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89 Questions around this concept.
The ratio of thermal energy released in two resistors R and 3R connected in parallel in an electric circuit is :
A cell of emf 6V and internal resistance $1 \Omega$ is connected $2 \Omega$ across. The power dissipated in $2 \Omega$ in watts is
A heater of 220 V heats a volume of water in 5 minute time. A heater of 110 V heats the same volume of water in
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A 100 V, 40 watt bulb is connected across 220 V DC supply. The power consumption by bulb is:
In the circuit shown in the figure, the heat produced in 5 ohm resistance is 10 calories per second. The heat produced in 4 resistance is
In the circuit as shown in the figure, the heat produced by 6 ohm resistance due to current flowing in it is 60 calorie per second. The heat generated across 3 ohm resistance per second will be
A 100 V, the 40walt bulb is connected across a 220V 0c supply. The power consumed by the bulb is
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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 $=\frac{\text { energy }}{\text { time }}=i^2 R=i R=\frac{V^2}{R} \quad$ (from Ohm's law)
Unit of heat is the joule (J)
Unit of power is watt (W)
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