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According to Ohm's law V=IR. This means that if the current is zero there is no voltage. Does this mean also that an unplugged battery has zero voltage? Then why does it say otherwise in the package.

In other words:

  1. What does it mean that a battery is 1.5 Volts if, after all, it depends on the resistance?

  2. What does a voltmeter measure if the voltage depends on resistance.

I'm sorry if this looks like a silly question. I'm not from electrical background.

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\$ V = I \cdot R = 0 \cdot \infty \$ Why do you think this is zero? \$ \infty \$ is not a number so \$ 0 \cdot \infty \$ is not defined and can be anything. In this case it's the batteries voltage. –  Warren Hill Jun 26 at 12:10
    
@WarrenHill For \$xy=c\$, where \$c\$ is fixed, if either \$x\$ or \$y\$ linearly approaches to zero, the other one logarithmically approaches to infinity. So, their values can be zero and infinite respectively, but their product is still not indefinite. –  hkBattousai Jun 26 at 12:15
    
@hkBattousai from a purely mathematical point of view \$ 0 \cdot \infty \$ is undefined as using your logic from \$ x \cdot y = c \$ and allowing \$ x \$ to go to zero y will go to \$ \infty \$ regardless of your initial choice of \$ c \$. –  Warren Hill Jun 26 at 12:20
    
@hkBattousai don't confuse limit and value of the function. The function here would be undefined, but the limit might exist. –  Ruslan Jun 26 at 14:19
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Ohm's law states that the current through a conductor between two points is directly proportional to the potential difference across the two points. (The equation V = IR simply re-states that, where R is the proportionality constant for the particular conductor.) What is the conductor in your example with a battery not hooked up to anything? If there is no conductor then Ohm's law doens't come into it! If there is a conductor then say what the conductor is. –  Eric Lippert Jun 26 at 15:37
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7 Answers 7

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Does this mean also that an unplugged battery has zero voltage?

No, Ohm's law is not universal; Ohm's law defines the ideal resistor circuit element.

But not all circuit elements are resistors and thus, not all circuit elements obey Ohm's law.

A battery is an approximate voltage source, not a resistor and thus does not obey Ohm's law.

For an ideal voltage source, the voltage across is independent of the current through; the voltage across is what it is regardless of the value of the current through.

Thus, if we connect an ideal voltage source and resistor together to form a circuit, the voltage across the resistor is fixed by the voltage source.

Since the voltage is fixed, the series current through the resistor and battery is determined by the resistance \$R\$ of the resistor:

$$I = \frac{1.5V}{R}$$

If we disconnect the battery, we have effectively made the resistance infinite and we get

$$I = \frac{1.5V}{\infty} = 0A$$

Physical voltage sources, such as a 1.5V battery, cannot supply unlimited current and, in fact, produce a finite current when short-circuited.

So, we typically model a physical battery by placing a resistor in series with the voltage source. But this is the topic of another question.

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The battery voltage is defined by its chemistry.

It is usually a redox reaction whose voltage can be determined by looking at electronegativity values of the used metals.

Why the Ohm law still apply?

Ohm law is a law that works given a certain model of the physical thing. Using that model, a battery is modeled as shown in this figure:

schematic

If there is no current flowing out of the battery, ohm law says that there is no voltage drop in R1. Thus the output voltage of the battery is V0: the nominal voltage of your battery.

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You are talking about a "singularity" here ...

Like the other answers state, this formula isn't applicable for an infinite value of R.

In practice, you will always have some "leakage" current in the range of nano- to micro-Ampere, because the resistance isn't really infinite. But this is hard to measure...

To answer your questions in detail:

  1. It does have 1.5 Volts. The voltage isn't "dependent" on resistance like you state. The formula U=I*R just gives the relationship between voltage, current and resistance. There is a good analogy for this: Water in a hose with a valve somewhere in this hose.

    • Voltage is the water pressure.
    • Current is the flow in the hose.
    • Resistance is the position of the valve. If the valve is closed, there is not flow. But there is still pressure in the hose.
  2. The voltmeter measure the voltage. In fact, it has a finite internal resistance, letting a small current pass through the voltmeter.

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As has been mentioned, the battery has the same voltage (difference in electrical potential) regardless of what is connected to it. A battery is a voltage source and will attempt to give the same voltage across different loads (by varying the current). By having nothing connected to the battery, you have an infinite resistance. Thus, I = V/R => That I = 0 as any V divided by an infinite resistance will approach 0. When you are measuring the actual voltage of the battery with say a voltmeter, it is actually running the current through a large value resistance (theoretically it would be infinite) and just calculating the voltage from the measured current through the resistor. Though, keep in mind that chemical and other types of batteries generally have an internal resistance that is built in due to the resistance the chemical elements themselves present. You can think of a battery as an ideal voltage source with a resistor in series.

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I =V/R. You have zero current because you have infinite resistance. V can be anything.

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Yes but my question is, what does it mean that the battery is 1.5 volts then? After all it depends on the resistance. –  Ambesh Jun 26 at 11:30
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Unplugging a battery is same as connecting an infinite resistor (the air between the batter terminals) across it. The Ohm's Law still applies. \$R=\infty\$, so \$I=0\$ regardless of the nominal battery voltage.

What does it mean that a battery is 1.5 Volts if, after all, it depends on the resistance?

1.5V is the open circuit voltage of the battery. It is not the voltage level under load. If the internal resistance of the battery is \$R_{int}\$ and the load resistance is \$R_{L}\$, the voltage on the load will be

$$ V_L = 1.5V \times \dfrac{R_L}{R_{int} + R_{L}}. $$

What does a voltmeter measure if the voltage depends on resistance.

The voltage on the battery can be found by the formula above. If there is no load resistor (i.e.; \$R_L=\infty\$), then according to the formula, the voltmeter will measure 1.5V.

For measuring the internal resistance of the battery, make \$R_L=0\$ (for a very short time!), and measure the current through the battery \$I_{sc}\$. You can find the internal resistance \$R_{int} = \dfrac{1.5V}{I_{sc}}\$

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The open circuit condition results in the maximum voltage that can be generated when looking at a battery. The resistance of the battery does not play a part in the battery voltage ( Chemestry is the most important thing here).... However as soon as an external resistance of significance is applied across the battery the internal resistance and external resistance will determine the voltage at the battery. –  Spoon Jun 26 at 12:04
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Voltage in real life terms, is called 'potential'.

  • A bucket of water has potential energy waiting to burst out of a leak. (Voltage)
  • The bucket material has the property to hold the water in the bucket. (Resistance)
  • Water flows out of a leaky bucket. (Current)

Bigger the leak (Lower the resistance) -> Stronger the flow (higher the current)

When there is no water leaking, the bucket of water still has its 'potential'. (1.5v)

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