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Why are the chemicals able to produce a unique potential difference? Quantum mechanical descriptions of molecules, which take into account the types of atoms and numbers of electrons in them, are able to predict the energy states they can have and the energies of reactions between them.

In the case of a lead-acid battery, an energy of 2 eV is given to each electron sent to the anode. Voltage is defined as the electrical potential energy divided by charge: V = P E q size 12{V= { {P rSub { size 8{E} } } over {q} } } {} . An electron volt is the energy given to a single electron by a voltage of 1 V. So the voltage here is 2 V, since 2 eV is given to each electron. It is the energy produced in each molecular reaction that produces the voltage. A different reaction produces a different energy and, hence, a different voltage.

Terminal voltage

The voltage output of a device is measured across its terminals and, thus, is called its terminal voltage     V size 12{V} {} . Terminal voltage is given by

V = emf Ir , size 12{V="emf" - ital "Ir"} {}

where r size 12{r} {} is the internal resistance and I size 12{I} {} is the current flowing at the time of the measurement.

I size 12{I} {} is positive if current flows away from the positive terminal, as shown in [link] . You can see that the larger the current, the smaller the terminal voltage. And it is likewise true that the larger the internal resistance, the smaller the terminal voltage.

Suppose a load resistance R load size 12{R rSub { size 8{"load"} } } {} is connected to a voltage source, as in [link] . Since the resistances are in series, the total resistance in the circuit is R load + r size 12{R rSub { size 8{"load"} } +r} {} . Thus the current is given by Ohm’s law to be

I = emf R load + r . size 12{I= { {"emf"} over {R rSub { size 8{"load"} } +r} } } {}
This schematic drawing of an electrical circuit shows an e m f, labeled as script E, driving a current through a resistive load R sub load and through the internal resistance r of the voltage source. The current is shown flowing in a clockwise direction from the positive end of the source.
Schematic of a voltage source and its load R load size 12{R rSub { size 8{"load"} } } {} . Since the internal resistance r size 12{r} {} is in series with the load, it can significantly affect the terminal voltage and current delivered to the load. (Note that the script E stands for emf.)

We see from this expression that the smaller the internal resistance r size 12{r} {} , the greater the current the voltage source supplies to its load R load size 12{R rSub { size 8{"load"} } } {} . As batteries are depleted, r size 12{r} {} increases. If r size 12{r} {} becomes a significant fraction of the load resistance, then the current is significantly reduced, as the following example illustrates.

Calculating terminal voltage, power dissipation, current, and resistance: terminal voltage and load

A certain battery has a 12.0-V emf and an internal resistance of 0 . 100 Ω size 12{0 "." "100" %OMEGA } {} . (a) Calculate its terminal voltage when connected to a 10.0- Ω size 12{"10" "." 0- %OMEGA } {} load. (b) What is the terminal voltage when connected to a 0 . 500- Ω size 12{0 "." "500-" %OMEGA } {} load? (c) What power does the 0 . 500- Ω size 12{0 "." "500-" %OMEGA } {} load dissipate? (d) If the internal resistance grows to 0 . 500 Ω size 12{0 "." "500 " %OMEGA } {} , find the current, terminal voltage, and power dissipated by a 0 . 500- Ω size 12{0 "." "500-" %OMEGA } {} load.

Strategy

The analysis above gave an expression for current when internal resistance is taken into account. Once the current is found, the terminal voltage can be calculated using the equation V = emf Ir size 12{V="emf" - ital "Ir"} {} . Once current is found, the power dissipated by a resistor can also be found.

Solution for (a)

Entering the given values for the emf, load resistance, and internal resistance into the expression above yields

I = emf R load + r = 12 . 0 V 10 . 1 Ω = 1 . 188 A . size 12{I= { {"emf"} over {R rSub { size 8{"load"} } +r} } = { {"12" "." 0" V"} over {"10" "." "1 " %OMEGA } } =1 "." "188"" A"} {}

Enter the known values into the equation V = emf Ir size 12{V="emf" - ital "Ir"} {} to get the terminal voltage:

Questions & Answers

Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point 1 the cross-sectional area of the pipe is 0.077 m2, and the magnitude of the fluid velocity is 3.50 m/s. (a) What is the fluid speed at points in the pipe where the cross
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Practice Key Terms 4

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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