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You take a lab battery and, when it isn't connected to a circuit, you measure a voltage of 4.00-V with a voltmeter. You then connect the battery to a 200-12 resistor and measure a current of 17.5-mA. What is the internal resistor of this battery?

Respuesta :

Answer:

[tex]r = 28.6 ohm[/tex]

Explanation:

As we know that when battery is in open circuit then the potential difference of the cell is known as EMF

so EMF is given as

[tex]EMF = 4.00 V[/tex]

now when the battery is connected across a resistance of 200 ohm then there is current flowing through the battery

it is given as

[tex]i = \frac{V}{R + r}[/tex]

[tex]17.5 \times 10^{-3} = \frac{4}{200 + r}[/tex]

[tex]200 + r = 228.6 [/tex]

[tex]r = 28.6 ohm[/tex]

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