5.2 PowerPoint

Report
Electricity and
Magnetism
Topic 5.2 Electric Circuits
Electromotive Force





Defining potential difference
The coulombs entering a lamp have electrical
potential energy;
those leaving have very little potential energy.
There is a potential difference (or p.d.) across
the lamp, because the potential energy of each
coulomb has been transferred to heat and light
within the lamp.
p.d. is measured in volts (V) and is often called
voltage.

The p.d. between two points is the
electrical potential energy transferred to
other forms, per coulomb of charge that
passes between the two points.
Resistors and bulbs transfer electrical
energy to other forms, but which
components provide electrical energy?
 A dry cell, a dynamo and a solar cell are
some examples.
 Any component that supplies electrical
energy is a source of electromotive
force or e.m.f.
 It is measured in volts.
 The e.m.f. of a dry cell is 1.5 V, that of a
car battery is 12 V





A battery transfers chemical energy to
electrical energy, so that as each coulomb
moves through the battery it gains electrical
potential energy.
The greater the e.m.f. of a source, the more
energy is transferred per coulomb. In fact:
The e.m.f of a source is the electrical
potential energy transferred from other forms,
per coulomb of charge that passes through
the source.
Compare this definition with the definition of
p.d. and make sure you know the difference
between them.
Internal Resistance
The cell gives 1.5 joules of electrical
energy to each coulomb that passes
through it,
 but the electrical energy transferred in the
resistor is less than 1.5 joules per coulomb
and can vary.
 The circuit seems to be losing
energy - can you think where?




The cell itself has some resistance, its internal
resistance.
Each coulomb gains energy as it travels through
the cell, but some of this energy is wasted or
`lost' as the coulombs move against the
resistance of the cell itself.
So, the energy delivered by each coulomb to the
circuit is less than the energy supplied to each
coulomb by the cell.
Very often the internal resistance is small
and can be ignored.
 Dry cells, however, have a significant
internal resistance.
 This is why a battery can become hot
when supplying electric current.
 The wasted energy is dissipated as heat.

Resistance Combinations
Resistors in series


The diagram shows three resistors connected in
series
There are 3 facts that you should know for a
series circuit:



the current through each resistor in series is the same
the total p.d., V across the resistors is the sum of the
p.d.s across the separate resistors, so: V = Vl + V2 +
V3
the combined resistance R in the circuit is the sum of
the separate resistors
R = Rl + R2 + R3
 Suppose we replace the 3 resistors with
one resistor R that will take the same
current I when the same p.d. V is placed
across it



This is shown in the diagram. Let's calculate
R.
We know that for the resistors in series:






V = Vl + V2 + V3
But for any resistor: p.d. = current x resistance
(V = I R).
If we apply this to each of our resistors, and
remember that the current through each
resistor is the same and equal to I, we get:
IR = IRl+IR2+IR3
If we now divide each term in the equation by
I,
we get:

R = R1 + R2 + R3
Resistors in parallel


We now have three resistors connected in
parallel:
There are 3 facts that you should know for a
parallel circuit:




the p.d. across each resistor in parallel is the same
the current in the main circuit is the sum of the
currents in each of the parallel branches, so:
I = I1 + I2 + I3
the combined resistance R is calculated from the
equation:

Suppose we replace the 3 resistors with
one resistor R that takes the same total
current I when the same p.d. V is placed
across it.









This is shown in the diagram. Now let's calculate R.
We know that for the resistors in parallel:
I = I1+I2+I3
But for any resistor, current = p.d. = resistance (I = V/R ).
If we apply this to each of our resistors, and remember
that the
p.d. across each resistor is the same and equal to V,
we get:V/R=V/R1 + V/R2 + V/R3
Now we divide each term by V, to get:
1/R=1/R1 + 1/R2 + 1/R3

You will find that the total resistance R is
always less than the smallest resistance in
the parallel combination.
Circuit Diagrams

You need to be able to recognize and use
the accepted circuit symbols included in
the Physics Data Booklet
Ammeters and Voltmeters




In order to measure the current, an ammeter is
placed in series, in the circuit.
What effect might this have on the size of the
current?
The ideal ammeter has zero resistance, so that
placing it in the circuit does not make the current
smaller.
Real ammeters do have very small
resistances - around 0.01 Ω.





A voltmeter is connected in parallel with a
component, in order to measure the p.d.
across it.
Why can this increase the current in the
circuit?
Since the voltmeter is in parallel with the
component, their combined resistance is less
than the component's resistance.
The ideal voltmeter has infinite resistance
and takes no current.
Digital voltmeters have very high resistances,
around 10 MΩ, and so they have little effect
on the circuit they are placed in.
Potential dividers

A potential divider is a device or a circuit
that uses two (or more) resistors or a
variable resistor (potentiometer) to provide
a fraction of the available voltage (p.d.)
from the supply.

The p.d. from the supply is divided across
the resistors in direct proportion to their
individual resistances.
Take the fixed resistance circuit - this is a
series circuit therefore the current in the
same at all points.
 Isupply = I1 = I2
 Where I1 = current through R1

I2 = current through R2


Using Ohm’s Law
Example
With sensors
A thermistor is a device which will usually
decrease in resistance with increasing
temperature.
 A light dependent resistor, LDR, will
decrease in resistance with increasing
light intensity. (Light Decreases its
Resistance).

Example
Calculate the readings on the meters
shown below when the thermistor has a
resistance of
 a) 1 kW (warm conditions) and b) 16 kW.
(cold conditions)


similar documents