Electrical Energy and Power

For a simple circuit consisting of a battery connected to a resistor R, the positive terminal of the battery (the longer plate) is at the higher potential while the negative terminal (the shorter plate) is at the lower potential. For a positive charge dQ moving around the circuit from point a

thru the battery and resistorand back to point a, point a is areference point that isgrounded and its potential is0 V.

As the charge moves from a tob thru the battery, its electricalpotential energy increases byan amount V·dQ while the chemical potential energy decreases by the same amount.

As the charge moves from c to d thru the resistor, it loses this electrical potential energy as it undergoes collisions with atoms in the resistor, producing thermal energy.

Neglect the resistance of the connecting wires, so there is no energy loss for paths bc and da.

When the charge returns to point a, it must have the same potential energy (O) as it had at the start.

The figure on the rightrepresents a circuit element with potentialdifference Va – Vb = Vab

between its terminals and current I passing thru it in the direction from a to b. The circuit element could be a battery, a

resistor, or something else. As charge passes thru the circuit element,

the electric field does work on the charge. As an amount of charge q passes thru the

circuit element, there is a change in potential energy equal to q·Vab.

If q > 0 and Vab = Va – Vb is positive, potential energy decreases as the charge falls from potential Va to lower potential Vb.

The moving charges don’t gain kinetic energy because the rate of charge flow (current) out of the circuit element must be the same as the rate of charge flow into the element.

Instead, the quantity q·Vab represents electrical energy transferred into the circuit element. This energy is often converted into another form of energy (thermal, EM).

If the potential at b is higher than the potential at a, Vab is negative and a net transfer of energy out of the circuit element occurs. The element acts as a source of energy (a battery).

The circuit element delivers electrical energy into the circuit to which it is attached.

A battery usually converts chemical energy into electrical energy and delivers it to the external circuit.

q·Vab.can denote either a quantity of energy delivered to a circuit element or a quantity of energy extracted from that element.

In electric circuits, the rate at which energy is delivered to or extracted from a circuit element is important. If the current thru the element is I, then in a time interval

dt an amount of charge dQ = I·dt passes thru the element.

The potential energy change for this amount of charge is Vab·dQ = Vab·I·dt.

Dividing by dt gives us the rate at which energy is transferred either into or out of the circuit element.

The time rate of energy transfer is power P: P = Vab·I.

Unit: Watt (J/s = V·A) Power Input to a Pure Resistance:

If the circuit element is a resistor,the potential difference is Vab = I·R.

The electrical power delivered tothe resistor by the circuit is:

The potential at a (where the current enters the resistor) is always higher than that at b (where the current exits).

Current enters the higher potential terminal of the device and transfers electric potential energy into the circuit element.

What happens to the energy? The moving charges collide with atoms in the

resistor and transfer some of their energy to these atoms, increasing the internal energy of the material.

Either the temperature of the resistor increases or there is a flow of heat out of it, or both.

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Energy is dissipated in the resistor at the rate I2·R. Every resistor has a power rating, the maximum

power the resistor can dissipate without becoming overheated and damaged.

Power Output of a Source: For the car battery connected to a headlight,

point a is at higher potential than pointb, so Va > Vb and Vab is positive.

The current I is leaving thesource at the higher potential terminal andenergy is being delivered tothe external circuit.

The rate of energy delivery to the circuit is: P = Vab·I.

For a source that can be described by EMF and an internal resistance r: Vab = EMF - I·r

Multiplying the equation by I: P = Vab·I = EMF·I - I2·r

EMF represents the work per unit charge the battery performs on the charges by the nonelectrical force in the battery that pushes the charges from point b to point a in the battery.

In a time dt, a charge dQ = I·dt flows thru the battery; the work done on it by the nonelectrostatic force is EMF·dQ = EMF·I·dt.

EMF·I is the rate at which work is done on the charges by the battery and represents the rate of conversion of nonelectrical energy to electrical energy within the battery.

I2·r is the rate at which electrical energy is dissipated in the internal resistance of the source.

The difference EMF·I - I2·r is the net electrical power output of the battery, the rate at which the battery delivers electrical energy to the external circuit.

Power Input to a Source: In a car, the alternator is

connected to the battery. The battery is charged by

the alternator.

The EMF of the alternator is larger than the EMF of the battery.

The current I in the circuit travels from the alternator to the battery, from the larger EMF to the smaller EMF.

Because of the reversal of current, for the battery:

Vab = EMF + I·r and P = Vab·I = EMF·I + I2·R Work is being done on the battery and there is

a conversion of electrical energy into nonelectrical energy in the battery at a rate EMF·I.

The I2·R term is the rate of dissipation of energy in the internal resistance of the battery.

The sum EMF·I + I2·R is the total electrical power input to the battery.

This also happens when a rechargeable battery (a storage battery)is connected to a charger.

The charger supplies electrical energy to the battery; part of it is converted to chemical energy, to be reconverted later, and the rest is dissipated (wasted) in the battery’s internal resistance, warming the battery and causing heat to flow out of it.

Power companies sell energy in kW·hr;Energy = power·time; E = P·t

Energy is usually expressed in Joules, but electric companies use kW·hr.

1 kW·hr = 3.6 x 106 J

The cost of operating an electrical device is:

Energy Transmission thru Power Lines: When transporting electrical energy thru

power lines, power companies want to minimize power transformed into internal energy in the lines and maximize the energy delivered to the other end of the transmission line.

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Because P = I·V, the same amount of power can be moved either at high currents and low potential differences or at low currents and high potential differences.

Power companies choose to move electrical energy at low currents and high potential differences for economic reasons:

Copper wire is very expensive and so it is cheaper to use high resistance wire (wire having a small cross-sectional area).

Power delivered to a resistor is I2·R, so a high resistance wire will result in a small current, reducing the I2·R loss (called Joule heating) in the wire.

Transformers are used to step-up and step-down the potential difference of the power in the transmission line.

Efficiency of transmission:

Power delivered to the line = power lost in line + power delivered by the line

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