You have seen that in an electric circuit, charges move and transfer their electric PE to the various devices connected in the circuit. If the moving charges have only one path to follow, then the value of the current will be the same everywhere in the circuit. However, if there are junctions in the circuit, that is, branches where the current can split up, then different amounts of current will flow in the various branches. This is much like the branching of a river or stream. In the river example, it is clear that the water is not lost: whatever mass of water that flows into the junction will also flow out of the junction into the various branches. Electric charge can neither be created or destroyed, a fundamental conservation law of nature. This means that whatever current flows into a junction will also flow out into the various branches. This is Kirchhoff’s 1 st law, also called the Junction Law: the sum off all current flowing into a junction will equal the sum of all current flowing out of the junction. Electric Circuits II -Kirchhoff’s Laws, pg 194

If you were to take a walk around the Wu building, perhaps visiting all three floors at some point, if you eventually returned to your starting point your overall change in height would be zero. While you increased your height some of the time, you decreased height at other times so that the sum of all height changes, Δy, was zero. Symbolically, you might represent this as: In electric circuits, when charges return to the same point in the circuit, they also return to the same voltage level. This means that as they traverse a circuit charges may go down in voltage level sometimes, but to return to the same point they must go up in voltage level the same amount. As the charge moves through devices in the circuit it will have both positive and negative ΔV’s. These will add to zero if the charge returns to the same point. This is Kirchhoff’s 2 nd law, also called the Loop Law.

Example: Use the loop law to calculate the battery voltage in the circuit below. The voltage drop across a resistor is always given by Proceeding clockwise from the negative terminal of the battery, you first go up an amount equal to the battery voltage, then down in voltage level as you go through each resistor. These must add to zero:

Example: a. Apply the Junction law to determine the value of i. b. Apply the loop law to the two “window pane” loops to obtain two equations involving the unknown battery voltage and the unknown resistance. c. Solve for the unknowns. a. Left loop Right loop Calculate the voltage difference between A and B.

Resistors in Series and in Parallel When several devices are connected in an electric circuit, the manner in which they are connected has a specific naming process. If two (or more) devices always have the same current established in them, irrespective of any changes that might occur elsewhere in the circuit, then you say the devices are connected in series. (Only one pathway through them) If two (or more) devices always have the same voltage drop across them, irrespective of any changes that might occur elsewhere in the circuit, then you say the devices are connected in parallel. (More than one pathway through them It is quite possible that some devices are neither in parallel nor in series. There is no special name given to this. Example: Identify the type of connection as series parallel, or neither in the following cases: 8Ω and 6Ω: 10Ω and 5Ω: 3Ω and 2Ω: 3Ω and 20Ω: battery and 6Ω:

When two or more resistors are in series, it is possible to replace them by a single resistor in such a way that the currents flowing in the circuit are not affected. This single resistor value is called the equivalent series resistance. The value of the equivalent series resistance can be calculated from the values of the individual resistance. It is a good Kirchhoff law activity to see how this works. Consider the circuit below with two resistors in series. The equivalent series resistor R s would replace the two resistors but leave the current unaffected. Apply the loop law to each circuit: Equating the two equations leads to Just add the resistor values. If there are more than two, just keep adding.

When two or more resistors are in parallel, it is possible to replace them by a single resistor in such a way that the currents flowing in the circuit are not affected. This single resistor value is called the equivalent parallel resistance. The value of the equivalent parallel resistance can be calculated from the values of the individual resistance. It is also a good Kirchhoff law activity to see how this works. Consider the circuit below with two resistors in parallel. Notice that each parallel resistor will have a different current. The equivalent parallel resistor R p would replace the two resistors, but leave other currents unaffected. Now apply the loop law to the inner and outer loop of the left figure. Do the same for the left figure:

Apply the junction law to the left figure: Finally, substitute your values for the currents derived from the loop law: If there are more resistors in parallel, just keep adding the inverses. Example: Calculate the equivalent resistance in the following networks.

In many more complex circuits, an overall equivalent resistance can be calculated by reducing all series and parallel networks systematically. This is much like removing parentheses in an algebraic expression: in a complex nest, work from the inside outward. Example: Determine the equivalent resistance of the network below: Start inside the nest. The 3Ω and the 2Ω are in series, adding to 5 Ω. This equivalent 5Ω resistor is in parallel with the 20Ω resistor. Find their equivalent parallel resistance Finally, the 8Ω, 4 Ω, and the 6 Ω are all in series, adding to 18 Ω

Exercise: Find the overall equivalent resistance:

Current Division Kirchhoff’s Junction law tells you that at a junction the sum of the currents entering the junction will equal the sum of the currents leaving the junction. However, this general law will not give you the specific values of the current flowing in each branch. These values depend on the resistance presented to the current in each path. When the paths leaving a junction are part of a network of parallel resistors, it is not difficult to calculate the specific value of each current. Consider the network below. 8 A flows into the junction. How do you determine the currents flowing through each individual resistor? The 20Ω resistor will have the least current, i 2. Since 2Ω is 1/10 the resistance, it will carry 10 times the current Since 5 Ω is 1/4 the resistance, it will carry 4 times the current The junction law says that all three currents must add to 8 A.

Exercises: Calculate the current in each resistor.

Simple Circuits A simple circuit is a collection of batteries and resistors that have two limitations on how they are connected. 1. All batteries are in series (i.e., they carry the same current). 2. All resistors can be combined to yield a single equivalent resistance using only the rules for series and parallel resistance. Since the batteries are in series, the equivalent battery voltage will be the algebraic sum of the individual battery values. Resistor networks will have to be reduced one at a time to eventually reach the single equivalent resistance. Once you have the equivalent battery voltage and the equivalent resistance, you can then calculate all the currents and voltage drops in the original circuit

Example: Calculate the following for the circuit below: a. equivalent resistance b. current in 26Ω resistor c. current in 5Ω resistor d. power consumed in 30Ω resistor e. voltage drop between points A and B. First reduce the parallel network. Now, all resistors are in series, so add them up. Since the batteries oppose each other, subtract the values: The equivalent circuit looks like this: Now find the current flowing in this “intermediate” circuit. Loop law: This is the key result. The current flowing in the intermediate circuit will always equal the current flowing in the batteries of the real circuit (and any elements that are in series with the batteries). Thus the current in the batteries and also the 20 Ω, the 26 Ω, and the 30 Ω resistors is 0.5 A.

To get the current in the 5 Ω resistor, use the current division techniques. Thus the current in the 5 Ω resistor is 0.4A. To get the power consumed in the 30 Ω resistor, just use the power equation for a device that obeys Ohm’s law: To find the voltage drop between A and B, start at one of the points, say A, and just add up the voltage drops of each device until you get to B. You get the same answer going counterclockwise through the 5Ω resistor also (careful of signs): What is the power supplied by the 60 V battery?

A Couple More Circuit Problems! 12 v 12  15  60  15  5  20  5  10  a. What is the equivalent resistance of the above circuit? b. What is the current through the 10  resistor? c. What is the potential difference across the 60  resistor? I. II. A 120 volt power line is protected by a 15 A fuse. How many 100 W bulbs can be connected to this line in parallel before the fuse is blown? III.You’ve probably tripped a breaker in your home. Describe the circumstances that lead to this inconvenience. IV. Why are the electric outlets near sinks different from other outlets?