Download Kirchhoff`s Rules - Haiku for Ignatius

KIRCHHOFF’S LAWS
1. Voltage Rule
2. Current Rule
1. Voltage Rule. This is based on the conservation of energy: “the sum
of voltages around a closed conducting loop (that is, a circuit) must be
zero”
Since voltage and work are related, we are saying the net work done
must be zero. Observe the diagram below:
Resistor
R1
Supplies energy for
work
Voltage
does work (uses energy)
+
Net Work = 0
The Resistor can’t do more work than is available
(it can’t do less either - if there’s too much energy generated, it burns
out)
2. Current Rule. This is based on the conservation of charge: “the sum
of currents entering a junction = the sum leaving”
[junction: a place where 2 or more conductors join]
I
So we see that
I1
I2
Resistor
R1
I = I1 + I2
This relationship is the
essence of the Current
Rule
How does this apply to Series and Parallel Circuits?
SERIES RESISTANCE
Let’s examine a Series Circuit First….
Resistor
R1
Resistor
R2
1.5 Volts +
1.5 Volts
+
Current, I: one path
start here: see below
1st: current rule = since there are no junctions this means that the
Current, I, never changes. So we only have one current.
2nd: voltage rule = the voltage inputs and decreases must add up to
zero!
To do this, one must make a trip around the circuit counting voltage
inputs and voltage drops. You can go clockwise or CCW but you MUST
maintain that direction in the loop.
Going Clockwise, we’ll start at the lower left corner. We add any
voltage inputs and any voltage drops:
+1.5 V + 1.5 V - V (lost to resistor 1) - V (lost to resistor 2) = 0
We can use OHM’S LAW, V = IR, to replace “V” in both terms above with
“IR”
+3.0 V - IR1 - IR2 = 0
Note that batteries in
series add their
voltages!!!
So we have:
+3.0 V = IR1 + IR2
+3.0 V = I (R1 + R2)
+3.0 V = I (Requivalent)
Therefore,
Requivalent = R1 + R2 + R3 + --- Rn
for series resistance
Example:
Problem:
a) Find the “equivalent” resistance
of the circuit shown
b) Find the current in the circuit
shown
c) Find the voltage (energy) loss
thru each resistor
a) Just add up all the resistances!
Rseries = R1 + R2 + R3 = 60 + 100 + 20 =
180 ohms
This is called the “Equivalent Circuit”
(because the power used is equivalent to
the original, complex circuit)
b) Now it’s easy to find the Current; just use OHM’S LAW:
V = IR
9 Volts = I (180 ohms)
I = 0.05 amperes
c) Now, since the current, I, is the same for all the resistors, then,
V(across resistor 3) = IR3
V3 = (0.05 amps)(20 ohms)
V3 = 1 volt
answer
Do the same for the other resistors. See if you get voltage drops
totaling 9 volts across all the resistors…
PARALLEL RESISTANCE
Note that the voltage is the same for all resistances!!!
(however, the work won’t be the same because it depends on the
current flow…recall W = -qV …so it depends on how much charge is
moving)
From the Current Rule we know that I = I1 + I2
Thus
I
V V
1
1

V(  )
R1 R2
R1 R2
I V(
1
REquivalent
)
So that in general,
1
1
1
1
1
 

 ... 
R1 R2 R3
Rn
Requivalent
For Parallel Resistors
Or, in a more useful format:
Requivalent 
1
( R11  R12  R13  ...  R1n )
Example:
Problem: Find
a) the equivalent resistance for this circuit
Solution:
Requivalent 
1
( R11  R12 )
1
1
1
Requivalent  1


 16.67 
1
( 100  20 ) 0.01  0.05 0.06
Notice that this is lower than either of the actual resistors.
b) Find the current I:
Solution: This result leaves us with an “equivalent” circuit:
For which we can easily find the general current using Ohm’s Law:
V = IR
V = IR
I = V/R = 3.0 volts/(16.67 ) = 0.18 amperes
c)Find the currents I1 and I2 !!
Solution:
Finding the individual currents in this basic parallel circuit is very
straightforward once we realize that ALL the resistors have 3.0 Volts
across them!
Therefore, using Ohm’s Law: V = IR we find…
through the 100 Ohm resistor:
through the 20 Ohm resistor:
I = V/R
= 3.0 / 100
= 0.03 amps
I = V/R
= 3.0 / 20
= 0.15 amps
Example 2:
What happens if we put another battery in the circuit?
Now it becomes a little more difficult to determine the current…we
have to use “loop” analysis (based on Kirchhoff’s voltage rule: the
voltage around a closed loop must add up to zero).
Let’s see how that works:
we have:
If we follow the arrow around we see we
increase by 3.0 volts and then pass
through the 100 ohm resistor, which is
going to use up voltage. How much?
Since V = IR, it will use I1(100) volts. So
3.0 - I1(100) = 0
or I1 = 3/100 = 0.03 amps going through the 100 ohm resistor.
Notice this didn’t change from the previous example. Now for the
second loop:
We follow this loop around and see that
1. we go up 3.0 volts again
2. we pass through the 20 ohm
resistor and use up voltage equal
to Vused = I2(20)
3. We go BACKWARDS through the
1.2 volt battery meaning…?
These must all cancel out to zero, so…
3 - I2(20) - 1.2 = 0
1.8 - I2(20) = 0
- I2(20) = -1.8
I2 = 1.8/20 = 0.09 amps
So
I1 = 0.03 amps
I2 = 0.09 amps
What does I = ???
Notice it’s lower than our first example because we’ve put in an
opposing battery.