# 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 model the use of Kirchhoff’s Laws to work the problem.

• Hint: When working a circuit problem with Kirchhoff's Laws, you need to keep track of all the different currents. If you recognize that current is the same through all resistors in series, you reduce the number of unknowns in your problem. I find that color-coding the circuit to show each region of “same” current helps to avoid record-keeping mistakes.

• In this example, I chose to model solving a circuit with Kirchhoff's Laws.

1. Current into a junction = current out of that junction
2. The change in voltage around any loop of the circuit = 0

Once you know current through a resistor, voltage drop is given by

ΔV = IR

• Step 1:

Now that you have found the current through each resistor, scroll down to find the voltage drop across each resistor

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Step 2:

• In this problem, you use Kirchhoff's Laws to find the current through each resistor in a circuit. Kirchhoff's Laws are Conservation of Charge and Conservation of Energy applied to circuits. Although you can use them without understanding the circuit, once you have also used ΔV = IR for each resistor, you should go back through the circuit to follow the current and track the energy.

In this case, there is a current of 1.7 A through the battery. At the junction to the parallel legs of the circuit, the current splits. The total current in the parallel section (0.79 A + 0.70 A + 0.16 A) = 1.7 A, or the current through the battery, as it should. .

No current is used up in resistors, but electric potential energy is converted to other forms in each resistor. In the steady state, the energy (per charge) provided by the battery is completely converted to other forms over any path that the current can follow. In this case, the battery provides a potential difference of 6.0 V. Through any branch of the parallel portion of the circuit, the current loses 0.8 V. Therefore, over any path the current loses (1.7 V + 0.8 V + 3.5 V) = 6.0 V as expected.