## Electrical energy and power

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 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.

#### 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:
(a) The power delivered to the city, Pdelivered, 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
Plost = I2R = (7.4*102 A)2*8 Ω = 4.4*106 W.
The power generated by the power plant therefore is
P = Pdelivered + Plost = 1.74*105 kW.
98% of the generated power is delivered to the city.

(b) The power delivered to the city, Pdelivered, 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
Plost = I2R = (1.6*106 A)2*8 Ω = 1.9*1013W.
The power generated by the power plant therefore is
P = Pdelivered + Plost = 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:
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.
U = (10-3 kg/cm3)(500 cm3)(4.19 kJ/(kgoC))(73 oC) = 153 kJ.
The power dissipated by the heater is P = IV = (5 A)(120 V) = 600 W.
The time t required to convert 153 kJ into thermal energy therefore is
t = U/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.