# University of Wisconsin Green Bay

What is the current through, and voltage drop across, each resistor in the following circuit?

• In this problem, you are asked to find current and voltage drop in an electric circuit. Any time that you are given an electric circuit and asked for properties associated with that circuit (current, resistance, voltage, charge, capacitance) you will work the problem in one of two ways.

If the circuit is straightforward and you can track the flow of current, the best approach is to use a basic understanding of circuits to work the problem. If, however, the circuit contains a Wheatstone Bridge or is, in general, too complicated to track the current, then you will need to use Kirchhoff's Laws.

In this example, I will use a basic understanding of circuits to work the problem.

• The first thing you need to do in a circuit problem is reduce the circuit to a single equivalent resistance (or capacitance.) In other words, once you know the effective resistance of the circuit as a whole, you can calculate how much current the battery generates. So you must begin by simplifying the circuit.

Hint: The most frequent mistakes made in these problems come about from skipping this step or doing too many things at once. If you carefully reduce the circuit and record your drawings along the way (I like to put each drawing in a column along the left side of my paper so that I can work my way back up), you may be able to take shortcuts in later steps. But don't take shortcuts now!

It is handy for these problems to use colors to track sections of the circuit where current is the same. So before I begin to reduce the circuit, I use colored pencils or highlighters to trace the currents. A single color means that current is the same throughout that path.

Once the parallel portions of the circuit have been reduced to a single resistor, you can see that the net resistance of the circuit is equal to that of a 3.0, a 1.0 and a 2.0 Ω resistor in series with each other. There will be the same amount of current through the battery if these three resistors are replaced by a single 6.0 Ω resistor.

• There are two ideas (charge is conserved, energy is conserved) and three equations that help you understand circuits.

As applied to circuits

Conservation of Charge gives: Current is the same through resistors in series.

Conservation of Energy gives: Voltage drop is the same across each leg of a parallel circuit.

The three equations are

Req = ΣiRi

1/Req = Σi1/Ri

ΔV = IR

You already used Conservation of Charge and the first two equations as you drew the pictures. To solve the problem, you will use Conservation of Energy, Conservation of Charge, and the third equation.

• Once you have reduced a circuit to a single voltage source and an equivalent resistance, you can find the current through, and voltage drop across, each resistor by starting with the simplest (equivalent) drawing and working your way back up.

1.5 A of current is produced by the battery. The current everywhere that is color-coded green is also 1.5 A.

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Current is the same through resistors in series.

We now know the current and voltage drop for the 3.0 Ω resistor in the lower leg of the circuit. That resistor requires no further calculation.

Check: 4.5 V + 1.5 V + 3.0 V = 9.0 V. The voltage gain across the battery is balanced by conversions across the resistors.

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Voltage drop is the same across each leg of a parallel circuit.

We now know the current and voltage drop for the 3.0 Ω resistors in the blue and yellow branches of the circuit. No further calculation is required for those two resistors. Note the color coding of the currents and their values corresponds to the colors used to code the circuit. In other words, everywhere that the circuit is orange, the current has the value 0.50 A.

Check: 0.5 A + 1.0 A = 1.5 A. Current is conserved in and out of the junction for each of the parallel elements of the circuit.

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For the orange leg of the left parallel element, note that current is the same through resistors in series.

for each of the 3.0 O resistors in this branch. Check that 1.5 V + 1.5 V = 3.0 V, giving the correct total voltage drop across the orange leg.

For the purple leg of the right parallel element, note that voltage drop is the same across resistors in parallel.

for each of the 3.0 O resistors in this parallel element. Check: 0.5 A + 0.5 A = 1.0 A. Current is conserved in and out of the junction.

Current and voltage drop for each resistor in the circuit are now known. No further calculation is necessary.

• In this problem, you first use the understandings that current is the same through all series portions of a circuit and the equations for equivalent resistance to replace the given circuit with one that has the same effect from the perspective of the battery.

This break down of the circuit into one equivalent resistance is the key to working circuit problems successfully. Start within parallel components of the circuit and work out, and don't take more than one step at a time.

Once you know the equivalent effect of the circuit on the battery, you can use the understanding that voltage drop is the same across parallel legs of a circuit (conservation of energy) and the equation defining the voltage drop across a resistor to find the current and voltage drop for each resistor in parallel, and the understanding that current is the same through all series portions of a circuit to find the voltage drop across resistors in series.

Keeping track of your information is key. At each step, if you fill what you know into the appropriate intermediate drawing you are far less likely to make mistakes. If you replaced a series of resistors with a single resistor, the current through all is the same. Likewise, if you replaced a parallel portion of the circuit with a single resistor, the voltage drop across all is the same.

In this case, there is a current of 1.5 A through the battery. At each junction, the current splits. The amount of current through the branches is inversely proportional to the relative resistance of the branches. No current is used up, but electric potential energy is converted to other forms in each resistor.