is designed to convert sound energy into electrical energy. When the plates of capacitor move closer, the capacitance increases so the charge increases and the p.d across it decreases.

But sorry I still don't know what is the reason why the period being 10 times the time constant making the microphone suitable to detect low frequency sound.

Thanks

 

[kuruman]

Microphone is designed to convert sound energy into electrical energy.

There is more to it than just energy. Suppose a microphone records a conversation between your mother and your father, both speaking equally loud. How can you tell them apart if energy is all that reaches your ears?

 

[songoku]

There is more to it than just energy. Suppose a microphone records a conversation between your mother and your father, both speaking equally loud. How can you tell them apart if energy is all that reaches your ears?

The frequency of their voice is different?

 

[kuruman]

Yes. So what is a microphone designed to do?

 

[songoku]

Yes. So what is a microphone designed to do?

To detect sound with different frequency

 

[kuruman]

That's only half of the story. What's the other half? Look at the drawing again. You see sound of different frequency come in on the left. What else do you see?

 

[songoku]

That's only half of the story. What's the other half? Look at the drawing again. You see sound of different frequency come in on the left. What else do you see?

The sound wave hits one plate of capacitor, causing it to vibrate then change the p.d on resistor so the output signal will be AC

 

[kuruman]

And what should the frequency of this AC signal be in an ideal situation?

 

[songoku]

And what should the frequency of this AC signal be in an ideal situation?

For this question, I think it should be 20 Hz

 

[kuruman]

Are you saying that the frequencies that vibrate the diaphragm are all converted to 20 Hz? The 20 Hz is just a lower limit for audio frequencies. What did you give for an answer in part (b) to "Explain how the sound wave produces an alternating output signal"?

 

[songoku]

Are you saying that the frequencies that vibrate the diaphragm are all converted to 20 Hz? The 20 Hz is just a lower limit for audio frequencies.

I thought since the microphone wanted to detect 20 Hz then the sound wave hitting the plate was at frequency of 20 Hz.

For normal audio sound range, it should be 20 Hz - 20 kHz.

What did you give for an answer in part (b) to "Explain how the sound wave produces an alternating output signal"?

When the plates of capacitor move closer, the capacitance increases so the charge increases and the p.d across it decreases. This will cause the p.d. across R increases.

When the plates of capacitor move further away, the capacitance decreases so the charge decreases and the p.d across it increases. This will cause the p.d. across R decreases.

Since the p.d. across R keeps changing, the output signal will be alternating

 

[kuruman]

Since the p.d. across R keeps changing, the output signal will be alternating

At what frequency? How do you know that the charge decreases and is not (nearly) constant?

 

[songoku]

At what frequency?

I am not sure but I think it would be in the range of audio sound 10 Hz - 20 kHz.

How do you know that the charge decreases and is not (nearly) constant?

Because if the capacitance increases, I think the ability of the capacitor to store charge increases so more charge would flow in the circuit

 

[kuruman]

I am not sure but I think it would be in the range of audio sound 10 Hz - 20 kHz.

Yes, you want the output voltage be in the audible range unattenuated. The job of a microphone is to convert air vibration frequencies to voltage frequencies.

Because if the capacitance increases, I think the ability of the capacitor to store charge increases so more charge would flow in the circuit

Why do you say that the capacitance increases? Imagine the diaphragm vibrating at 1000 HZ. In one second the capacitance increases and decreases 2000 times per second. It gets more complicated if you have a superposition of frequencies instead of a pure tone.

Let's try another approach. If the input voltage is $V_{\text{in}}$ and the output voltage $V_{\text{out}}$, can you find an expression for the ratio $V_{\text{out}}/V_{\text{in}}$? It should be frequency-dependent. Plot the ratio as a function of frequency and see what you can say about it.

 

[songoku]

Why do you say that the capacitance increases? Imagine the diaphragm vibrating at 1000 HZ. In one second the capacitance increases and decreases 2000 times per second. It gets more complicated if you have a superposition of frequencies instead of a pure tone.

Because of the statement from the question: "The sound waves cause the flexible front plate to vibrate and change the capacitance. Moving the plates closer together increases the capacitance. Moving the plates further apart decreases the capacitance"

So I thought since the capacitance is changing, the charge and p.d also changing.

Let's try another approach. If the input voltage is $V_{\text{in}}$ and the output voltage $V_{\text{out}}$, can you find an expression for the ratio $V_{\text{out}}/V_{\text{in}}$? It should be frequency-dependent. Plot the ratio as a function of frequency and see what you can say about it.

By input voltage, do you mean the voltage of the battery?

Thanks

 

[kuruman]

So I thought since the capacitance is changing, the charge and p.d also changing.

Yes it is changing. You said "increasing" and I wanted to be sure that you understood that it is also decreasing.

By input voltage, do you mean the voltage of the battery?

I mean the voltage across the capacitor. Assume that it's an AC signal of some frequency $\omega.$

 

[songoku]

I mean the voltage across the capacitor. Assume that it's an AC signal of some frequency $\omega.$

This is what I can come up with (although I am pretty sure this is not what you want):

Let $V_{\text{in}}=A \sin (\omega t)$ and $V_{\text{battery}} = \varepsilon$

$$V_{\text{in}} + V_{\text{out}} = \varepsilon$$
$$V_{\text{out}} = \varepsilon - V_{\text{in}}$$
$$\frac{V_{\text{out}}}{V_{\text{in}}}=\frac{\varepsilon}{A \sin (\omega t)}-1$$

 

[kuruman]

You can't have $A$ on the right hand side because it is unknown. The battery keeps the capacitor charged and the only AC component comes from $V_{\text{in}}.$ Consider that as your AC source and do a proper analysis taking into account the impedance of the circuit.

 

[Steve4Physics]

Apologies for interrupting the flow of this conversation.

Using the values given, $\frac 1{RC}=$ 200Hz – a whole order of magnitude greater than 20Hz.

This tells us that the circuit is certainly suitable for frequencies greater than 200Hz. But the question asks why the circuit is suitable for frequencies of only "about... 20Hz” - this is potentially confusing.

The original question – worth a meagre 4 marks – might be expecting application of some basic knowledge of high-pass RC filters.

If the OP has covered high-pass RC filters they might want to revise the topic first.

The -3dB cut-off frequency is 32Hz, still above 20Hz. But, using this cut-off frequency (which requires prior knowledge of high-pass RC filters), it makes a bit more sense in the context of the question.

 

[songoku]

You can't have $A$ on the right hand side because it is unknown. The battery keeps the capacitor charged and the only AC component comes from $V_{\text{in}}.$ Consider that as your AC source and do a proper analysis taking into account the impedance of the circuit.

$$\frac{V_{\text{out}}}{V_{\text{in}}}=\frac{R}{X_C}$$
$$=R \omega C$$

Not sure how to draw the graph of $\frac{V_{\text{out}}}{V_{\text{in}}}$ against $\omega$ because C is not constant.

$C=\epsilon_o \frac{A}{d}$ and the value of $d$ is changing. Is it also frequency-dependent?

Using the values given, $\frac 1{RC}=$ 200Hz – a whole order of magnitude greater than 20Hz.

This tells us that the circuit is certainly suitable for frequencies greater than 200Hz.

Sorry I don't understand this part. Why the circuit is suitable for frequencies greater than 200 Hz?

The original question – worth a meagre 4 marks – might be expecting application of some basic knowledge of high-pass RC filters.

If the OP has covered high-pass RC filters they might want to revise the topic first.

The -3dB cut-off frequency is 32Hz, still above 20Hz. But, using this cut-off frequency (which requires prior knowledge of high-pass RC filters), it makes a bit more sense in the context of the question.

I don't study high-pass RC filters. This question is in chapter "Capacitor" and the things I learn in the chapter is about $C = \frac{Q}{V}$ , charging and discharging and equation of $x=x_o e^{-\frac{t}{\tau}}$

Thanks

 

[Steve4Physics]

Sorry I don't understand this part. Why the circuit is suitable for frequencies greater than 200 Hz?

It was using prior knowledge. I recognised the circuit to be a high-pass filter and knew that the cut-off frequency was of order $\frac1{RC}$ which is 200Hz.

I don't study high-pass RC filters. This question is in chapter "Capacitor" and the things I learn in the chapter is about $C = \frac{Q}{V}$ , charging and discharging and equation of $x=x_o e^{-\frac{t}{\tau}}$

It is extremely difficult (IMO) to answer the question with only this knowledge. As already noted by @kuruman, you could do an analysis based on impedance/reactance. I'll give you a (possibly too big) start....

$R$ and $C$ form a potential divider with $V_R$ and $V_C$ the amplitudes of the alternating voltages. We want $V_R$ to be large, as this is the output voltage.

We apply sound of frequency $f$. This results in an alternating current of the same frequency in the circuit.

a) If $V_R > V_C$ we have a 'strong' output.
b) If $V_R < V_C$ we have a 'weak' output.
c) If $V_R=V_C$, this is the ‘cut-off’ condition.

Since we have a potential divider, to meet condition a) we require $R$ to be greater than than $X_C$. I.e. $R>\frac 1 {2\pi fC}$. Solving that gives us the frequency range for a strong output. (Frequencies well outside this range will produce only very weak outputs, so the microphone would be unsuitable for such frequencies).

Find the frequency range of the microphone and show us your working/result.

 

[songoku]

$R$ and $C$ form a potential divider with $V_R$ and $V_C$ the amplitudes of the alternating voltages. We want $V_R$ to be large, as this is the output voltage.

We apply sound of frequency $f$. This results in an alternating current of the same frequency in the circuit.

a) If $V_R > V_C$ we have a 'strong' output.
b) If $V_R < V_C$ we have a 'weak' output.
c) If $V_R=V_C$, this is the ‘cut-off’ condition.

Since we have a potential divider, to meet condition a) we require $R$ to be greater than than $X_C$. I.e. $R>\frac 1 {2\pi fC}$. Solving that gives us the frequency range for a strong output. (Frequencies well outside this range will produce only very weak outputs, so the microphone would be unsuitable for such frequencies).

Find the frequency range of the microphone and show us your working/result.

For cut-off condition:

$$R=X_C$$
$$R=\frac{1}{2\pi f C}$$
$$f=32 Hz$$

So for f = 20 Hz, the output will be 'weak' and the microphone is not suitable to detect the frequency?

 

[songoku]

Sorry @Steve4Physics but question (c) asks for a reason why the circuit is suitable, not why it is not suitable. Or in your opinion the question is wrong?

Thanks

 

[Steve4Physics]

So for f = 20 Hz, the output will be 'weak' and the microphone is not suitable to detect the frequency?

As the frequency falls below 32Hz, there isn’t a sudden drop in output – it’s a gradual reduction referred to as ‘roll-off’.

There will still be some output at 20Hz. In fact the output will be about half of what you would get at high frequencies. So at 20Hz there is a reduced, but still significant, output.

You can't (IMO) be expected to realise this without doing the necessary calculations.

 

[Steve4Physics]

Sorry @Steve4Physics but question (c) asks for a reason why the circuit is suitable, not why it is not suitable. Or in your opinion the question is wrong?

The question is not wrong. But to answer it requires you to work out the circuit's output at 20Hz. You can't do this with only a basic knowledge of RC circuits. You need to have covered the relevant AC circuit theory (e.g. adding impedances); unless you have covered this, the question can't be answered properly. Have you covered AC theory?

Edit - typos' fixed.

 

[songoku]

Have you covered AC theory?

More like I read it by myself. I know the formula of impedance
$$Z=\sqrt{R^2 + (X_C)^2}$$

Actually the answer has just been sent to students

I don't understand the part "capacitor completes discharging / charging during cycle of signal"

1) Does the capacitor discharge when the capacitance changes from increasing to decreasing so the charge also changes from increasing to decreasing?

2) When capacitor discharges, there will be current flows from the capacitor. Is this current in the opposite direction of current from the battery?

3) The capacitor will be completely discharged or charged after around 5 time constant. But why is this important to determine that the circuit is suitable for 20 Hz?

Thanks

 

[hutchphd]

This problem is a bit of a hash. It is important but badly presented. The last line of the supplied solution is dimensionally wrong.
3)The crux of the Physics is in your third question. The net signal power from the microphone circuit depends upon volts and amps. If the capacitor is not given sufficient time to discharge at least halfway, the signal will be much diminished.
2)The current flows both directions as the microphone oscillates about equilibrium
1)The capacitor will either charge or discharge depending upon whether squeezed or pulled. The battery will maintain this on average

 

[Steve4Physics]

I don't understand the part "capacitor completes discharging / charging during cycle of signal"

The wording is misleading because it could be interpreted as 'complete charging' and 'complete discharging' - whic is wrong. (It would be apporoximately true if the signal were a low frequency square wave - a bit like connecting a battery to R and C in series, then reversing the battery polarity, and keep repeating this.)

The model anwer gives no explanation of why the circuit operates as a high-pass filter (rather than a low-pass one). To me, the model answer is muddled and unsuitable. And, as already noted by @hutchphd, the last line is wrong.

1) Does the capacitor discharge when the capacitance changes from increasing to decreasing so the charge also changes from increasing to decreasing?
2) When capacitor discharges, there will be current flows from the capacitor. Is this current in the opposite direction of current from the battery?
3) The capacitor will be completely discharged or charged after around 5 time constant. But why is this important to determine that the circuit is suitable for 20 Hz

During each sound-cycle, there are 4 stages, corresponding to the 4 parts ( +-+-) of a sine curve:

i) While the pressure increases, plate-separation decreases, capacitance increases, stored charge increases, current flows (in, let’s say, the +ve direction).

ii) While the pressure decreases, returning to P_0, plate-separation increases, capacitance decreases, stored charge decreases, current flows (in the -ve direction).

iii) While the pressure continues to decreases (below $P_0$, plate-separation increases, capacitance decreases, stored charge decreases, current flows (in the -ve direction).

iv) While the pressure increases back to $P_0$, plate-separation decreases, capacitance increases, stored charge increases, current flows (in the +ve direction).

Note that the charge on the capacitor will oscillate about its rest-value. We can not think in terms of charging and discharging the capacitor in simple DC terms. Treating the capacitor as fully charged or fully discharged after 5 time-constants is irrelevant - we are not dealing with the DC case.

I don't like either the question or the model answer.

EDIT: Sorry - various problems with partially completed reply getting posted.

 

[kuruman]

Going back to the idea of potential divider suggested by @Steve4Physics in post #23, consider the following:
You already know that the circuit impedance is $Z=\sqrt{R^2+\dfrac{1}{(\omega C)^2}}.$
Find the AC current in the circuit assuming that AC emf is $V_{\text{in}}.$
Knowing the AC current, find the voltage $V_{\text{out}}$ across the resistor.
Find the ratio $V_{\text{out}}/V_{\text{in}}.$ It should be a function of $\omega.$
Plot the ratio against the frequency $f$ IN Hz for the numbers that are given.

Note that the charge on the capacitor will oscillate about its rest-value. We can not think in terms of charging and discharging the capacitor in simple DC terms. Treating the capacitor as fully charged or fully discharged after 5 time-constants is irrelevant - we are not dealing with the DC case.

I agree. High fidelity means that one can view the fluctuations of the capacitor voltage above and below the fully-charged value as a faithful reproduction of the air fluctuations above and below the ambient pressure caused by the traveling sound waves.

 

[songoku]

3)The crux of the Physics is in your third question. The net signal power from the microphone circuit depends upon volts and amps. If the capacitor is not given sufficient time to discharge at least halfway, the signal will be much diminished.

Why the capacitor need to discharge at least half of the charge for the signal to be significant?

2)The current flows both directions as the microphone oscillates about equilibrium

During each sound-cycle, there are 4 stages, corresponding to the 4 parts ( +-+-) of a sine curve:

i) While the pressure increases, plate-separation decreases, capacitance increases, stored charge increases, current flows (in, let’s say, the +ve direction).

ii) While the pressure decreases, returning to P_0, plate-separation increases, capacitance decreases, stored charge decreases, current flows (in the -ve direction).

iii) While the pressure continues to decreases (below $P_0$, plate-separation increases, capacitance decreases, stored charge decreases, current flows (in the -ve direction).

iv) While the pressure increases back to $P_0$, plate-separation decreases, capacitance increases, stored charge increases, current flows (in the +ve direction).

We take conventional current to flow from positive to negative terminal of battery but in this case there is an instant where the current flow from negative to terminal positive of battery. What does this mean?

Going back to the idea of potential divider suggested by @Steve4Physics in post #23, consider the following:
You already know that the circuit impedance is $Z=\sqrt{R^2+\dfrac{1}{(\omega C)^2}}.$
Find the AC current in the circuit assuming that AC emf is $V_{\text{in}}.$
Knowing the AC current, find the voltage $V_{\text{out}}$ across the resistor.
Find the ratio $V_{\text{out}}/V_{\text{in}}.$ It should be a function of $\omega.$
Plot the ratio against the frequency $f$ IN Hz for the numbers that are given.

Do you mean this?
$$I=\frac{V_{\text{in}}}{\sqrt{R^2+\frac{1}{(\omega C)^2}}}$$
$$V_{\text{out}}=\frac{V_{\text{in}}}{\sqrt{R^2+\frac{1}{(\omega C)^2}}} \times R$$
$$\frac{V_{\text{out}}}{V_{\text{in}}}=\frac{R}{\sqrt{R^2+\frac{1}{(\omega C)^2}}}$$

I got the graph to be like this:

Do we need to consider the emf of battery when finding the current?

Thanks

 

[kuruman]

That's the general idea behind the graph but you need to make a better plot. There is a lot of space that is not used vertically and there are no units or lables on the horizontal axis. Remember that the audible frequency is from 20 Hz to 20,000 Hz and that should be on your plot.

You don't have to worry about the battery because it doesn't provide AC. When there is no sound, there is no current in the circuit.

 

[Steve4Physics]

I deleted my previous reply as it was too confusing.

We take conventional current to flow from positive to negative terminal of battery but in this case there is an instant where the current flow from negative to terminal positive of battery. What does this mean?

With no sound, the charge on the capacitor is steady ($Q_0$).

When there is a sound wave, the following happens repeatedly:
- the capacitor's charge increases to $Q_0+\delta Q$; the current flows one way during this process;
- the capacitor's charge decreases to $Q_0-\delta Q$; the current flows the opposite way during this process.

This results in an alternating current with the same frequency as the sound wave.

 

[songoku]

That's the general idea behind the graph but you need to make a better plot. There is a lot of space that is not used vertically and there are no units or lables on the horizontal axis. Remember that the audible frequency is from 20 Hz to 20,000 Hz and that should be on your plot.

You don't have to worry about the battery because it doesn't provide AC. When there is no sound, there is no current in the circuit.

Just neglect the part where f is negative.

At f = 20 Hz, the value of Vout / Vin would be around 0.53. How to interpret this value?

I deleted my previous reply as it was too confusing.

With no sound, the charge on the capacitor is steady ($Q_0$).

When there is a sound wave, the following happens repeatedly:
- the capacitor's charge increases to $Q_0+\delta Q$; the current flows one way during this process;
- the capacitor's charge decreases to $Q_0-\delta Q$; the current flows the opposite way during this process.

This results in an alternating current with the same frequency as the sound wave.

Is it correct to say the capacitor undergoes the charging and discharging process? Something like when the capacitor's charge increases, it means the capacitor is being charged and when the capacitor's charge decreases, it means the capacitor is discharging?

What is the function of the battery in this circuit?

Thanks

 

[Steve4Physics]

Is it correct to say the capacitor undergoes the charging and discharging process

It is misleading because it suggests that the capacitor alternates between 'fully' charged and 'fully' discharged. That's not what happens.

The capacitor charges slightly and discharges slightly. Call these 'charge oscillations'.

The charge on the capacitor when there is no sound is $Q_0$. With sound, the charge might oscillate between $0.999Q_0$ and $1.001Q_0$ for example. The changes are small.

What is the function of the battery in this circuit?

When there is sound, there are charge-oscillations in the capacitor. As a result, there is alternating current through the resistor. Without a battery, the capacitor would be uncharged - there would be no charge oscillations. So there would be no current.

So the function of the battery is to maintain a charge on the capacitor.

Edit - minor.

 

[songoku]

So the function of the battery is to maintain a charge on the capacitor.

From the point of view of the battery, what is the meaning (or maybe consequence) of the alternating signal produced in the circuit?

At one time, the current will flow clockwise, and at the other time, it would be anticlockwise. Let say when the current flows clockwise the capacitor is slightly charging so it means that the capacitor is being charged by the battery. When the current flows anticlockwise, the capacitor is slightly discharging so it means the capacitor discharges through the resistor and battery? Is the battery being charged by the capacitor?

Thanks

 

[hutchphd]

I believe this is an electromechanical problem, with most of the physics (at least low end response) governed by the coupling of the pressure wave into the microphone diaphragm. Then the unknown mechanical inertia makes it a parametric oscillator and any simple "RC" analysis is likely misguided.

 

[Averagesupernova]

...and any simple "RC" analysis is likely misguided.

Absolutely. This is what is known as a condenser mic. The capacitor is considered the source. So as a source it's output impedance is likely lowered as the frequency increases. Simple analysis shows that around 30 hertz is the knee and the frequency response is flat as the frequency increases.
-
The OP claims to have no interest in RC high pass at this point.

I don't study high-pass RC filters.

I don't understand how all of the questions can be answered without having a grasp of a basic RC filter.

 

[Steve4Physics]

From the point of view of the battery, what is the meaning (or maybe consequence) of the alternating signal produced in the circuit?

I don't understand your question.

At one time, the current will flow clockwise, and at the other time, it would be anticlockwise.

Correct. This happens repeatedly. That's why you get an alternating current through the resistor (and therefore an alternating voltage across it) when there is sound.

Let say when the current flows clockwise the capacitor is slightly charging so it means that the capacitor is being charged by the battery.

Ok - for half of each cycle.

When the current flows anticlockwise, the capacitor is slightly discharging so it means the capacitor discharges through the resistor and battery? Is the battery being charged by the capacitor?

Yes - for the other half of each cycle.

But thinking in those terms will not help you understand part c) of the original question (if that's what you are trying to do). E.g. see @Averagesupernova comments in Post #39.

 

[songoku]

Yes - for the other half of each cycle.

But thinking in those terms will not help you understand part c) of the original question (if that's what you are trying to do). E.g. see @Averagesupernova comments in Post #39.

I am not trying to understand question (c) since from all the replies I know I won't be able to understand without knowing some other things outside what I have learned.

The best I can guess would be from the graph suggested by @kuruman and also your reply. The ratio of Vout / Vin is around 0.5 from the graph, matching what you said that at 20 Hz the output will be half of what it would be at high frequency and it is still significant (although I don't understand why the ratio of 0.5 is still significant)

I am trying to clear some doubts in my mind about the circuit. What happens to the battery when it is being charged by the capacitor? I think the voltage of the battery is constant so when it is being charged, what does the effect?

Thanks

 

[Tom.G]

Yes, an 'ideal' (or perfect) battery has a constant voltage. If you have ever had to replace batteries in a flashlight when the the light gets dim, you have seen that a real battery is not 'ideal'.

There are also batteries that can be recharged, they are used in cars and in cell phones.

Even a flashlight battery can be slightly recharged a few times.

Recharging a battery reverses the chemical process that supplies power used in normal use.

So the answer to your question is the power from the capacitor very slightly recharges the battery.

I hope this helps!

Cheers,
Tom

 

[Steve4Physics]

although I don't understand why the ratio of 0.5 is still significant

The filter doesn't have a sharp cut-off: it doesn't completely block all frequencies below some exact value and it doesn't completely transmit all frequencies above that value. For a signal of frequency, $f$, the fraction of the signal transmitted is a smoothly varying function of $f$.

So, for convenience, we can call the cut-off frequency the value of $f$ when some fraction of the signal's amplitude gets transmitted. We could choose, for example, 0.1 or 0.9 as the fraction - but 0.5 is a ‘natural’ choice.

I am trying to clear some doubts in my mind about the circuit. What happens to the battery when it is being charged by the capacitor? I think the voltage of the battery is constant so when it is being charged, what does the effect?

While a (real) battery is being charged, the voltage is greater than the battery's 'own voltage' (greater than the battery's emf to use the correct terminology).

 

[songoku]

Thank you very much for all the help and explanation kuruman, Steve4Physics, hutchphd, Averagesupernova, Tom.G