Terminal Voltage

The voltage output of a device is measured across its terminals and, thus, is called its terminal voltage \(\phantom{\rule{0.25em}{0ex}}V\). Terminal voltage is given by

\(V=\text{emf}-\text{Ir},\)

where \(r\) is the internal resistance and \(I\) is the current flowing at the time of the measurement.

\(I\) is positive if current flows away from the positive terminal, as shown in this figure. 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 \({R}_{\text{load}}\) is connected to a voltage source, as in this figure. Since the resistances are in series, the total resistance in the circuit is \({R}_{\text{load}}+r\). Thus the current is given by Ohm’s law to be

\(I=\cfrac{\text{emf}}{{R}_{\text{load}}+r}.\)

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 \({R}_{\text{load}}\). Since the internal resistance \(r\) 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 \(r\), the greater the current the voltage source supplies to its load \({R}_{\text{load}}\). As batteries are depleted, \(r\) increases. If \(r\) becomes a significant fraction of the load resistance, then the current is significantly reduced, as the following example illustrates.

Example: 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 \(0\text{.}\text{100}\phantom{\rule{0.25em}{0ex}}\Omega \). (a) Calculate its terminal voltage when connected to a \(\text{10.0-}\Omega \) load. (b) What is the terminal voltage when connected to a \(0\text{.}\text{500-}\Omega \) load? (c) What power does the \(0\text{.}\text{500-}\Omega \) load dissipate? (d) If the internal resistance grows to \(0\text{.}\text{500}\phantom{\rule{0.25em}{0ex}}\Omega \), find the current, terminal voltage, and power dissipated by a \(0\text{.}\text{500-}\Omega \) 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 \(V=\text{emf}-\text{Ir}\). 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

\(I=\cfrac{\text{emf}}{{R}_{\text{load}}+r}=\cfrac{\text{12}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{V}}{\text{10}\text{.}\text{1}\phantom{\rule{0.15em}{0ex}}\Omega }=1\text{.}\text{188}\phantom{\rule{0.25em}{0ex}}\text{A}.\)

Enter the known values into the equation \(V=\text{emf}-\text{Ir}\) to get the terminal voltage:

\(\begin{array}{lll}V& =& \text{emf}-\text{Ir}=\text{12.0 V}-(\text{1.188 A})(\text{0.100 Ω})\\ & =& \text{11.9 V.}\end{array}\)

Discussion for (a)

The terminal voltage here is only slightly lower than the emf, implying that \(\text{10}\text{.}0\phantom{\rule{0.25em}{0ex}}\Omega \) is a light load for this particular battery.

Solution for (b)

Similarly, with \({R}_{\text{load}}=0\text{.}\text{500}\phantom{\rule{0.15em}{0ex}}\Omega \), the current is

\(I=\cfrac{\text{emf}}{{R}_{\text{load}}+r}=\cfrac{\text{12}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{V}}{0\text{.}\text{600}\phantom{\rule{0.15em}{0ex}}\Omega }=\text{20}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{A}.\)

The terminal voltage is now

\(\begin{array}{lll}V& =& \text{emf}-\text{Ir}=\text{12.0 V}-(\text{20.0 A})(\text{0.100 Ω})\\ & =& \text{10}\text{.}\text{0 V.}\end{array}\)

Discussion for (b)

This terminal voltage exhibits a more significant reduction compared with emf, implying \(0\text{.}\text{500}\phantom{\rule{0.15em}{0ex}}\Omega \) is a heavy load for this battery.

Solution for (c)

The power dissipated by the \(0\text{.}\text{500 - Ω}\) load can be found using the formula \(P={I}^{2}R\). Entering the known values gives

\({P}_{\text{load}}={I}^{2}{R}_{\text{load}}={(\text{20.0}\phantom{\rule{0.25em}{0ex}}\text{A})}^{2}(0\text{.500 Ω})=2.00×{10}^{2}\phantom{\rule{0.25em}{0ex}}\text{W}.\)

Discussion for (c)

Note that this power can also be obtained using the expressions \(\cfrac{{V}^{2}}{R}\) or \(\text{IV}\), where \(V\) is the terminal voltage (10.0 V in this case).

Solution for (d)

Here the internal resistance has increased, perhaps due to the depletion of the battery, to the point where it is as great as the load resistance. As before, we first find the current by entering the known values into the expression, yielding

\(I=\cfrac{\text{emf}}{{R}_{\text{load}}+r}=\cfrac{\text{12}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{V}}{1\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\Omega }=\text{12}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{A}.\)

Now the terminal voltage is

\(\begin{array}{lll}V& =& \text{emf}-\text{Ir}=\text{12.0 V}-(\text{12.0 A})(\text{0.500 Ω})\\ & =& \text{6.00 V,}\end{array}\)

and the power dissipated by the load is

\({P}_{\text{load}}={I}^{2}{R}_{\text{load}}={(\text{12.0 A})}^{2}(0\text{.}\text{500}\phantom{\rule{0.25em}{0ex}}\Omega )=\text{72}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{W}.\)

Discussion for (d)

We see that the increased internal resistance has significantly decreased terminal voltage, current, and power delivered to a load.

Battery testers, such as those in this figure, use small load resistors to intentionally draw current to determine whether the terminal voltage drops below an acceptable level. They really test the internal resistance of the battery. If internal resistance is high, the battery is weak, as evidenced by its low terminal voltage.

The first photograph shows an avionics electronics technician working inside an aircraft carrier, measuring voltage of a battery with a voltmeter probe. The second photograph shows the small black battery tester which has an LED screen that indicates the terminal voltage of four batteries inserted into its case.

These two battery testers measure terminal voltage under a load to determine the condition of a battery. The large device is being used by a U.S. Navy electronics technician to test large batteries aboard the aircraft carrier USS Nimitz and has a small resistance that can dissipate large amounts of power. (credit: U.S. Navy photo by Photographer’s Mate Airman Jason A. Johnston) The small device is used on small batteries and has a digital display to indicate the acceptability of their terminal voltage. (credit: Keith Williamson)

Some batteries can be recharged by passing a current through them in the direction opposite to the current they supply to a resistance. This is done routinely in cars and batteries for small electrical appliances and electronic devices, and is represented pictorially in this figure. The voltage output of the battery charger must be greater than the emf of the battery to reverse current through it. This will cause the terminal voltage of the battery to be greater than the emf, since \(V=\text{emf}-\text{Ir}\), and \(I\) is now negative.

The diagram shows a car battery being charged with cables from a battery charger. The current flows from the positive terminal of the charger to the positive terminal of the battery, through the battery and back out the negative terminal of the battery to the negative terminal of the charger.

A car battery charger reverses the normal direction of current through a battery, reversing its chemical reaction and replenishing its chemical potential.

Terminal Voltage

Terminal Voltage

Example: Calculating Terminal Voltage, Power Dissipation, Current, and Resistance: Terminal Voltage and Load

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