Transcript ppt - Animated Science

Kirchoffs Laws
Physics – 12.1.7 Kirchhoff’s laws
Conservation of charge and energy in simple d.c. circuits. The relationships
between currents, voltages and resistances in series and parallel circuits;
questions will not be set which require the use of simultaneous equations to
calculate currents or potential difference
Mr Powell
Kirchoffs Law I
 The “current law” states that at a
junction all the currents should add
up.
I3 = I1 + I2 or I1 + I2 - I3 = 0
•
currents towards a point is designated
as positive
•
those away from a point as negative.
•
In other words the sum of all currents
entering a junction must equal the
sum of those leaving it.
•
Imagine it like water in a system of
canals!
Kirchoffs Law I
Examples;
If I1 = 0.1A, I2 = 0.2A, I3 = 0.3A
If I1 = -0.1A, I2 = -0.2A, I3 = -0.3A
If I1 = 2A, I2 = 3A, I3 = 5A
There are some important multipliers for current:
1 microamp (1 A) = 1 x 10-6 A
1 milliamp (mA) = 1 x 10-3 A
Also remember to make sure you work out current in Amps and time in
seconds in your final answers!
Kirchoffs Law I – Test it out…
 You can test out this theory
by making a resistor network
and connecting two power
sources which you alter to
change the currents.
Kirchoffs Law I – Questions?
 Work out the currents and directions missing on these two
junctions?
7A
3A
Kirchoff Law I – Complex Questions

Given the following circuit can you pick out how the current might behave
using kirchoffs current laws. We are looking to find the current in the main
branch or 0.75 resistor and also through the 1 resistor

You may want to redraw the circuit then apply simple ideas of additive
resistance and ratios to find out the currents?
Kirchoff Law I - Answers

This example is more simple than it looks. In fact you have one resistor on
its own (0.75)

Then the other three are in parallel with each other. With the 1  on one
branch and the two 1.5  resistors on the other.

You can then simply use resistance ratios to determine the current flow.

The ratio for the parallel part is 1:3 so we say that the least current flows
through the most resistive part. Hence: current through 1  resistor must
be 0.75A.

The current through main branch must be the sum of these i.e. 1A
Kirchoff Law I - Answers

In detail this means can rewrite
the circuit and fill in the values
for V and I as such…

Energy (p.d.) is shared simply
according to resistance.
Kirchoffs Law II
 The “voltage law” states that the sum of e.m.f’s around a circuit or loop is
equal to zero. i.e. all energy is transferred before you return back to the cell.
 In other words the sum of all voltage sources must equal the sum of all
voltages dropped across resistances in the circuit, or part of circuit.
 Think of it like walking around a series of hills and returning back to point of
origin – you are then at the same height!
For more complex examples we must note the following rules;
 There is a potential rise whenever we go through a source of e.m.f from the
– to the + side.
 There is a potential fall whenever we go through a resistance in the same
direction as the flow of conventional current. i.e. + to NB. Both laws become obvious when you start applying them to problems.
Just use these sheets as a reference point.
Simple e.m.f example
1) If I = 100 mA what is that in amps?
100 mA = 0.1 A
2) What is the current in each resistor?
0.1 A
3) Work out the voltage across each resistor.
V = IR so
V1 = 0.1A x 30  = 3 V
V2 = 0.1A x 40  = 4 V ;
V3 = 0.1A x 50  = 5V
4) What is the total resistance?
RT = 30 + 40+ 50 = 120
5) What is the battery voltage?
V = 0.1A x 120  = 12V
Complex Examples

Hint1: This is complex and you must
try and be consistent in your
calculations in direction and which
way the current is flowing or p.d. is
lost!
If we apply Kirchhoffs laws
(previous slide) about current we
can say that;
I1 = I2 + I3 or 0 = I2 + I3 - I1


Hint2: This type of circuit is the top
end of AS and you will not be asked
to solve one using this Simultaneous
Eq method unless they give you
some values to help you out!
If we apply Kirchhoffs laws
(previous slide) about pd = 0 in
closed loop we can say the
following two things
Starting at Point “P” and going
clockwise around the left-hand
loop;
-I3R2 + E1 - I1R1 = 0

Starting at Point “P” and
going clockwise around the
right-hand loop;
-E2 - I2R3 + I3R2 = 0
EXTENSION WORK
R1
P
I3
E1
I2
E2
R2
I1
R3
Hint: consider only E direction not I’s
Complex Examples
Starting at Point “P” and going clockwise around the left-hand loop;
-I3R2 + E1 - I1R1 = 0
Starting at Point “P” and going clockwise around the right-hand loop;
- E2 - I2R3 +I3R2 - = 0
p.d. is lowered
with flow I1
R1
P
I3
E1
p.d. is
increased with
flow I1
PD against flow of
current
I2
E2
R2
I1
p.d. down
with flow I3
left loop
R3
p.d. is up with
flow I3 right
loop
PD against flow of
current
EXTENSION WORK
Complex question….

Use the theory from the previous slide to answer the question below
working out the current flow we have called I3.

Hint use the technique shown on the previous slide. But try and
reason it out yourself with the rules you have been given. This may
take some time!

Your answer should include the following;
1.
2.
3.
4.
Reference to the rules of current flow & p.d
Explanation as to why each contribution is + or –
Equation for each loop
Answer for I3
E1
E1 = 6V
E2 = 2V
R1=10
I1
R2=10 
R3=2 
Answer: I3 = 0.4A
R1
P
I3
I2
E2
R2
R3
Complex example..


Kirchoffs 1st Law; I1 = I2 + I3
This worked example relies on two equations
found from two loops. Each defined for a
separate power source.
Kirchoffs 2nd law;
Solve simultaneously to find I’s
30V = 20I3 + 5I1
Sum PD Loop AEDBA
- Eqn 1
Sum PD Loop FEDCF
10V = 20I3 - 10I2
10V = 20I3 - 10(I1 - I3)
10V = 30I3 - 10I1
- Eqn 2
Add 2 x Eqn 1 + Eqn 2
70V = 70 I3
I3 = 1 A
Substitute this in Eqn 1
EXTENSION WORK
I1 = 2A so I2 = 1 A