Physics · Ch 2 — Current Electricity

Determination of Internal Resistance

2.4.2

Determination of Internal Resistance

Once an external resistance R is switched into the circuit and a current I actually flows (Figure 2.20(b)), the voltmeter no longer reads the full emf $\varepsilon$ -- it instead reads a smaller value V, because a portion of the emf, equal to $Ir$ , is unavoidably used up driving the current through the battery's own internal resistance r. The potential drop across the external resistance itself is $V=IR$ (2.35), and comparing this to the emf gives

V = \varepsilon - Ir, \qquad\text{i.e.}\qquad Ir = \varepsilon - V \qquad (2.36)

Dividing equation (2.36) by equation (2.35) ( $IR=V$ ) gives $\dfrac{Ir}{IR}=\dfrac{\varepsilon-V}{V}$ , which rearranges to

r = \left(\dfrac{\varepsilon}{V}-1\right)R \qquad (2.37)

Since $\varepsilon$ , V and R are all measurable, this single relation lets the internal resistance r be determined directly from the two voltmeter readings (open-circuit emf, and terminal voltage with R connected) together with the known value of R.

Because of this internal resistance, the power a battery actually delivers to the useful, external part of a circuit is always somewhat less than its ideal rating. The total power the battery supplies is $P=I\varepsilon = I(V+Ir)$ (from 2.36), and since $V=IR$ , this expands to

$P = I(IR+Ir) = I^2R + I^2r \qquad (2.38)$ …

Figure 2.20Internal resistance of the cell

What this figure shows. Panel (a) shows the same open-circuit voltmeter setup as Figure 2.19, cell of emf $\varepsilon$ and internal resistance r with only a voltmeter connected, reading the full emf. Panel (b) shows the circuit completed with an external resistance R now included and current I flowing; the voltmeter now reads a smaller terminal voltage V (equal to IR), because part of the emf, equal to Ir, has been used up driving the current thro …

Misc Example 2.17Terminal voltage, internal resistance and power split from a real battery

Worked out. A battery of emf 12 V connected to a $3\ \Omega$ resistor drives a current of 3.93 A; the terminal voltage, internal resistance, and power delivered by the battery and to the resistor are all required. (a) The terminal voltage equals the voltage across the resistor, $V=IR=3.93\times3=11.79$ V. The internal resistance is $r=(\varepsilon-V)/I=(12-11.79)/3.93\approx0.05\ \Omega$ . (b) The power delivered by the battery is $P=I\varepsilon=3.93\times12=47.1$ W, while the power actually delivered to the external resistor is $I^2R\approx46.3$ W; the remaining $47.1-46.3=0.8$ W (equal to $I^2r$ ) is dissipated uselessly inside the battery's own internal resistance and can never be de …

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