An external resistor R is connected between the terminals of...

An external resistor $R$ is connected between the terminals of a battery. The value of $R$ varies. For each $R$ value, the current $I$ in the circuit and the terminal voltage $V_{a b}$ of the battery are measured. The results are plotted in Fig. $\mathbf{P} 25.72,$ a graph of $V_{a b}$ versus $I$ that shows the best straight-line fit to the data. (a) Use the graph in Fig. $\mathrm{P} 25.72$ to calculate the battery's emf and internal resistance. (b) For what value of $R$ is $V_{a b}$ equal to $80.0 \%$ of the battery emf?

Key Concepts

Graphical Analysis of Circuit Parameters

Graphical analysis, such as plotting terminal voltage against current, provides a visual and mathematical method to extract key battery parameters. The y-intercept of the plot corresponds to the battery's EMF, while the slope (when taken with the correct sign) is related to the internal resistance. This technique underscores the application of linear regression in experimental physics and engineering to determine circuit characteristics.

Voltage Divider Principle

The voltage divider principle explains how the total voltage in a series circuit divides among the components according to their resistances. In the context of a battery with both internal and external resistances, this principle is used to determine the proportion of the battery's EMF that appears across the external resistor. This concept is essential when setting conditions such as the terminal voltage being a specific fraction of the EMF.

Internal Resistance

Internal resistance is an inherent property of a battery that causes a voltage drop when current flows. It is represented in circuit equations as the resistance in series with the EMF. This resistance affects the terminal voltage and can be determined from the slope of a terminal voltage versus current graph, which reflects how voltage decreases as current increases.

Battery Electromotive Force (EMF)

The EMF is the ideal voltage provided by a battery when no current flows. It is a measure of the chemical energy available for conversion into electrical energy. In circuit analysis, the EMF appears as a constant value that is reduced by any voltage drop due to internal resistance when the battery is under load.

Terminal Voltage

The terminal voltage is the actual voltage measurable at the battery’s output terminals when it is connected to a load. It is lower than the EMF by the voltage drop across the battery's internal resistance. The relationship is given by the formula V = EMF - I·(internal resistance), which is central to understanding battery performance under load.

Transcript

00:01 Hi, in the given problem here, this is the graphical representation between the variation of terminal voltage of the cell with the current passing through the circuit.

00:26 Here this is terminal voltage vab, shown along y -axis and current in the circuit, shown.

00:36 Shown along x -axis.

00:40 If we look for the values of current and terminal voltages, then we find that at 6 .00 ampere, the terminal voltage across the resistance, external resistance was 24 .0 volt, whether we can say this is the potential drop across external resistance or terminal voltage across the cell.

01:14 The same thing.

01:16 And then if you look at the value of current to be 3 .00 ampere corresponding to this value of current, the terminal voltage across the cell is found to be 30 .0 volt.

01:38 Now, using the expression for terminal potential difference, vab, that is given as emf minus i into r, where this epsilon is emf of the cell, and this r is its internal resistance.

02:08 So in order to find the value of emf and internal resistance, first of all, we look, we put the values of vab and i using 6 amp as current and 24 volt as vab.

02:28 So we get 24 .0 volt is equal to emf is missing minus 6 .00 into r, internal resistance, which is also missing.

02:43 Then for the second set of values, 30 .0 volt for terminal potential difference and emf epsilon minus for current, this is 3 .00r, internal resistance.

03:01 Now, if we subtract these two equations, here this is minus, minus, plus, we will get here, this is minus 6 .0 is equal to canceling this here we will get minus 3 .00 r.

03:22 So finally, canceling this negative sign from both the sides, the value of internal resistance of the cell comes out to be 2 .00 oom, which is one of the answer of this first part of the problem.

03:40 Now, putting this value of internal resistance of the cell in this one of the equations, 24 .0 is equal to e minus 6 .00 r...

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