0
\$\begingroup\$

I have a query.

Suppose, I have a battery.

Can some battery have enough voltage but not deliver the required current? How is this possible?

My question comes from car batteries but it is not limited to automotive. It is batteries in general.

I have seen, some car batteries holding like 10V, but the car does not start-up.

Similarly, does this scenario arise in other fields also? Like having some voltage, but the load doesn't get the required current? How is this possible?

Request you to provide some clarity on my misunderstanding.

\$\endgroup\$
4
\$\begingroup\$

The internal resistance of the battery plays a larger role in the maximum current it can provide.

Suppose you have the circuit shown below:

Battery Circuit The boxed area represents the battery and the resistor RL represents the load you are trying to power. Inside the battery, you have a voltage source and the internal resistance of the battery, which may be in the range of milliohms or less. Now imagine RL was very large, say 100kOhms. Then the total resistance of the circuit would be dominated by RL. Using Ohm's law, we can calculate the current through the circuit as $$ I = \frac{V_o}{R_s+R_L} $$ Since Rs is much smaller than RL, we can say $$ I = \frac{V_o}{R_L} $$ This means that the battery can supply any current that RL allows, as long as RL is much larger than Rs.

But now suppose RL is of the same order of magnitude as Rs, or even less than Rs. Then We can no longer ignore Rs in the current equation. Let's take the extreme case that RL is much smaller than Rs. Then we can ignore RL in the current equation, and we get $$ I = \frac{V_o}{R_s} $$ This is the maximum current that the battery can supply (we essentially short the terminals of the battery together). This is because Rs is internal to the battery and we cannot change it.

In the case of a 12V car battery, a huge current is needed to start the car's engine. Thus, the internal resistance of the battery needs to be small. A brief internet search says that the internal resistance for some car batteries is roughly 20milliOhm. This would correspond to a max current (short circuit current) of 12/0.02=600A -- the current needed to start a car engine.

Now take an ordinary 9V battery. Again, a brief internet search says that 9V batteries have internal resistances somewhere around 1 to 2Ohm. Let's take 1Ohm. The max (short circuit) current a 9V can provide is 9/1=9A -- way less than the car battery.


Another thing to consider is that as you draw more current from a battery (by decreasing the load resistance), more voltage will be dropped across the internal resistance, Rs. The total voltage drop across the internal resistance is again given by Ohm's law: $$ V_s = V_0/R_s $$ The net effect of this is that the total voltage you'll see across the terminals of the battery will drop as you draw more current from the battery. At some point, the voltage will reach zero -- this is when you short the battery terminals together.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Thank you very much for the detailed answer. In the last line, why do you say, "this is when you short the battery terminals together." And in any case, for an example, if a 5V battery, drops to 3V due to prolonged use, will it still be able to provide current to the load? I think I am just confusing these terms and misunderstand them \$\endgroup\$ – Newbie Mar 8 at 7:47
  • 1
    \$\begingroup\$ Shorting the terminals together is the same thing as saying connect a RL=0Ohm resistor across the terminals. Doing so will result in current flowing in the circuit. But the voltage across the 0Ohm resistor, given by Ohm's law, is 0Ohm * I = 0. Since we have zero volts across the RL, we also have zero voltage across the batteries terminals. \$\endgroup\$ – LetterSized Mar 8 at 7:50
  • 1
    \$\begingroup\$ Regarding the second question, if the voltage drops to 3V, then the max current will also drop. Again, the max current is related to V0/Rs. If V0 drops, then the max current drops. Another thing that I didn't mention is that as you consume the capacity of batteries, sometimes the internal resistance of the battery will start increasing, which will reduce the max current even further. To answer will it provide current to the load depends on the particular load in question. If the load is just sipping a tiny amount of current, then the answer is probably yes. But there is no guarantee. \$\endgroup\$ – LetterSized Mar 8 at 7:51
  • 1
    \$\begingroup\$ The voltage decrease over time varies from battery to battery. In general though, the voltage will drop rapidly at first, then reach some steady state which it will hold for a long time. Once the battery's capacity has been depleted, the voltage will rapidly drop to 0V. As an example, here is the datasheet for an Energizer AA battery: data.energizer.com/PDFs/E91.pdf At the bottom of the datasheet there are four graphs which show how the voltage decreases over time. \$\endgroup\$ – LetterSized Mar 8 at 8:06
  • 1
    \$\begingroup\$ Regarding the second question, you are talking about extremes. In practice, you'd never short a battery with copper wire, since it might cause a fire or at least heat up the battery a lot! But yes, if you were to short the battery with copper wire and assume nothing bad happened, the voltage would very quickly drop to zero as the battery capacity is depleted. In the beginning, most of the voltage would be dropped across the internal resistance, but eventually the voltage source in the battery would drop to 0. \$\endgroup\$ – LetterSized Mar 8 at 8:07
1
\$\begingroup\$

If a batteries mA capacity is so low, that it can't even power the smallest loads without fully discharging immediately, then this could probably happen. Two scenarios where this could happen are when the battery is dead, so it can't keep a charge, or when you're using a small battery that isn't designed to have a large capacity. So pretty much at this point, the battery is more like a regular capacitor since it can't hold a charge.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ Thank you. Could you just provide me an example in our day to day life regarding this battery having enough, but unable to provide current? Any example please \$\endgroup\$ – Newbie Mar 8 at 6:56
  • \$\begingroup\$ Suppose, I have a 5V battery and a load of 1A. But the battery capacity is 200mAH. What will happen in this case? Will it provide for the 1A load? Suppose if the battery provides for 1A load, what will happen or what will be the voltage of the battery after providing the 1A for sometime \$\endgroup\$ – Newbie Mar 8 at 7:06
  • 1
    \$\begingroup\$ I don't have any particular examples for batteries in the uA range, but I think Graphite Batteries, which are pretty new to my understanding, fit in that range. If you draw 1 A from a battery rated for 200mAH, it will still be able to provide 1A (assuming that it's rated for that). But that means that your 200mAH battery will discharge 1A for 20 minutes (in theory). In practice, the current will decrease as the battery capacity decreases. Since when a battery loses it's charge, the voltage on it decreases. So over time, your 4.2V LiPo cell will drop to 2.9V as the capacity drops. \$\endgroup\$ – Jay Mar 8 at 7:35
  • \$\begingroup\$ Thank you. Could you please tell me how you got the 2.9V from the 4.2V value? Please help me understand the calculation \$\endgroup\$ – Newbie Mar 8 at 7:38
  • 1
    \$\begingroup\$ The battery capacity does not directly determine the maximum current of a battery. It only determines how long the battery can supply a current for (that is, how much energy is can output over a period of time). The max current is determined by it's internal resistance. Many 4.2V lipo batteries can supply much more current than 9V batteries since they tend have lower internal resistances. That being said, the maximum current you can safely draw from a battery is often related to its capacity (see C ratings), but this varies battery to battery. \$\endgroup\$ – LetterSized Mar 8 at 7:40
1
\$\begingroup\$

Batteries has a parameter that is called "C rate". This parameter simply determines how much current It can give in time. If too much current is required from battery then It's chemistry is corrupted.

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.