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Kirchhoff’s Rules Script
Kirchhoff’s Rules
Kirchhoff’s rules make it possible to analyze more complex circuits. Kirchhoff gave
us the junction rule and the loop rule. We will use both in solving complex circuits.
Junction Rule
The junction rule states that the sum of the current entering and leaving at any
junction in a circuit must equal zero. Stated as an equality the sum of the current
equals zero. This is a statement of conservation of charge. The charge that enters a
junction must exit the junction. So current that flows in must also flow out.
Junction Rule
As you can see in the diagram current I sub 1 goes into the junction and I sub 2
and I sub 3 come out.
I sub 1 – I sub 2 – I sub 3 equals zero so I sub 1 equals I sub 2 plus I sub 3.
Using the Junction Rule
You may use the junction rule as often as you need it as long as you have a new
current that has not been in the other times you used it. The limit is one less than
the number of junctions you have in the circuit.
Loop Rule
The loop rule states the voltage around any closed loop of a circuit must add to be
zero. Stated as an equation the sum of voltage equals zero. You must choose a
closed loop in the circuit and think of the potential differences across each device as
you solve the problem using the loop rule.
Loop Rule Application
Current flows through a resistor from the high potential end to the low potential
end. If you move through a resistor with the current the electric potential difference
is – Current times resistance. If you move through the resistor against the current
the electric potential difference is + current times resistance.
Loop Rule Application
As you move through the source emf if you move from negative to positive the
potential difference is + emf. IF a source is moved through from positive to
negative then the potential difference is –emf. As you can see below as you move
to the right through the battery the emf is positive. As you move to the left through
the battery the emf is negative.
Using the Loop Rule
You can use the loop rule as often as you need to as long as there is a new circuit
element introduced in the new loop, a new resistor, battery or a new current. You
can set up as many equations as you like from the junction and loop rules, however
they are only useful if the number of equations match the number of unknown
currents in the circuit.
Problem One
In the circuit below what is the total resistance? What is the current through each
resistor? What is the potential drop across each resistor?
Problem One Solution
View the problem below.
(We first find the total resistance. We add 3 plus 6 Ohms and get 9 Ohms. We add
this to the 18 Ohms in parallel and get 6 Ohms for the total resistance. Using Ohm’s
Law, 10 volts equals the total current times 6 Ohms. The total current is 5 Amps.
This is the current leaving the source and entering the branches of the parallel
portion of this circuit. Next we find the current through the 18 Ohm resistor. Since
it is in a loop with the source the potential difference across the 18 Ohms resistor is
30.0 volts. 30 volts divided by 18 Ohms gives us 1.67 amps. Now the total current
minus 1.67 amps equals the current through the other branch. This is 3.33 amps.
We use this to find the potential difference across each of the other resistors. The
potential difference across the 3 ohm resistor is 3.33 amps times 3 Ohms which is
9.99 volts. The potential difference across the 6 Ohm resistor is 19.98 volts. When
you add these two together you get 9.99 V + 19.98 V= 30.0 volts which supports
the loop rule.)
Adding Resistors in Parallel Problem
Two resistors both 8 Ohms are connected in parallel. A 12.0 volt battery is
connected across the resistors. What is the total resistance? When the switch is
closed what is the voltage across each resistor? What is the current through each
resistor?
Adding Resistors in Parallel Solution First we add the resistors in parallel and
get the equivalent resistance of 4 Ohms. Using the Loop Rule the voltage drop
across the resistors is 12.0 volts. Using Ohm’s Law we get the current. 12.0 volts
equals current times 8 ohms. The current is 1.5 amps.
Problem Two
Find each current if each light bulb is 3.0 Ω and the battery supplies 6.0 V.
Problem Two Solution
Now to solve this I will use Kirchoff’s loop rule. I choose to do the first loop with I
sub 1. The voltage of the battery equals the voltage drop across the light bulb, so it
is 6.0V. That makes the current through that bulb 6.0 Volts divided by 3.0 Ohms
which is 2.0 Amps.
Now we turn to I sub 2. This light bulb makes the same loop as the first light bulb
so it has a voltage drop of 6.0 volts as well and a current of 2.0 Amps.
Now we turn to the two light bulbs in series. We set up Kirchhoff’s Loop rule and
get 6.0 V = V3 + V4 . We know that V sub 3 equals I sub 3 times 3.0 Ohms and V
sub 4 equals I sub 3 times 3.o Ohms. We then solve for I sub 3 and get 1.0 Amps.
Now we can find the total resistance and divide 6.0 Volts by the total resistance.
This gives us 5.0 Amps for the total current. We could also use the junction rule
and add I sub 1 plus I sub 2 plus I sub 3 and get 5.0 Amps for the total current.
Problem Three
In the circuit below what is the total resistance? What is the current through each
resistor if the battery is 30.0 V? What is the potential drop across each resistor?
Problem Three Solution
View the problem below.
(First we add the 3 and 6 ohm resistors in parallel. This is 2 ohms. Then we add
that to the 7 Ohm resistor in series with the 2 ohm equivalent. This is 9 ohms. Now
we add this to the 18 ohms in parallel with the three we just added together. This
gives us 6 ohms. Then we add the 3 ohms to it giving us 9 ohms. We add this to
the 18 ohm resistor in parallel giving us 6 ohms. The total resistance is 6 ohms plus
9 ohms which is 15 ohms total. Then we look at current and potential across each
resistor. V = IR. 30.0 V =I (15 Ohms). I = 2.0 amps.
We can determine the voltage through the 9 ohm resistor which is 9x2 which is 18
volts. Now the 18 ohm resistor is 30.0 volts – 18.0 volts which is 12.0 volts. The
current through the 18 ohm resistor is 12 volts divided by 18 ohms which is 0.67
amps. The current in the 3 ohm resistor is 2 amps – 0.67 amps = 1.33 amps. The
potential drop is 3 ohms times 1.33 amps which is 3.99 volts. The voltage through
the 18 ohm resistor is 12 volts – 3.99 volts =8.0 volts. The current through the 18
ohm resistor is 8.0 volts divided by 18 ohms is 0.44 amps. The current through the
7 ohm resistor is 1.33 amps – 0. 44 amps = 0.89 amps. The potential drop across
the 7 ohm resistor is 0.89 amps times 7 ohms which is 6.23 volts. The potential
through the 3 ohm resistor and the 6 ohm resistor equal and are 12 volts – 3.99
volts – 6.23 volts =1.78 volts. The current through the 3 ohm resistor is 1.78 V/3
ohms = 0.59 amps. The current in the 6 ohm resistor equals 1.78 volts/ 6 ohms =
0.30 amps.)
Wrap it up
Solving complex combinations of resistors becomes like a puzzle. You must first
figure out the smaller pieces to determine the potential drop and current through
each resistor. Find the total resistance first and then using Ohm’s Law and
Kirchoff’s rules you can solve complex circuits.