Terminal Voltage
The voltage output of a device is measured across its terminals and, thus, is called its terminal voltagenull Terminal voltage is given by
4.44 null
where null is the internal resistance and null is the current flowing at the time of the measurement.
null is positive if current flows away from the positive terminal, as shown in Figure 4.9. 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 null is connected to a voltage source, as in Figure 4.12. Since the resistances are in series, the total resistance in the circuit is null Thus, the current is given by Ohm’s law to be
4.45 null
We see from this expression that the smaller the internal resistance null the greater the current the voltage source supplies to its load null As batteries are depleted, null increases. If null becomes a significant fraction of the load resistance, then the current is significantly reduced, as the following example illustrates.
Example 4.4 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 null (a) Calculate its terminal voltage when connected to a null load. (b) What is the terminal voltage when connected to a null load? (c) What power does the null load dissipate? (d) If the internal resistance grows to null find the current, terminal voltage, and power dissipated by a null 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 null 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
4.46 null
Enter the known values into the equation null to get the terminal voltage.
4.47
Discussion for (a)
The terminal voltage here is only slightly lower than the emf, implying that null is a light load for this particular battery.
Solution for (b)
Similarly, with null the current is
4.48 null
The terminal voltage is now
4.49
Discussion for (b)
This terminal voltage exhibits a more significant reduction compared with emf, implying that null is a heavy load for this battery.
Solution for (c)
The power dissipated by the null load can be found using the formula null Entering the known values gives
4.50 null
Discussion for (c)
Note that this power can also be obtained using the expressions null or null where null 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
4.51 null
Now the terminal voltage is
4.52
and the power dissipated by the load is
4.53 null
Discussion for (d)
We see that the increased internal resistance has significantly decreased terminal voltage, current, and power delivered to a load.
Applying the Science Practices: Internal Resistance
The internal resistance of a battery can be estimated using a simple activity. The circuit shown in the figure below includes a resistor R in series with a battery along with an ammeter and voltmeter to measure the current and voltage, respectively.
The currents and voltages measured for several R values are shown in the table below. Using the data given in the table and applying graphical analysis, determine the emf and internal resistance of the battery. Your response should clearly explain the method used to obtain the result.
Resistance |
Current (A) |
Voltage (V) |
RI |
3.53 |
4.24 |
RII |
2.07 |
4.97 |
RIII |
1.46 |
5.27 |
RIV |
1.13 |
5.43 |
Table 4.1
Answer
Plot the measured currents and voltages on a graph. The terminal voltage of a battery is equal to the emf of the battery minus the voltage drop across the internal resistance of the battery, or V = emf – Ir. Using this linear relationship and the plotted graph, the internal resistance and emf of the battery can be found. The graph for this case is shown below. The equation is V = –0.50I+6.0, and hence the internal resistance will be equal to 0.5 Ω and emf will be equal to 6 V.
Battery testers, such as those in Figure 4.15, 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.
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 Figure 4.16. 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 null and null is now negative.