208l10.ppt

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If you want to make electrons flow through a resistor, you must produce a potential

difference between the ends of the device. To maintain a steady flow, a charge pump is

needed. This is a device that does work on charges while maintaining a steady potentialdifference across a pair of terminals. Such a device is referred to as an emfdevice.

Pumping Charges

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Work, Energy and EMF

An emf device performs two functions: it maintains a potential

difference and it moves charge from one terminal to the other inside thedevice

An ideal emf device lacks any internal resistance to the internal

movement of charges from terminal to terminal. For example, an ideal

battery with an emf of 12V always has a voltage of 12V between its

terminals

A real emf device, such as real battery, has internal resistance to the

internal movement of charge. When a real emf device is not connected

to circuit, the potential difference between its terminals is equal to its

emf. But when it has current through it, the potential difference

between its terminal, is different that the emf.

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Work, Energy and EMF

An emf device does positive work on the charge passing

through it if the current is from the negative to positive terminal

and negative work if the current is in the other direction

Positive work done by an emf device results in a decrease in thestore of energy of the device, chemical energy in the case of a

battery. Negative work results in an increase in the store of

energy of the device. If the device is a battery, then in the first

case it is discharging and in the second it is charging.

The potential difference across an ideal emf device does not

change with direction of current through the device.

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Calculating the Current in a Single Loop

There are two ways to calculate the current

in the simple single-loop circuit; one

method is based on energy conservation and

the other is based on potential. The circuit

consist of an ideal battery with emf, a

resistor R, and two connecting wires (zero

resistance).Energy method

The equation, P=i2 R, tells us that in a time interval dt an amount of energy given by i2 Rwill appear in

the resistor as thermal energy. During the same interval, a charge dq = idt will have move the battery, and

the battery will have done work on this charge, according to the definition ofemf, equal to

Form the principle of conservatio

n of energy, the work done by the battery must equal the

thermal energy in the resistor

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Energy method

Emf is the energy per unit charge transferred to the

moving charges by the battery. The quantity iRis the

energy per unit charge transferred from the moving

charge to thermal energy within the resistor. The energy

per unit charge transferred to the moving charges is

equal to the energy per charge transferred from them.Solving forI, we find

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Potential Method

Suppose we start at any point in the circuit

to the right and mentally proceed around the

circuit, adding algebraically the potential

difference that we encounter. When we

arrive at our starting point, we must havereturned to our starting potential. This is

called the loop rule in circuits

This is often referred to as Kirchoffs loop rule or voltage law (Kirchoff is a german physicist).

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Potential Method

Let us starting a point a, whose potential is Va, and

mentally walk clockwise around the circuit until we

are back at a, keeping track of the voltages along the

way. Our starting point is at the low-potential

terminal of the battery. Since the battery is ideal, the

potential difference between its terminals is zero. So

when we pass through the battery to high potential

terminal, the change in potential is +.

As we walk along the top wire to the top end of the resistor, there is not potential change because the

wire has negligible resistance so it is same as the high-potential terminal of the battery . When we

pass through the resistor, however the change in potential isiR.

We return to point a along the bottom wire. Again, since the wire has negligible resistance, we again

find no potential change. Because we transverse a complete loop, our initial potential, as modifiedfor potential changes along the way, must equal to our final potential; that is

Since Va cancels we can

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Potential Method

To prepare for circuits of greater complexity, let us lay down two global rules

for finding potential differences as we move around the a loop:

(a) rightward (b) All the same (c) b , a and c tie (d) b , a and c tie

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Internal Resistance

The figure shows a real battery with an

internal resistance r, wired to an externalresistor of resistance R. The internal

resistance of the battery is the electrical

resistance of the conducting materials of the

battery and is an intrinsic characteristic of

the battery which is not a removable part. In

the figure, however the battery is drawn as if

it could be separated into an ideal battery

with emfand a resistance r. The order ofthese separated parts drawn does not matter.

If we apply the loop rule clock wise we have

Solving for the currentIf r= 0, we have the ideal

battery relationship

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Resistors in Series

Figure (a) shows three resistances connected

in series to an ideal battery of emf. The

battery applies a potential difference V=across the three resistor combination.

In other words, the resistors occur one after another along a single path for the current.

We seek the resistance Req that is equivalent to the three-resistance series. By

equivalent, we mean that Reqcan replace the combination without changing the

current i through the combination or the potential difference between a and b. Let

us apply the loop rule, starting at terminal a and going clockwise around the circuit.

We findor

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Resistors in Series

If we replaced the three resistances with a

single equivalent resistance, we have thecircuit shown in (b) and we would find

Comparing the two equations we find

The extension to n resistances is straightforward

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Potential Differences

We often want to find the potential between

two points in a circuit. In the figure what is

pontential difference between pointsb and a ?

To find out, let us start at at pointb and

transverse the circuit clockwise to point a,passing through resistor R. IfVa and Vb are the

potentials at a and b, respectively, we have

because we experience a decrease in

potential in going through a resistance in the

direction of the current. We rewrite

Recalling ris the

internal resistance

We can rewrite the

difference between

point a and b in

terms of

To explicitly show the effects

of the internal resistance

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Potential Differences

Let us again calculate Va-Vb starting at

pointb, but this time proceeding

counterclockwise topoint a through the

battery. We have

or

Combining withWith some algebra we

once again obtain

Suppose we set=12V, R=10 and r= 2.0,The potential difference across the terminals is Va-Vb = 12V x 10 /(10 + 2.0) = 10V