To practice Problem-Solving Strategy 25.1 Power and Energy in Circuits. A device

Question

To practice Problem-Solving Strategy 25.1 Power and Energy in Circuits. A device for heating a cup of water in a car connects to the car's battery, which has an emf E = 10.0 V and an internal resistance rint = 0.08 . The heating element that is immersed in the cup of water is a resistive coil with resistance R. David wants to experiment with the device, so he connects an ammeter into the circuit and measures 10.0 A when the device is connected to the car's battery. From this, he calculates the time to boil a cup of water using the device. If the energy required is 100 kJ , how long does it take to boil a cup of water?

D) The energy delivered to the resistive coil is dissipated as heat at a rate equal to the power input of the circuit. However, not all of the energy in the circuit is dissipated by the coil. Because the emf source has internal resistance, energy is also dissipated by the battery as heat. Calculate the rate of dissipation of energy Pbat in the battery.

Explanation / Answer

E = 10 volts

r = 0.08 ohms

Let R is the resiatnce of the heating element.

we know, Rnet = r + R

I = E/Rnet

Rnet = E/I

r + R = E/I

R = E/I - r

= 10/10 - 0.08

= 0.92 ohms

so, power dissipeted at heating element, P = I^2*R

= 10^2*0.92

= 92 W

so, time taken to produce 100kJ thermal energy = thermal energy/Power

= 100*10^3/92

= 1087 s or 18.12 minutes <<<<<<----------Answer

D) Power dissiapted in battery, P_bat = I^2*r

= 10^2*0.08

= 8 Watts <<<<<<----------Answer

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