On this page** I've referred to relative
ammeter readings as a1, a2 **etc., but **on all other pages I**_{1}, I_{2}
etc. might well be used since I is the equation symbol for electric current**.**

These are relatively simple circuit
diagrams copied and I just want you to think in 'simple'
conceptual way e.g. which bulbs light up and how brightly AND compare current
flow in different parts of the circuits.

I've rarely included the rectangular resistor
circuit symbol here, but don't forget
**a lamp (e.g. a filament bulb) is a resistor**.

These circuit diagrams in include ammeters,
switches and a simple battery power supply.

Wiring in series or parallel in the circuits
is discussed.

Assume all ammeter readings e.g. a1, a2 etc.
are in amperes (A).

No specific resistors or voltmeters are included at the moment
and no
calculations yet either!.

**1.**
**Circuit diagram 01**:

1 ammeter, 1 switch, 1 cell, and 1 bulb all wired in
**series** in a simple single loop.

Assume bulb glows with normal brightness,
so **1 cell powers 1 bulb correctly** - not too dim or 'blows' the bulb!

**In a
series
circuit, all the components are wired together end to end**, non in a
separate loop.

**2.**

**Circuit diagram 02**: 1 ammeter, 1 switch, 2 cells and 2 bulbs all in
**series**.

Here we have doubled the potential
difference (p.d.), but we have also doubled the resistance, the effects
cancel each other out, therefore the lamp will glow with normal brightness
AS IN CIRCUIT 01.

**3.**

**Circuit diagram 03**: 1 ammeter, 1 switch, 2 cells in series with 1 bulb all wired
in series.

Here, doubling the p.d. will double the
current and the bulb will glow more brightly than in circuits 01 and 02
(will probably blow the bulb!).

**
4.**

**Circuit diagram 04**: 1 ammeter, 1 switch, 1 cell and 2 bulbs all wired in series.

Here, doubling the resistance will halve
the current and the bulbs will glow dimmer than in circuits 01 and 02.

5.

**Circuit diagram 05**: 1 ammeter, 1 switch, 3 cells and 3 bulbs all wired in
series.

Here we have tripled the p.d., but also
tripled the resistance, so the bulbs will glow normally as in circuits 01
and 02.

6.

**Circuit diagram 06**: 1 ammeter, 1 switch, 3 cells and 2 bulbs all wired in
series.

Here the bulbs will glow a little more
brightly than in circuits 01 and 02. Can you figure out why?

7.

**Circuit diagram 07**: 1 ammeter, 1 switch, 3 cells and 1 bulb all wired in series.

Here the bulb will glow VERY bright for a
few seconds and then likely to burn out!

Compared to circuits like 01 and 02, you have tripled the p.d. but kept the
minimum of one resistor (1 bulb), too much current flowing for the bulb filament!

**
8.**

**Circuit diagram 08**: 1 ammeter, 1 switch, 1 cell and 3 bulbs all wired in series.

Compared to circuit 07, here the bulbs
will glow very dimly, much less so than in circuits 01 and 02.

You have tripled the resistance and kept
the minimum p.d.

Therefore the current flow is much lower
than in circuit 07, less electrical energy to light the bulbs.

9.

**Circuit diagram 09**: 1 ammeter, 1 switch, 1 cell and 3 bulbs all wired in series.

Here the bulbs will glow a little bit
dimmer than their 'normal' brightness compared to circuit 05. Can you see why?

**When components are wired in **
**parallel****,
each one is in a separate loop (or branch), effectively both ends of each
loop of components are connected together.**

Note the two slightly different styles of
drawing the circuit - they both amount to the same things.

Here things are getting a bit more
complicated and I'm introducing what the relative ammeter readings might be.

From now on, I'm less interested in how
bright the bulbs glow, but what might the relative ammeter readings be?

Circuits 01 to 09 were simple loops and
the current flow is identical at any point in the circuit.

However, here, the current is split to
power each bulb individually in the **parallel** sections of the circuit.

The ammeter current readings a1 + a2 MUST
equal ammeter reading a3 because the current flowing from the battery, even
if it is split, it must be the same in total.** You can't lose or gain
electrons!**, so **a1 + a2 = a3**.

Also ammeter readings **a1 = a2**,
assuming the bulbs have the same resistance, so the same current will flow
through them equally as they both experience the same p.d.

In section 3.
Ohm's Law we will look at these
situations in a quantitative way.

**
12.**

Circuit diagram 12: Here everything is wired in a simple
**series** loop.

The bulbs b1 and b2 will glow normally
and with equal brightness, assuming they are of **equal resistance**.

Since everything is wired in series, all
the ammeter readings will be the same,

therefore the ammeter reading a1 = a2 = a3.

13.
** 14. **

**Comparing circuit diagrams 13 and 14:
They are actually the same
circuit**

Same as circuits 10 and 11 except nothing
happens until you close the switches.

To light a bulb you must close switch s3
and either or both switches s1 and s2.

Here you can** light each bulb
individually, which you cannot do if they were wired in series**, as in
circuit 12 above.

**
15.**

**Circuit diagram 15**: Everything wired in series.

Same as circuit 12 except nothing happens
until you close all three switches, and then all three bulbs will light up.

16.

**Circuit diagram 16**: The bulbs will glow very brightly and the filaments will
probably burn out!

Can you see why the lamps might just
light for a few seconds before going out!?

Just compare circuits 16 and 02.

**
17.**

**Circuit diagram 17**: The bulbs will glow very dimly, the 4 bulbs equate to a high
total resistance.

When resistances e.g. lamp bulbs are
**wired in series**, you **add them up to get the total resistance**.

Again, just compare circuits 16 and 02.

This circuit has double the
resistance of circuit 02.

18.
**Circuit diagram 18**:

1 ammeter, 1 switch, 2 cells wired in series with
**3 pairs
of ammeters and bulbs wired in parallel**.

If you followed the arguments for
circuits 11 and 12 you should be able to deduce the following:

All three bulbs b1 to b3 will glow
with the same brightness - all subjected to the same p.d.

Relative ammeter readings:

a1 = a2 = a3 (assuming all bulbs
have the same resistance).

Total current flowing in the
circuit = a4 = a1 + a2 + a3

For example, if a4 read 12A, a1, a2 and a3 meters
would all read 4A because all the bulbs are wired in parallel.

Or you could deduce that if any of the meters a1 to a3 read 4A,
then you could deduce that a4 meter would read 12A.

19.
**Circuit diagram 19**:

This simple loop circuit includes a variable resistor ().

By varying the resistance, you can vary
the current flow and control how brightly the bulb glows.

This is the simplest circuit to
illustrate how a dimmer switch works.

The greater the resistance, the lower the
current, the dimmer the bulb lights up.

In terms of ammeter readings and bulb
brightness:

a4 = a1 + a2 + a3, but a1, a2 and a3
ammeter readings will all be different because of the different numbers
of bulbs, that is each sequence of bulbs equates to a different
resistance for the same potential difference.

When you have bulbs wired in series
you add up the individual resistances to get the total resistance.

So, in circuit 21, for the bulbs, we
have relative resistance ratio values of 1 : 2 : 3 (left to right).

The bigger the resistance, the lower
the current, so the relative ammeter readings will be a1 > a2 > a3,

and the brightness sequence for the
bulbs is b1 > b2/b3 > b3/b4/b5.

If the bulbs have identical resistances you can deduce
that in terms of ammeter readings

a2 = 2 x a2 = 3 x a3

**
22.**
**Circuit diagram 22: **

This is a two-way switch system e.g. for a landing light in
a house.

You can switch the light on from two
different locations e.g. the ground floor and first floor of a house.

For the lamps light up, switches s1a and s2a must be both
closed, OR, both switches s1b and s2b must be closed to give a complete
circuit.

**
25.**
**26.**

**Circuit diagrams 25 and 26**:

When you close the switch s, only bulb b2 will light up.

The extra wire 'short circuits' and
bypasses bulb b1 - virtually no current will flow through it.

The extra wire with no resistor in it will have offer less
resistance than the thin bulb filament.

In circuit 26 it is the same situation
and only bulb b2 lights up AND you don't even have to close the switch.

**
27.**
**Circuit diagram 27:**

Following on from circuits 25 and 26, when you close the
switch only bulb b1 will be lit.

Virtually no current will flow through
bulb b2.