# Wright State University Lake Campus/2017-9/Phy2410/Chapter 10: Circuits

In lieu of individual labs, students are expected to contribute to this project. During the writing phase, students are all asked to propose a sentence or paragraph and record it in their notes. After solicitation from a a few students, the instructor types in some wikitext. Editing by students from home is strongly encouraged, but not required. Whenever possible, we shall create an activity or lab using either PhEt or materials on hand in the lab. The project consists of developing a workable outline from Chapter 10 of Openstax University Physics. See also:

## Chapters

### Electromotive force

The emf, or ε, is actually a type of voltage, not unlike the electric potential. Here, we use the term to describe a simple model for an actual battery. If the load resister, R in this circuit is removed, the current vanishes, and the measured terminal voltage equals the emf: Vterminal = ε. Neither the ideal battery of voltage ε nor the internal resistor r actually exist as physical entities. But they adequately describe how the voltage typically drops as current is drawn from a battery. If a current is flowing through the battery, the voltage measured across it's two terminals is,

$V_{\text{terminal}}=\varepsilon -Ir$ ### Resistors in Series and Parallel ${\frac {1}{R_{\mathrm {total} }}}={\frac {1}{R_{1}}}+{\frac {1}{R_{2}}}+\cdots +{\frac {1}{R_{n}}}$ . $R_{\text{total}}=R_{1}+R_{2}+\cdots +R_{n}$ For only two resistors in parallel, the unreciprocated expression is reasonably simple:

$R_{\mathrm {total} }={\frac {R_{1}R_{2}}{R_{1}+R_{2}}}.$ This sometimes goes by the mnemonic "product over sum".

The resistances add when they are connected in series (shown at right).

#### Activity

calculate and simulate real battery with a load

Draw a circuit diagram for an real battery with an emf of 10 volts, internal resistance of one ohm, and a load of two ohms. Find:

1. The power developed in the internal resistor and the load. Total current 2.5A. Power in internal resistor 4.25W. Power in load was 12.75. The power delivered by the ideal emf source was (10V)(2.5A)=25W
2. The power delivered by the emf in the battery.
3. How do they compare?

What was the total current you observed, and why was it different?

We set up the circuit and measured 2.38 Amps for the current, which was a bit less than the 2.5A that we expected. It was possible to add an internal resistance to the 10V battery, and this was done by the instructor, who set it to 0.2Ω.

Discussion: We can use two different arguments to show that the emf is delivering less power due to this internal resistance. One uses P=Iε, and since I is reduced while ε remains at 10V, we see that P is reduced. We can get the same answer using P=ε2/R: When the denominsator increases the expression decreases.
Further discussion: One would think that using a battery with internal resistance would be less productive. At the next lab, we will use the laptops to set up the same circuit and the voltmeter to "measure" the power devolped in the 3Ω resistor. As expected, (to be continued)

## Equation sheets

### Expanded

• $V_{\text{terminal}}=\varepsilon -Ir$ is the terminal voltage where ε is the emf, I is the current, and r is the internal resistance.
• $R_{s}=\Sigma R_{i}$ is the equivalent resistance of resistors in parallel. If they are in series, the equivalent resistance is $R_{eq}^{-1}=\Sigma R_{i}^{-1}$ • Kirchoff's node rule is $\Sigma I_{\text{out}}=\Sigma I_{\text{in}}$ . The loop rules is $\Sigma \Delta V=0$ . The voltage drop $\Delta V>0$ if the path crosses through a voltage source from the negative to the positive terminals, or if it travels opposite the current when passing through a resistor.
• $I=(\varepsilon /R)\exp(-t/\tau )$ is the exponentially decaying current if a resistor R is placed in series with an emf ε that is used to charge a capactor C. The RC decay time is $\tau =RC$ 