DC circuits

eggplantfarflungElectronics - Devices

Oct 7, 2013 (4 years and 1 month ago)

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Created by YJ Soon April 20
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DC Circuits

Written by YJ Soon (
yjsoon@ri.sch.edu.sg
)




This work is licensed under the Creative Commons Attribution
-
NonCommercial
-
ShareAlike 2.5 License.
To view a copy of this license, visit
http://creativecommons.org/licenses/by
-
nc
-
sa/2.5/

or send a letter to
Creative Commons, 543 Howard Street, 5th Floor, San Francisco, California, 94105, USA.



Introduction


In this chapter, we move on fro
m the basics
covered in the last chapter (what is current, what
is voltage, what is resistance, what is the
meaning of life
1
) and actually construct and
solve simple D.C. circuits.


1.

Drawing Circuits

2.

Kirchhoff’s Laws

3.

Resistor Arrangements

4.

Special Resistors

5.

Potential Divider


1. Drawing Circuits


You’ll need to know how to draw circuit diagrams. The following diagram, adapted
2

from the
textbook
Jacaranda Physics 1
3
, has almost everything you need:




Make use of the first symbol i
f multiple symbols are presented for the same component.




1

42

2

“Adapte
d” == “taken wholesale”. Also == “shrunk by a nonsensical amount to fit on the page”.

3

Written by

Lofts, O’Keeffe, Robertson, Pentland, Hill and Pearce. Published by Wiley, 1997.


A simple circuit. This will be tested. Really, really.

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2.
Kirchhoff’s Laws


Kirchhoff’s Current Law


Kirchhoff’s Current Law states:


The current flowing into any junction in an electrical circuit is equal to the
current flowing out of it.



In the jun
ction below, fill in your own arrow heads pointing inwards for 1, 2 and 3; and arrow
heads pointing outwards for 4, 5 and 6 (my WordArt kung
-
foo is not strong):


I
1

+ I
2

+ I
3

= I
4

+ I
5

+ I
6


This is a direct result of the princi
ple of conservation of charge. This is because current is just
the rate of flow of charge, and the amount of charge entering a junction must equal the amount
of charge leaving. This also implies that
current anywhere along a series circuit is constant
.


Ki
rchhoff’s Voltage Law


Kirchhoff’s Voltage Law states:


In any closed loop of a circuit, the sum of the
potential

drops must equal the
sum of e.m.f.s in the loop
.

or:

The change in potential in any closed loop of a circuit will sum to zero.


In the two
loops below:


V
0

= V
1

+ V
2
+ V
3

+ V
4

(outer loop)

V
0

= V
1

+ V
5

(smaller left loop)


I
1

I
2

I
3

I
4

I
5

I
6

+ V
1



+ V
2




V
4
+

+
V
3


+
V
5


V
0

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This is a direct result of the principle of conservation of energy. Voltage is defined as energy per
unit charge; hence, in any closed loop, t
he amount of energy provided (electromotive force)
must equal the amount of energy expended (potential drop).

This also implies that
potential
difference across the separate branches of a parallel circuit is the same.

3.
Resistor Arrangements


Resistors in

Series



Let R
eff

be the effective resistance of the resistor network, i.e. the above set of resistors can be
substituted by a single resistor of resistance R
eff
.


R
eff

= R
1

+ R
2

+ R
3

(+ more if there are more resistors in para
llel)


Deriving the
equation
:




Let there be a potential difference of
V

across the ends of the above resistor network. Let
the potential difference across each
resistor
R
i

be
V
i
.



Then
, applying Kirchhoff’s Voltage Law,
V = V
1

+ V
2

+ V
3



Let the effective r
esistance be R
eff
, the current through the circuit be
I
, and the current
through each resistor
R
i

be
I
i
.

Apply Ohm’s Law (
V=RI
):
IR
eff

= I
1
R
1

+ I
2
R
2

+ I
3
R
3



However, current in a series circuit is the same, i.e.
I = I
1

=

I
2

=

I
3



Substituting:
IR
eff

=

IR
1

+
IR
2

+ IR
3



Dividing by
I: R
eff

= R
1

+ R
2

+ R
3



w00t!


Resistors in Parallel



(+ more if more resistors in parallel)


Deriving the equation from Kirchhoff’s Laws:




Let there be a current of
I

entering th
e junction of the above resistor network. Let the
current across each
resistor
R
i

be
I
i
.



Then
, applying Kirchhoff’s Curent Law,
I

=
I
1

+
I
2

+
I
3



Let the effective resistance be R
eff
, the voltage across the circuit be
V
, and the voltage
across each resistor

R
i

be
V
i
.

Apply Ohm’s Law (
I=V/R
):
V/
R
eff

=
V
1
/
R
1

+ V
2
/
R
2

+
V
3
/
R
3



However, the potential difference across the separate branches of the parallel circuit is
the same, i.e.
V

=
V
1

=

V
2

=

V
3



Substituting:
V/
R
eff

=
V
/
R
1

+ V
/
R
2

+
V
/
R
3

R
1

R
2

R
3

R
1

R
2

R
3

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Dividing by
V
:
1/
R
eff

=
1
/
R
1

+ 1
/
R
2

+
1
/
R
3



Are you satisfied now?



How about now?







Now?



Hey, look, behind you, a three
-
headed monkey!


N
ote
: For two resistors
R
1

and
R
2

in parallel:
. This is a simple derivation from
the above equation, and can be easily

remembered as the product of resistances over the sums (it
can’t be the other way around


think of the resultant units).


Another note
: The combined resistance of a parallel resistor network is always less than that of
any individual resistor in the netw
ork. Think of it this way: The parallel resistors provide
alternate paths for the current to flow through. More paths means less “resistance to movement”,
hence less overall resistance.


Yet another note
: Look! Behind you! A three
-
headed monkey!


Simplify
ing complex circuits


Try to redraw any complicated
-
looking resistor network by labelling the junctions and resistors
on the circuit, and collapsing junctions into single nodes or expanding single nodes into multiple
junctions (whichever is clearer) until
you see the series or parallel connections.


For example:




Another example:


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which is hence equal to:



How about this next famous example, the resistor cube? Given that each of the resistors is 10
k
Ω, can you show that the combined resistance of t
he circuit between point A and point B is
8.33 kΩ? Hint: Check which points are equipotential and connect them together in your newly
-
drawn circuit.





4.
Special Resistors


Voltmeter and Ammeter


The resistance of a voltmeter should be very
high. Since

it is placed in parallel with the
component being measured, it has to draw as
little current as possible. (more resistance means
less current, because more resistance means
harder for charge to move through, hence charge
moves slower through it, hence cur
rent (= charge
/ time) is less).


The resistance of an ammeter should be very
low, as it is placed in series with the circuit. If its
resistance is high, it will result in the entire
circuit drawing less current and hence affect the
reading of any circuit

it is placed in.


Resistance is futile
.

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Thermistor


Thermistors are thermally
-
sensitive resistors that can either increase or decrease in resistance as
the temperature changes. Check to make sure which you’re dealing with.


Light
-
dependent resistor


The resistance of a light
-
dependent resistor (or photoresistor)
decreases

when the amount of
illumination it receives
increases.
Light energy is used to move electrons in a semiconductor to
the conduction band, hence reducing resistance.


5.
Potential divider



The potential divide
r, or voltage divider, is an example of resistors in series, used to divide the
voltage (really?!) of an input voltage, V
in
, to a value needed, V
out
. This could be useful for logic
circuits, automatic switches, brightness tuning, or setting silly questions

to befuddle students.




Proof: The current
I

flowing through R1 and R2 is the same, since they are in series. Ignore the
extra lines coming out to the right, they’re just for reference. Hence:




Which can be rearranged to obtain the given formula.


Thermistors and light
-
dependent resistors can be used in place of R
1

or R
2
. The output voltage of
the circuit will hence depend on the temperature or illumination of the resisto
r.


For example, if a thermistor (with a negative temperature coefficient) is replaces R
2
, as
temperature increases, its resistance would decrease. Hence, from the above equation, output
V
in

V
out

R
2

R
1

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voltage V
out

would decrease (also logically


lower resistance dissi
pates less energy; hence,
there is less voltage across it). However, if the same thermistor replaces R
1
, it would result in the
output voltage V
out
increasing. This can be proven using Kirchhoff’s Voltage Law.


Conclusion


That’s all.


end.