6.5: Real Batteries
- Page ID
- 122199
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\dsum}{\displaystyle\sum\limits} \)
\( \newcommand{\dint}{\displaystyle\int\limits} \)
\( \newcommand{\dlim}{\displaystyle\lim\limits} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\(\newcommand{\longvect}{\overrightarrow}\)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)By the end of this section, you will be able to:
- Define the internal resistance of a battery.
- Calculate the terminal voltage of a real battery based on its source voltage and internal resistance.
Internal Resistance and Terminal Voltage
Most real sources of potential will have some internal losses. For a real battery, the amount of resistance to the flow of current within the voltage source is called the internal resistance. The internal resistance \(r\) of a battery can behave in complex ways. It generally increases as a battery is depleted, due to the oxidation of the plates or the reduction of the acidity of the electrolyte. However, internal resistance may also depend on the magnitude and direction of the current through a voltage source, its temperature, and even its history. The internal resistance of rechargeable nickel-cadmium cells, for example, depends on how many times and how deeply they have been depleted. A simple model for a battery consists of an idealized source voltage \(\varepsilon\) and an internal resistance \(r\) (Figure \(\PageIndex{1}\)).
Suppose an external resistor, known as the load resistance \(R\), is connected to a voltage source such as a battery, as in Figure \(\PageIndex{2}\). The figure shows a model of a battery with a source voltage \(\varepsilon\), an internal resistance \(r\), and a load resistor \(R\) connected across its terminals. Using conventional current flow, positive charges leave the positive terminal of the battery, travel through the resistor, and return to the negative terminal of the battery. The terminal voltage of the battery depends on the source voltage, the internal resistance, and the current, and is equal to
\[V_{terminal} = \varepsilon - Ir\]
For a given source voltage and internal resistance, the terminal voltage decreases as the current increases due to the potential drop \(Ir\) of the internal resistance.
A graph of the potential difference across each element of the circuit is shown in Figure \(\PageIndex{3}\). A current \(I\) runs through the circuit, and the potential drop across the internal resistor is equal to \(Ir\). The terminal voltage is equal to \(\varepsilon - Ir\), which is equal to the potential drop across the load resistor \(IR = \varepsilon - Ir\). As with potential energy, it is the change in voltage that is important. When the term “voltage” is used, we assume that it is actually the change in the potential, or \(\Delta V\). However, \(\Delta\) is often omitted for convenience. In this example, the connecting wires from Point a to Point c and from Point e to Point a are modeled as ideal wires meaning that they have negligible resistance and therefore negligible voltage change.
The current through the load resistor is \(I = \frac{\varepsilon}{r + R}\). We see from this expression that the smaller the internal resistance \(r\), the greater the current the voltage source supplies to its load \(R\). 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.
A given battery has a 12.00-V emf and an internal resistance of \(0.100 \, \Omega\). (a) Calculate its terminal voltage when connected to a \(10.00\text{-}\Omega\) load. (b) What is the terminal voltage when connected to a \(0.500\text{-}\Omega\) load? (c) What power does the \(0.500\text{-}\Omega\) load dissipate? (d) If the internal resistance grows to \(0.500 \, \Omega\), find the current, terminal voltage, and power dissipated by a \(0.500\text{-}\Omega\) load.
Solution
PLAN
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 by using the equation \(V_{\mathrm{terminal}} = \varepsilon - Ir\). Once current is found, we can also find the power dissipated by the resistor. The connecting wires to the resistor will be modeled as ideal (negligible resistance).
SKETCH
The circuits under consideration have the form shown in Fig. \(\PageIndex{4}\)
CALCULATE
- Entering the given values for the emf, load resistance, and internal resistance into the expression above yields \[I = \frac{\varepsilon}{R + r} = \frac{12.00 \, \mathrm{V}}{10.10 \, \Omega} = 1.188 \, \mathrm{A}. \nonumber \] Enter the known values into the equation\(V_{\mathrm{terminal}} = \varepsilon - Ir\) to get the terminal voltage: \[V_{\mathrm{terminal}} = \varepsilon - Ir = 12.00 \, \mathrm{V} - (1.188 \, \mathrm{A})(0.100 \, \Omega) = 11.90 \, \mathrm{V}. \nonumber \] The terminal voltage here is only slightly lower than the emf, implying that the current drawn by this light load is not significant.
- Similarly, with \(R_{\mathrm{load}} = 0.500 \, \Omega\), the current is \[I = \frac{\varepsilon}{R + r} = \frac{12.00 \, \mathrm{V}}{0.600 \, \Omega} = 20.00 \, \mathrm{A}. \nonumber \] The terminal voltage is now \[V_{\mathrm{terminal}} = \varepsilon - Ir = 12.00 \, \mathrm{V} - (20.00 \, \mathrm{A})(0.100 \, \Omega) = 10.00 \, \mathrm{V}. \nonumber \] The terminal voltage exhibits a more significant reduction compared with emf, implying \(0.500 \, \Omega\) is a heavy load for this battery. A “heavy load” signifies a larger draw of current from the source but not a larger resistance.
- The power dissipated by the \(0.500 \, \Omega\) load can be found using the formula \(P = I^2R\). Entering the known values gives \[P = I^2R = (20.0 \, \mathrm{A})^2(0.500 \, \Omega) = 2.00 \times 10^2 \, \mathrm{W}. \nonumber \] Note that this power can also be obtained using the expression \(\frac{V^2}{R}\) or \(IV\), where \(V\) is the terminal voltage (10.0 V in this case).
- 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 \[I = \frac{\varepsilon}{R + r} = \frac{12.00 \, \mathrm{V}}{1.00 \, \Omega} = 12.00 \, \mathrm{A}. \nonumber \] Now the terminal voltage is \[V_{\mathrm{terminal}} = \varepsilon - Ir = 12.00 \, \mathrm{V} - (12.00 \, \mathrm{A})(0.500 \, \Omega) = 6.00 \, \mathrm{V}, \nonumber \] and the power dissipated by the load is \[P = I^2R = (12.00 \, \mathrm{A})^2(0.500 \, \Omega) = 72.00 \, \mathrm{W}. \nonumber \] We see that the increased internal resistance has significantly decreased the terminal voltage, current, and power delivered to a load.
CHECK
The internal resistance of a battery can increase for many reasons. For example, the internal resistance of a rechargeable battery increases as the number of times the battery is recharged increases. The increased internal resistance may have two effects on the battery. First, the terminal voltage will decrease. Second, the battery may overheat due to the increased power dissipated by the internal resistance.
If you place a wire directly across the two terminals of a battery, effectively shorting out the terminals, the battery will begin to get hot. Why do you suppose this happens?
- Solution
-
If a wire is connected across the terminals, the load resistance is close to zero, or at least considerably less than the internal resistance of the battery. Since the internal resistance is small, the current through the circuit will be large, \(I = \frac{\varepsilon}{R + r} = \frac{\varepsilon}{0 + r} = \frac{\varepsilon}{r}\). The large current causes a high power to be dissipated by the internal resistance \((P = I^2r)\). The power is dissipated as heat.
Battery Testers
Battery testers, such as those in Figure \(\PageIndex{5}\), use small load resistors to intentionally draw current to determine whether the terminal potential drops below an acceptable level. Although it is difficult to measure the internal resistance of a battery, battery testers can provide a measurement of 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 an appliance. This is done routinely in cars and in batteries for small electrical appliances and electronic devices (Figure \(\PageIndex{6}\)). The voltage output of the battery charger must be greater than the source voltage of the battery to reverse the current through it. This causes the terminal voltage of the battery to be greater than the source voltage, since \(V = \varepsilon - Ir\) and \(I\) is now negative.
It is important to understand the consequences of the internal resistance of devices which provide source voltage, such as batteries and solar cells, but often, the analysis of circuits is done with the terminal voltage of the battery, as we have done in the previous sections. The terminal voltage is referred to as simply as \(V\), dropping the subscript “terminal.” This is because the internal resistance of the battery is difficult to measure directly and can change over time.
Contributors and Attributions
Samuel J. Ling (Truman State University), Jeff Sanny (Loyola Marymount University), and Bill Moebs with many contributing authors. This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).