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 (note: this link will take you to different problem).

    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.)

    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!

  • 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

    1.) Req = ∑i Ri
    2.) 1/Req = ∑i (1/Ri)
    3.) 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.

  • Note: In this example, I use color in branches of the circuit, calculations, and descriptions. Note that all elements in a given color relate to each other.

    1.74 A of current come through the battery. The current everywhere that is color-coded green and bolded is also 1.74 A (there are no junctions in the green portion of the circuit.)

    We now know the current and voltage drop across the 1.0 and 2.0 Ω resistors in the series portion of the circuit, so those resistors require no further calculation.

    Check: 1.7 V + 3.5 V + 0.79 V = 6.0 V. The voltage gain across the battery is balanced by conversions across the resistors.

    Check: 0.79 A + 0.79 A + 0.16 A = 1.74 A. Current is conserved in and out of the junction.

    Check: 0.32 V + 0.47 V = 0.79 V = drop across parallel branch

  • In this problem, you first use the understanding 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.

    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 1.7 A through the battery. At the junction to the parallel legs of the circuit, the current splits. Just over 45% of the current goes through each of the 1 Ω resistors, and 9.1% of the current goes through the 5 Ω resistor. No current is used up, but electric potential energy is converted to other forms in each resistor.