An ideal battery would produce an extraordinarily large current if “shorted” by connecting the positive and negative terminals with a short

Question

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=____

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Kim Cúc 3 years 2021-09-03T18:26:49+00:00 1 Answers 99 views 0

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    2021-09-03T18:28:38+00:00

    Answer:

    Resistance, R=0.529\ \Omega

    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 :

    V=IR

    R=\dfrac{V}{I}

    R=\dfrac{9\ V}{17\ A}

    R=0.529\ \Omega

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

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