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15 December, 13:09

An unknown resistor is connected between the terminals of a 3.00 V battery. Energy is dissipated in the resistor at the rate of 0.560 W. The same resistor is then connected between the terminals of a 1.50 V battery. At what rate is energy now dissipated?

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  1. 15 December, 16:44
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    The energy dissipate at a rate of 11.34 W when the resistor is connected between the terminals of a 1.50 V battery.

    Explanation:

    the rate at which energy dissipeted is given by power and we know that when given resistance, R and voltage, V then power is given by:

    P = V^2/R

    R = V^2/P

    = (3.00) ^2 / (0.560)

    = 5.04 ohms

    then when the resistor is now connected between the terminals of a 1.50 V battery, then the power is given by:

    P = V^2/R

    = (1.50) ^2 / (5.40)

    = 11.34 W.

    Therefore, the energy dissipate at a rate of 11.34 W when the resistor is connected between the terminals of a 1.50 V battery.
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