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An ideal battery would produce an extraordinarily large current if "shorted" by connecting the positive and negative terminals with a short wire of very low resistance. Real batteries do not. The current of a real battery is limited by the fact that the battery itself has resistance.

What is the resistance of a 9.0 V battery that produces a 17 A current when shorted by a wire of negligible resistance?

R=____

Respuesta :

Answer:

Resistance, [tex]R=0.529\ \Omega[/tex]

Explanation:

Given that,

Voltage of the battery, V = 9 volts

Current produced in the circuit, I = 17 A

We need to find the resistance when shorted by a wire of negligible resistance. It is a case of Ohm's law. The voltage is given by :

[tex]V=IR[/tex]

[tex]R=\dfrac{V}{I}[/tex]

[tex]R=\dfrac{9\ V}{17\ A}[/tex]

[tex]R=0.529\ \Omega[/tex]

So, the resistance in the circuit is 0.529 ohms. Hence, this is the required solution.

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