# Labs from Chicago, Winter 1994 : Parallel Circuits.

Dr. Rich Kron, Dr. Heidi Newberg, and Luisa Rebull
Labs written for the CARA Space Explorers, Winter 1994

This is meant to be handed out to the students.

## I. Introduction

In this lab we will construct a more complicated circuit than we did last week. At the end of last week's lab we measured the voltage, current, and resistance in a circuit that looked like this:

Figure 1

First, we will build this series circuit as a review of the concepts, and later we will consider a more complicated system that looks like this:

Figure 2

Note that in the first circuit there is only one path for the current, while in the second circuit there are two paths, one through resistor R2 and one through resistor R3.

## II. Review of a Series Circuit

The diagrams just given use a symbol for a resistor, but in fact any electrical component that has resistance will do - a light bulb, a buzzer, or a motor, for example. The brightness of a light bulb, the loudness of a buzzer, and the speed of a motor all depend on how much current is flowing through the circuit.

### Step 1.

Select three resistors, all with resistances within a factor of ten of each other. For example, if you find three resistors with a red band in the third position, you are assured that they will be OK. Use your multimeter set to W to measure the resistance of each of these three resistors:

```resistor		color code		resistance
(ohms)

R1			___________		____________

R2			___________		____________

R3			___________		____________
```
These three resistors are going to be used in the same positions marked in Figure 1 and Figure 2, so be sure to keep track of which one is which! Do this by writing in the resistances next to the resistor symbols R1, R2, and R3 on Figures 1 and 2.

### Step 2

Turn the dial on your multimeter to DC volts and measure the exact voltage of your 1.5 V or 3 V battery.

voltage of battery: Vbat = _____________ V

Now, build a circuit using the battery and the two resistors R1 and R2, as in Figure 1 (we'll get to resistor R3 later). To avoid running the battery down, we'll keep the switch open until we want to measure something.

### Step 3

Close the switch in the circuit you made in Step 2, and touch the probes to the points marked A and B (the red probe should be on point A). The reading is the "voltage drop across resistor R1." We also call this VAB as in the table below. Enter your measured value in the table. Now, repeat the experiment with the probes touching points B and C (this time, the red probe should be on point B), and enter the value for VBC in the table.

```	voltage		resistance			current
V			R				   I

AB	VAB = _______		R1 = ______		IAB = _______

BC	VBC = _______		R2 = ______		IBC = _______

```

### Step 4

Now we are going to analyze this circuit. We'll make a claim, then check it.

Claim: the current through R1 (IAB) has to be equal to the current through R2 (IBC), otherwise current would pile up somewhere.

Check: Since you have measured both the resistances (Step 1) and the voltages (step 3), and since you know Ohm's law, you can calculate the currents. Do this, and enter the calculated currents I into the table above. Is the claim verified? That is, does IAB = IBC?

We can also predict the values for VAB and VBC. Your battery has a voltage of Vbat; then,

```VAB  =  Vbat  x   R1 .
(R1+R2)
and

VBC  =  Vbat  x   R2 .
(R1+R2)

```
Calculate both of these numbers and fill out the table below:

```		measured		calculated

VAB		_________	__________

VBC		_________	__________

```

## III. Parallel Circuit

### Step 5

Now add your resistor called R3 to the circuit so that the circuit looks like Figure 2. The two resistors R2 and R3 are said to be in parallel. We'll make another claim and another check:

Claim: the combination of the two resistors called R2 and R3 acts like a single resistance R with a value of

```R  =  	 _____1______
(1/R2 + 1/R3)  .
```
Check: Now measure VBC again; if this claim is correct, then

```VBC  =  Vbat  x  R/(R1+R)  .
```
Fill out the table below:

```		measured		calculated

VBC		_________	__________

```

### Step 6

Remember, the current can't pile up anywhere in a circuit. This means that, for example, the current flowing into point B (see Figure 2) has to be equal to the current flowing out of point B. In other words, the current flowing through resistor R1 (I1) is equal to the combination (sum) of the two currents flowing through resistors R2 (I2) and R3 (I3).

Write down an equation that relates I1, I2, and I3:

I1 = _____________ .

Review: if two resistors R1 and R2 are connected in series as in Figure 1, then their combined resistance is

R = R1 + R2 (series)

If two resistors R2 and R3 are connected in parallel as in Figure 2, then their combined resistance is

```R  =  	______1______
(1/R2 + 1/R3)  		(parallel)

IV. Power

Build a circuit as in Figure 1, except replace the two
resistors with a light bulb.  If you have been using a
1.5 V battery, use one of the small, round light
bulbs.  If you have been using a 3 V battery, use one
of the larger, oblong bulbs.  The light bulb acts just
like a resistor:  it limits the flow of current from
the positive terminal of the battery to the negative
terminal.  The current heats the filament to a high
temperature, so high that it emits light.

Heating something up takes energy, which is defined as
the ability to do work.  The brighter a light bulb
looks, the more energy (in the form of heat and light)
is being produced per second or per hour.  Electrical
energy has been converted into heat and light; that
is, electrical energy is consumed and heat/light
energy is produced.  The "Watt" is a unit of a
physical quantity called power.  Power is the rate of
consuming or producing energy, so it has units

power  =  P  =  energy/time

The higher the Wattage of a light bulb, the faster it
uses electrical energy.  To calculate the power that
formula

P  =  V  x  I

where V is the voltage of your battery and I is the
current flowing in the circuit.  Remembering that
Ohm's Law is

V  =  I  x  R  ,

write down an expression for the power P in terms of
just I and R.

P  =  _____________

If one of your resistors used in the previous Steps
was getting hot, then you were putting too much power
into it.

Important Disclaimers and
Caveats

Go back to the Chicago Winter 1994 Electronics home
page.

```