The amount of work done by an emf in moving an amount of charge ΔQ through a potential difference V is ΔW = ΔQ*V.  (Voltage is defined as work per unit charge.)  The potential energy gained by ΔQ is ΔU = ΔW.  The rate at which the emf does work is ΔW/Δt = (ΔQ/Δt)*V = I*V.  This is the power P supplied by the emf.

P = I*V.

The emf provides the potential energy of the separated charges.  The electric field then does work on the separated charges, converting this potential energy into some other form.  For a current flowing in a wire with resistance R, the voltage drop across the wire is  ΔV = IR.  Potential energy is converted into thermal energy through frictional forces.  The power dissipated by the wire is

P = I*V = I²R = V²/R,

using our definition of resistance.
For a given problem, which equation should you use?  Consider which quantities are given and which you would have to calculate first.

Problem:

A high-voltage transmission line that connects a city to a power plant consists of a pair of copper cables, each with a resistance of 4 Ω.  The current flows to the city along one cable, and back along the other.
(a)  The transmission line delivers to the city 1.7*10⁵ kW of power at 2.3*10⁵ V.  What is the current in the transmission line?  How much power is lost as Joule heat in the transmission line?
(b)  If the transmission line delivers the same 1.7*10⁵ kW of power at 110 V, how much power would be lost in Joule heat?  Is it more efficient to transmit power at high voltage or at low voltage?

Solution:

• Reasoning:
  The power (energy (J) per second) delivered to the city is P = IV.  The higher the voltage at which the power is delivered, the lower is the current that flows to the substation.  To get to the substation, the current has to flow through the transmission line wires, which have a fixed resistance R.  The power dissipated by the wires is I²R, so the lower the current, the lower is the amount of energy per second dissipated by the wires.
• Details of the calculation:
  (a) The power delivered to the city, P_delivered, is equal to 1.7*10⁵ kW.  The voltage V is equal to 2.3*10⁵ V.  The current through the cables therefore is I = P/V = 7.4*10²A.  This current is flowing through the transmission cables and through all the equipment hooked up to the transmission lines in the city.  Some of the electrical energy is converted into thermal energy in the transmission lines.  The electric energy dissipated in the cables per unit time is equal to
  P_lost = I²R = (7.4*10² A)²*8 Ω = 4.4*10⁶ W.
  The power generated by the power plant therefore is
  P = P_delivered + P_lost = 1.74*10⁵ kW.
  98% of the generated power is delivered to the city.
  
  (b) The power delivered to the city, P_delivered, is equal to 1.7*10⁵ kW.  The voltage V is equal to 110 V.  The current through the cables therefore is I = P/V = 1.6*10⁶ A.  This current is flowing through the transmission cables.  The electric energy dissipated in the cables is equal to
  P_lost = I²R = (1.6*10⁶ A)²*8 Ω = 1.9*10¹³W.
  The power generated by the power plant therefore is
  P = P_delivered + P_lost = 1.9*10¹³ kW.
  Only 0.0009% of the generated power is delivered to the city.

Transmission of electrical energy at high voltage is much more efficient that the transmission at low voltage.

The voltage of the transmission line is therefore reduced to 110 V as close as possible to the house of the customer.

Problem:

We want to heat up 0.5 liter of water with a resistive heater that carries 5 A at 120 V.  How long will it take to bring the water from 27 ^oC to the boiling point?  How much does this cost at $0.10/kWh?  The specific heat of water is 4.19 kJ/(kgC). 

Solution:

• Reasoning:
  The energy required to raise the temperature of the water by ΔT is ΔU = mcΔT, where m is the mass and c is the specific heat of the water.
  The power dissipated by the heater is P = IV = energy/time.  The time t required to convert an amount of electrical energy ΔU into thermal energy is t = ΔU/P.
• Details of the calculation:
  ΔU  = (10^-3 kg/cm³)(500 cm³)(4.19 kJ/(kg^oC))(73 ^oC) = 153 kJ.
  P = IV = (5 A)(120 V) = 600 W.
  The time t required to convert 153 kJ into thermal energy therefore is
  t = ΔU/P = (153000/600) s = 255 s = 4.3 minutes.
  The electrical energy converted into heat in units of kWh is
  U = (0.6 kW)(255 s)(1 hour/3600 s) = 0.043 kWh.
  At $0.10/kWh, we spend $0.0043 to heat the water.