Question

An auto mechanic needs to determine the emf and internal resistance of an old battery. He performs two measurements: in the first, he applies a voltmeter to the battery's terminals and reads 11.9 V;11.9 V; in the second, he applies an ammeter to the terminals and reads 16.1 A.16.1 A.
What are the battery's emf E and internal resistance r?

Answer 1

Answer:

Hence the battery's emf E is ε = 11.9 V.

The internal resistance is r = 0.739 ohms.

Explanation:

Now we know that

Voltage V = 11.9 V.

Current I = 16.1 A.

Hence this is an ideal voltmeter there are no current flows when the Voltmeter is applied.

ε = V + I r

∵ I = 0

ε = V

ε = 11.9 V

Then the ammeter is applied.

Let's take ( r ) to be the total resistance which is equal to internal resistance.

V = I r

r = $\frac{V}{I}$

 $= \frac{11.9}{16.1}$

r = 0.739 ohms

Answer 2

The battery's emf (E) and internal resistance (r) are 11.9 Volts and 0.739 Ampere respectively.

Given the following data:

• Voltage = 11.9 Volts.

• Current = 16.1 Amperes.

To determine the battery's emf (E) and internal resistance (r):

How to calculate emf (E).

For an ideal voltmeter, there isn't a flow of current and as such the current is equal to 0.

Mathematically, emf (E) is given by this formula:

$E = V + IR$

Substituting the given parameters into the formula, we have;

$E = 11.9 + 0R\\\\E = 11.9 + 0$

E = 11.9 Volts.

For the internal resistance (r):

Note: The total resistance is equal to internal resistance.

Applying Ohm's law, we have:

$R = \frac{V}{I} \\\\R = \frac{11.9}{16.1}$

R = r = 0.739 Ampere.

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