TEST OF PRACTICAL KNOWLEDGE QUESTION You are provided with a battery of e.m.f. E, a key K, a voltmeter, a standard resistor R0 = \(2\Omega\), a resistance b...

Determination of the e.m.f. and internal resistance of a battery

The e.m.f. of the battery, measured with the key open, is:

\(E=2.00\ \text{V}\)

With \(R_0=2\ \Omega\), the resistance box was set successively to \(2, 5, 10, 12, 15\) and \(20\ \Omega\). The voltmeter was connected across the standard resistor \(R_0\).

Table of readings

Graph of \(R\) against \(V^{-1}\)

The graph is a straight line. Using two well-separated points on the line, \((V^{-1}=1.25\ \text{V}^{-1},\ R=2.0\ \Omega)\) and \((V^{-1}=5.75\ \text{V}^{-1},\ R=20.0\ \Omega)\):

\[s=\frac{\Delta R}{\Delta(V^{-1})}=\frac{20.0-2.0}{5.75-1.25}=\frac{18.0}{4.50}=4.0\ \Omega\,\text{V}.\]

From \(R=sV^{-1}+c\),

\[c=R-sV^{-1}=2.0-(4.0\times1.25)=-3.0\ \Omega.\]

Hence,

\[\alpha=\frac{s}{R_0}=\frac{4.0}{2.0}=2.0\ \text{V},\]

\[\beta=-c-R_0=-(-3.0)-2.0=1.0\ \Omega.\]

Therefore, \(\alpha\), the e.m.f. of the battery, is \(2.0\ \text{V}\), while \(\beta\), the internal resistance of the battery, is \(1.0\ \Omega\).

Precautions

1. The key was closed only while taking a reading, to prevent heating and polarization of the battery.
2. All connections were clean and tight, and the voltmeter was read at eye level to avoid parallax error.