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    Heat And Power Developed In A Resistor MCQ - Practice Questions with Answers

    Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

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    • 60 Questions around this concept.

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    The resistive network shown below is connected to a D.C. source of 16 V. The power consumed by the network is 4 watts. The value of R (in $\Omega$) is :

    In the circuit shown, the resistance r is a variable resistance. If for $r=f R$, the heat generation in r is maximum then the value of $f$ is:

    If the voltage across a bulb rated 220 volt - 100-watt drops by 2.5% of its rated value, the percentage of the rated value by which the power would decrease is:

    A filament bulb (500 W, 100 V) is to be used in a 230 V main supply. When a resistance R is connected in series, it works perfectly and the bulb consumes 500 W. The value of R

    The charge flowing through a resistance R varies with time t as Q = at - bt2, where a and b are positive constants. The total heat produced in R is:

    Two electric bulbs have tungsten filaments of the same length. If one of them gives 60 \mathrm{~W} and the other 100 \mathrm{~W}, then

    A heater coil is cut into two equal parts and only one part is now used in the heater. The heat generated will now be

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    A line having a total resistance of 0.5\Omega delivers \mathrm{15\: kW} at \mathrm{240W} to a small factory. The efficiency of transmission is near -

    A capacitor of capacitance \mathrm{C} has \mathrm{Q}. It's connected to an identical capacitor through a resistance. Find heat produced in the resistance.
     

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    An electric bulb rated for \mathrm{500 \mathrm{~W} \: at \: 100 \mathrm{~V}} is used in a circuit having \mathrm{200 \mathrm{~V}} supply. Find the resistance \mathrm{R} that must be put in series with the buck, so that the bulb draws \mathrm{500 \mathrm{~W}}.
     

    Concepts Covered - 1

    Heat and power developed in a resistor
    1. Heat and Power developed in a resistor

    Heat developed in a resistor: When a steady current flows through a resistance R for time t , the loss in electric potential energy appears as increased thermal energy(Heat H) of resistor and $H=i^2 R t$

    The power developed $=\frac{\text { energy }}{\text { time }}=i^2 R=i R=\frac{V^2}{R}$
    (from Ohm's law)
    Unit of heat is the joule ( J )
    Unit of power is watt (W)

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    Heat and power developed in a resistor

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