6
\$\begingroup\$

suppose a 9v battery is connected to a load which draws 2 amps of current. so how does the battery determines that load requires this much current ? I mean if the battery throws about 3 amps, then it would just blow the load, so how the battery just supplies 2 amps and not just any other value?

\$\endgroup\$
5
  • 1
    \$\begingroup\$ The battery does not determine anything in this instance. The load takes what the load wants. Look up Ohm's Law. \$\endgroup\$
    – Trevor_G
    Jan 28 '18 at 19:21
  • 1
    \$\begingroup\$ @Trevor_G - although of course, with a real battery, there is an internal resistance that will limit the maximum current that can flow... \$\endgroup\$
    – Jules
    Jan 29 '18 at 1:43
  • 1
    \$\begingroup\$ It is a negotiation process. The battery offers 9V. The load says, if you give me 9V, I will take 2A. Then the battery says, well, if you are going to take 2A, then I can only give you 8V. Then the load says, well, if you are only giving me 8V, then I guess I will take 1.8A. Then the load says, well, at 1.8A I can give you 8.1V. And then the load says, well, at 8.1V, I would actually like 1.81A, and so on. Eventually, they find an operating point where they are both happy. \$\endgroup\$
    – mkeith
    Jan 29 '18 at 4:09
  • 1
    \$\begingroup\$ When you deal with power sources that are not simple voltage or current sources, sometimes the best thing to do is draw a load line. You draw one for the source, and another for the load, and where they cross, that is what the voltage and current will be. \$\endgroup\$
    – mkeith
    Jan 29 '18 at 4:11
  • 1
    \$\begingroup\$ en.wikipedia.org/wiki/Load_line_(electronics) \$\endgroup\$
    – mkeith
    Jan 29 '18 at 4:12
8
\$\begingroup\$

Does the battery determine the amount of current flowing in the circuit?

Well... yes and no. The battery will try and give the load whatever it asks for not the other way round. This is true for any voltage source not just batteries (current sources will try and push a set current through a circuit but voltage sources will just sit there and do as they're told).

If the load wants 16.73 micro amps, that's what it will get, if it wants 500mA, that's what it will get. All normal batteries along with almost all power supplies you will normally encounter (except for LED drivers) try only to keep their output at a constant voltage (even if you're drawing 0A, although some ancient PC power supplies go a bit funny with no load).

The problems start when heavy loads are connected to light-duty sources, just like how I'll struggle to lift an 80kg railway sleeper, small power supplies will struggle to maintain their designed output voltage under heavy load.

Everything that isn't superconducting has resistance, everything. Small things generally have more, big currents and high resistances turn into big voltage drops (and lots of internal heating). So if you try and pull lots of current from a small battery, you might find that its output voltage drops right down and keeps dropping until either the load turns off or it stops trying to draw full current. Electronics are not that tough, keep pushing them and eventually they just give up trying (it'll probably get pretty warm too).

\$\endgroup\$
2
\$\begingroup\$

suppose a 9v battery is connected to a load which draws 2 amps. so how does the battery determines that load requires this much current?

It works in the same way as the national electricity grid does. My house is powered at 230 V by the Irish national grid which has a peak supply of about 5,000 MW (5 GW). When I switch on a 25 W light bulb it just draws that much power from the grid and not the whole 5 GW (which would burn out my house in a spectacular flash).

The current drawn from the grid or from your battery depends on the resistance of the device connected.

$$ Current = \frac {voltage}{resistance} = \frac {V}{R} $$

$$ Power = \frac {voltage^2}{resistance} = \frac {V^2}{R} $$

or $$ Power = volts \ \times \ amps = VI $$

\$\endgroup\$
2
\$\begingroup\$

A battery has no such ability as push certain current through a load regardless what a load wants and loads generally have no such ability as suck a certain current regardless what a battery offers. The current is a result, the found balance between the voltage and resistances in the circuit.

We can construct control circuits which try to regulate the current, but they fail as soon as some basic laws of the electricity must be violated. For example many led light drivers try to keep a constant current current through a led chain. The led driver succeeds if it has enough voltage for the number of the leds. It fails if there are too much leds or other resistance.

The most basic laws which cannot be violated are Ohms law, Kirchoff's laws and the energy principle. You should learn these if you try to calculate something about the electricity.

The situation is same in water supplying systems. There are no such systems in use which try to push say 2 liters/second. The system has certain pressure. You have not a faucet which unavoidably takes say 3 liters/second. You turn the faucet to have some resistance and the flow is the resulted balance between the pressure and flow resistance.

You can construct a controller which opens the faucet until the flow is 3 liters/second. It has a flow measurement sensor. Your controller fails if the water supplying system haven't enough pressure to push that much water/second through all the pipes between you and the pump station.

Voltage in a battery or pressure in the water supplying system can be so high that a breakdown occurs. But that's due a tendency to push a certain current or water flow only if there's some current or flow control system which can rise the voltage or pressure over the breakdown limit.

\$\endgroup\$
1
\$\begingroup\$

Does the battery determine the amount of current flow in the circuit?

yes and no

For the most part no, and this would be because most times a given circuit is deigned or wired properly as well as the battery being sized properly for the given power requirements.

Remember a battery is a chemical device, and it is the chemical reaction within the battery that is important to know about regarding whatever circuit the battery is going to power.

YES a battery could determine the amount of current flowing in the circuit. In this kind of scenario the battery would basically be undersized or already discharged to where the elctro-chemical reaction within the battery cannot keep up where the limiting factor would normally be load resistance; in this case it would be the chemical reaction within the battery or possibly the batteries internal resistance becoming greater as it gets hotter trying discharge more current than it was designed to. A good example would be using a 12 volt lantern battery (or two 6v lantern batteries) to start your car; a typical lead-acid car battery rated at 500 cold cranking amps versus the lantern batteries (which do not have a CCA rating but let's say it has 1 amp rating). Try and use the lantern batteries to power a car starter (DC motor) which are rated at around 1.2 Kw or 1.5 hp and generally draw around 500 amps and it would be the lantern batteries internal resistance and chemical reaction which determines how much current flows in that circuit not the resistance of the DC starter motor.... just replace the DC motor with a 0.00? ohm simple resistor (for argument's sake) and hopefully you get the idea.

I mean if the battery throws about 3 amps, then it would just blow the load, so how the battery just supplies 2 amps and not just any other value?

ANY battery does not just throw amperage or current; it will first discharge as much current as physically possible based on its chemistry and chemical reaction,

and then what limits this discharged current is the resistance of the circuit it is connected to. And then if the battery in undersized where the circuit resistance is so low that will allow whatever battery to discharge at an extreme rate causing it to first overheat which is sometimes described as that battery's internal resistance increasing which then would be the limiting factor in the amount of current, if the battery doesn't catch fire and explode.

Really need to consider the make/model and chemistry of the battery, you mention a 9v battery and a 3 amp discharge which is not realistic, those batteries do milliamps at best. And then you have alkaline, Ni-Cd, Ni-MH, various lithium chemistries, lead-acid and so on. Some are quite safe, some can be quite powerful and dangerous. But for example if a circuit designed for 12 volts having a resistance or 360 ohms and an expected current draw of 0.033 amps then it makes no difference if you use a little duracel 12v type 21/23 battery, your car battery; the limiting factor for battery discharge would be the circuit resistance and not the battery's physical capability, chemistry, and electrical capacity.

\$\endgroup\$
0
\$\begingroup\$

suppose a 9v battery is connected ... if the battery throws about 3 amps

A battery supplies electric power within some limits, and there's an equation for its output, characterized by the terminal voltage and the output current. The battery is characterized by an equation with voltage and current variables, plus constants (which are the datasheet entries for the battery you choose).

A '9V battery' is not completely defined; there's more to it than that 'terminal voltage under zero load' value. The 'throws 3 amps' scenario might be another part of the puzzle, like a short-circuit current. If so,(9V, 0 A) is one point on a curve, and (0V, 3A) is a second point. Draw a line between those points, and THAT might be the battery characteristic function.

Similarly, a load draws electric power within some limits, and there's an equation for its input current and terminal voltage.

See where this is headed? Two equations, and two unknowns, is often a soluble mathematical problem. So, if you know both the battery and the load, you have enough data to predict, mathematically, what electric power transfer happens when they are connected. You just solve for voltage and current.

It's a little more complex, because a battery charge state, or load temperature, or even barometric pressure, can be involved.

\$\endgroup\$
0
\$\begingroup\$

The amount of current the battery will provide is going to rely on the circuit equivalent resistance. Batteries can usually hold up to a certain value, which after such its output voltage will drop due to its internal resistance as more current will be flowing, more voltage is dropped on this internal resistance. To control the current you´d need a separate circuit to do so. A battery is a constant voltage source, and that´s what it´s going to do: provide a constant voltage to the circuit, regardless of current.

\$\endgroup\$
0
\$\begingroup\$

your battery never determine the amount of current throw to the load, rather the load resistance and operating voltage of the load determine the amount of current. For two or more load resistance (Vs= Vr1+Vr2+Vr3...+Vrn) and each voltage drop (Vr1=IR1, Vr2=IR2, ..., Vrn=IRn). Therefore; for a closed loop of single load resistance, the voltage drop(Vr) is the same as source voltage/battery voltage(Vs) and current can be determined by the equation (Vr or Vs=IR). this implies if the operating load resistance is 4.5ohm it is possible to draw 2 amp unless it will below or above the required amount. To draw the desired amount of current; the load resistance (designed resistance or designed load of an equipment) multiplied by desired current should equal with the battery voltage. Le's assume the load resistance is 4.5ohm and battery voltage is 9v, so current flow through the loop is 2amp; for the same load resistance(not be changed in any variation of voltage and current), if the battery voltage is 18v the current flow through the loop becomes 18v/4.5ohm=4amp. if I am wrong please give me feed back.

\$\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.