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Simplifying Circuits
A circuit is any closed loop between two or more points through which electrons may flow from a voltage or current source. Circuits range in complexity from one, basic component to a variety of components arranged in different ways. This handout will discuss the basics of circuits and the associated laws required to analyze and simplify them. The following table defines key terms needed to work with circuits.
Basic Terms Definition SI Units Formula
Resistance “R”
The ratio of voltage (V) across a conductor to the current (I) in the conductor.
Ohms (Ω) R = V/I
Current “I”
The amount of charge passing through a particular region over a set amount of time.
Amperes (A) I = V/R
Voltage “V”
A measure of potential difference/electric potential across a circuit.
Volts (V) = (^1 CoulombSecond ) V = I*R
Power “P”
The rate at which electric energy travels through a circuit to a given point.
Watt (W) = (Second^1 Joule) P = I*V
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Series and Parallel There are two basic configurations of resistors within circuits: series and parallel. In a series configuration, the resistors are connected in a single path so that the charge must travel through them in sequence.
For circuits containing resistors in a series configuration, the same amount of current will flow through every component, but the voltage will change. The equivalent resistance (represented as RE or RT if there is only one resistance remaining) is calculated by applying the following equation:
A parallel configuration of resistors, however, allows multiple paths for the charge to travel throughout the circuit.
The resistors in the circuit shown on the right are in a parallel configuration, and the voltage will remain the same across each resistor. The current will change. The equivalent resistance is calculated using the following formula:
RT = R 1 + R 2 + ⋯ + RN
RT^ =
R 1 +
R 2 +^ ⋯^ +^
Rn
Resistors in Series
Resistors in Parallel
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Example Find the current and voltage across each resistor of the following circuit, if ΔV = 18 V. At first glance, this circuit falls under neither of the two configurations discussed earlier—series nor parallel—rather it contains a combination of the two. In order to find the current and voltage across each resistor, simplify the circuit to a basic state (containing only a single resistor). Then, reconstruct it step-by-step. Following the aforementioned rules, the first step is to analyze the circuit. To do this, find a section where all resistors are in either series or parallel and that is furthest from the voltage source.
Step 1 – Where to Start By looking at the circuit shown below, resistors R 3 and R 4 are the best fit for the previously stated rule regarding where to begin analyzing. Since these two resistors are in a series configuration, combine them as follows and calculate their equivalent resistance using the series equation. Recall the equation for resistance in a series configuration from earlier:
RT = R 1 + R 2 + ⋯ + RN
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When simplifying into equivalent resistances, it is necessary to add a new row in the chart for each RE created within the circuit. For example, since RE1 was just calculated, there should be a new row added to the bottom of the chart as follows:
Component Resistance (Ω) Current (mA) Voltage (V) R 1 25 R 2 60 R 3 5 R 4 15 R 5 20 RE1 20
Step 2a - Simplify
By simplifying the resistors in series, R 3 and R 4 become one equivalent resistance, labeled RE with a value of 20 Ohms. Now, repeat the process, but this time using resistors R 2 and the newly created RE1.
Recall the equation for resistance in a parallel configuration from earlier: 1 RT^ =
R 1 +
R 2 +^ ⋯^ +^
Rn 𝐄𝐄 𝐑𝐑𝐄𝐄𝟐𝟐^ =^
𝟔𝟔𝟐𝟐 +^
𝟐𝟐𝟐𝟐 =^
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Because there is only one resistor in the circuit, the voltage flowing though the resistor must be equivalent to the amount coming through the voltage source (18V). With the resistance and voltage known, there is only one unknown value in The Ohm’s Law equation (V = I*R), so the current (I) may now be calculated:
RT: R = 60 Ω V = 18.0 V I = VR = 1860 = .3 A = 300 mA
Now voltage (V), current (I), and resistance (R) are known for RT (or RE3), and the circuit can be rebuilt. The Ohm’s Law equation will be used during this process to evaluate the other components within the circuit. At this point, the chart should have all resistance values filled in along with the voltage and current for RT as follows:
Component Resistance (Ω) Current (mA) Voltage (V) R 1 25 R 2 60 R 3 5 R 4 15 R 5 20 RE1 20 RE2 15 RE3 = RT 60 300 18.
Current is often calculated to be a decimal when solving circuits, so it is common practice to write the value in terms of milliamps (mA).
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Step 3 – Reconstruct & Solve
To solve for the current and voltage across all of the resistors, undo the most recent change made when simplifying the circuit, in this case steps 2b and 2c. In the process of undoing a step, first determine whether the resistors are in parallel or series configuration. This will determine which value from the simplified resistor will remain constant and carry over, in this case, RT = R 1
- RE2 + R 5. Because these three resistors are in a series setup, their current equals the current flowing through RT, which is 300mA. Using V=I*R, the voltage for each resistor can be solved using their current (300mA) and their resistance given at the beginning of the problem.
R 1 : R = 25 Ω
I = .3 A = 300 mA V = (.3)*(25) = 7.5 V
RE2: R = 15 Ω
I = .3 A = 300 mA V = (.3)*(15) = 4.5 V
R 5 : R = 20 Ω
I = .3 A = 300 mA V = (.3)*(20) = 6 V
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Component Resistance (Ω) Current (mA) Voltage (V) R 1 25 300 7. R 2 60 75 4. R 3 5 225 1. R 4 15 225 3. R 5 20 300 6 RE1 20 225 4. RE2 15 300 4. RE3 = RT 60 300 18.
As the completed chart above shows, the voltage, current, and resistance of each resistor within the system are now known. Using this method of simplifying circuits is helpful in determining the properties of individual resistors within a complex circuit. For more practice with this method, see the following pages containing example problems.
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Practice Problems: Simplifying Circuits Problem 1:
Find the voltage and current (in mA) across resistors 1-5 as well as the total resistance.
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Solutions:
Component Resistance (Ω) Current (mA) Voltage (V) R 1 90 200 18 R 2 45 300 13. R 3 15 300 4. R 4 25 600 15 R 5 5 600 3 RT 16.36 1100 18
Component Resistance (Ω) Current (mA) Voltage (V) R 1 97 202.4 19. R 2 27 101.1 2. R 3 54 50.6 2. R 4 31 50.6 1. R 5 23 50.6 1. R 6 13 202.4 2. RT 123.5 202.4 25 Equivalent Resistances have been omitted due to the existence of multiple possible answers
