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 = 0.04 Ω . 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 12.0 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?

The first thing we must do is calculate the external resistance (R) of the circuit, from the description we notice that it is a series circuit, by which the resistors are added

V = i (r + R)

We replace we calculate

r + R = V / i

R = v / i - r

R = 10/12 -0.04

R = 0.793 Ω

We calculate the power supplied

P = V i = I² R

P = 12² 0.793

P = 114 W

This is the power dissipated in the external resistance

We use the relationship, that power is work per unit of time and that work is the variation of energy

P = E / t

t = E / P

t = 100 10³/114

t = 877 s

The time to boil the water is 877 s