Question

Andy wants to measure the internal resistance of a 9V battery. When he connects the battery directly to a voltmeter, the meter measure 9.0V. When he connects the battery with a 10ohm resistor and use a voltmeter to measure the voltage across the resistor, the meter measure 8.5V. What is the resistance of the internal resistor? What are current going through the battery in both situations? How would you connect the ammeter in order to measure the current going through the internal resistor?

Answer 1

Answer on Question #71500, Physics / Molecular Physics | Thermodynamics

Question. Andy wants to measure the internal resistance of a $9V$ battery. When he connects the battery directly to a voltmeter, the meter measure $9.0V$ . When he connects the battery with a 10 ohm resistor and use a voltmeter to measure the voltage across the resistor, the meter measure $8.5V$ .

What is the resistance of the internal resistor?

What are current going through the battery in both situations?

How would you connect the ammeter in order to measure the current going through the internal resistor?

Given.

$U = 9V; U_{0} = 9.0V; U_{1} = 8.5V; R = 10\, ohm\, v;$

Find.

$r, I_0, I - ?$

Solution.

When the voltmeter is connected to the battery and it shows $U_{0} = 9.0V$ , then its resistance $R_{V}\rightarrow \infty$ and $\mathcal{E} = U_0 = 9.0V$ ( $\mathcal{E}$ is the electromotive force) and accordingly $I_0 = 0$ .

The current across the resistor

$$I = \frac {U _ {1}}{R} = \frac {8 . 5}{1 0} = 0. 8 5 A.$$

So

$$\mathcal {E} = U _ {0} = 9. 0 V.$$ $$I = \frac {\mathcal {E}}{R + r} \rightarrow I R + I r = \mathcal {E} \rightarrow r = \frac {\mathcal {E} - I R}{I} = \frac {9 . 0 - 0 . 8 5 \cdot 1 0}{0 . 8 5} = 0. 5 8 8 o h m$$

In order to measure current through the battery, the ammeter should be connected in series.

Answer. $r = 0.588 \, ohm$ ; $I_0 = 0$ ; $I = 0.85 \, A$ ; the ammeter should be connected in series.

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