Physics 2: Resistor Reduction
Kalamazoo Valley Community College
John Stahl
The point of reducing resistors is the same as
capacitors. We want to reduce the resistors to a
single resistance sensed by the voltage source.
We will focus on circuits that contain series and
parallel elements. Resistors are insulators and
the more resistors in series the larger the net
resistance. In parallel there are more paths for
charges to flow through and the net resistance is
smaller.
Reduce the resistors to the single net resistance seen by the source. Solve for the
voltage across R4.
The problem starts by identifying the paths for current flow. This will help us find the
relationships between the different resistors.
R2
R3
R1
R5
Vs
R6
R4
I2
Current flows from the battery into a
central junction (I1) which splits into
three currents (I2, I3, and I5).
R2
Two of the currents (I2 and I3) meet
up to form the last current (I4).
I1
We want to reduce the series
resistors in the separate branches.
R3
R1
I3
R5
Vs
R6
R4
I5
I4
R1
R2
R3
R4
R5
R6
200W
150W
175W
300W
250W
150W
In the branch for I2, both R2 and R3 have the same current flowing through them.
The two resistors are combined.
𝑅𝑒𝑞 = 𝑅2 + 𝑅3
𝑅𝑒𝑞 = 150Ω + 175Ω
𝑅𝑒𝑞 = 325Ω
I2
Replace R2 and R3 with the Req then
redraw the circuit.
325W
Take a moment and look for the next set
of resistors we will reduced.
R5 and Req both share a common points
in the circuit. These two resistors are in
parallel.
R1
Vs
I1
R5
R6
R4
I5
1
1
1
=
+
𝑅𝑒𝑞 𝑅5 325Ω
1
1
1
=
+
𝑅𝑒𝑞 250Ω 325Ω
𝑅𝑒𝑞 = 141Ω
I3
I4
We combine the two parallel paths into
a single path. The consequence of
doing this is the current I2 and I3 are
now a single current I23.
I23
I1
Vs
141W
R1
R6
Again. Take a moment to look for the
next resistors we are going to reduce.
R4
I5
Notice I23 and I4 are now the same arrow. We done really need to label the new
current, since we already have a name for it.
A single current passes through Req and R4. These two are in series.
𝑅𝑒𝑞 = 𝑅4 + 141Ω
𝑅𝑒𝑞 = 300Ω + 141Ω
𝑅𝑒𝑞 = 441Ω
I4
We are down to three resistors. R6 and
Req both start and end at the same
points, which makes them parallel. We
can clearly see I1 enters the junction
and splits into I5 and I4.
I1
Vs
R1
R6
441W
I5
1
1
1
=
+
𝑅𝑒𝑞 𝑅6 441Ω
1
1
1
=
+
𝑅𝑒𝑞 150Ω 441Ω
𝑅𝑒𝑞 = 112Ω
Redraw the circuit again with the parallel resistors swapped for the single
equivalent resistor.
I4
We are finally down to our last reduction. Like
before, by combining the two parallel paths we
no longer have I4 and I5. The combination of the
currents is actually I1.
I1
R1
Vs
112W
The last two resistors are in series and we can
make the final reduction.
I45
𝑅𝑒𝑞 = 𝑅1 + 112Ω
𝑅𝑒𝑞 = 200Ω + 112Ω
𝑅𝑒𝑞 = 312Ω
I1
Vs
With the single equivalent resistance we can now
trace all the currents and voltages in the circuit. This
will be shown in the follow up set.
312W