When a 20.0-ohm resistor is connected across the terminals of a 12.0-V battery, the voltage across the terminals of the battery falls by 0.3

Question

When a 20.0-ohm resistor is connected across the terminals of a 12.0-V battery, the voltage across the terminals of the battery falls by 0.300 V. What is the internal resistance of this battery

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Yến Oanh 4 years 2021-08-13T03:48:58+00:00 1 Answers 110 views 0

Answers ( )

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    2021-08-13T03:50:25+00:00

    Answer:

    The  internal resistance is  r =  0.5 \ \Omega

    Explanation:

    From the question we are told that the resistance of

       The  resistance of the resistor is  R  =  20.0\  \Omega

        The  voltage is V  = 12.0 \ V

         The magnitude of the voltage fall is  e   =  0.300\  V

    Generally the current flowing through the terminal due to the voltage of the battery  is  mathematically represented as

            I  =  \frac{V}{R}

    substituting values

            I  =  \frac{12.0 }{20 }

           I  =  0.6 \ A

    The internal resistance of the battery is mathematically represented as

          r =  \frac{e}{I}

    substituting values

         r =  \frac{0.300}{ 0.6 }

        r =  0.5 \ \Omega

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