Java Security Update: Oracle has updated the security settings needed to run Physlets.
Click here for help on updating Java and setting Java security.
Illustration 30.7: The Loop Rule
Please wait for the animation to completely load.
Kirchhoff's loop rule states that the sum of all the potential differences around a closed loop equals zero. In other words, Σ ΔV = 0 for a complete loop. Restart.
In the circuit shown, current from the battery flows through the resistors before returning to the battery. This illustration follows a hypothetical charge as it flows through the upper of the parallel resistors. This is, of course, just a simulation. Current also flows through the lower resistor, and the current through these two resistors is not the same. In fact, an accurate microscopic simulation would need ~1020 electrons moving counterclockwise around the circuit. Because an electron-flow representation of current would be awkward, we use the standard definition of current and show a hypothetical unit of positive charge flowing out of the positive battery terminal, through the resistors, and into the negative terminal.
In a circuit, there are charges moving through the potential differences from the battery and across the resistor. So another way to state the loop rule is that, when a charge goes around a complete loop and returns to its starting point, its potential energy must be the same. Positive charges gain energy when they go through batteries from the - terminal to the + terminal, and they give up that energy to resistors as they pass through them.
Use the loop rule to determine the current through the battery in a circuit consisting of a 16-volt battery connected to a set of three resistors: a 2-Ω resistor in series with a parallel combination of a 2-Ω resistor and a 3-Ω resistor.
Now consider a Kirchhoff loop consisting of the battery and the two 2-Ω resistors. It doesn't matter where we start, as long as we come back to the same spot. Let's go clockwise around the loop starting at the bottom left corner.
+16 V - (2 Ω)*I - (2 Ω)*3I/5 = 0
+16 V = (10 Ω)*I/5 + (6 Ω)*I/5
+16 V = (16 Ω)*I/5.
This gives I = 5 A.
Run the animation and follow the energy of the unit charge as it passes through each circuit element. Each voltage drop represents the amount of energy that is lost or gained when the charge passes through a circuit element. This demonstrates that, as charge flows around a complete loop, the gains in energy are always offset by the losses. The total change in energy is zero.
Illustration authored by Andrew Duffy.
Script authored by Andrew Duffy.