# 9/28 Group Activity

ghw@udel.edu if you find typos or mistakes in this material.

New Generation

• Easy:
According to the circuit shown below, will this light the bulb? Why?
[Drawing of bulb with base touching bottom of battery and contact wired to other end of battery.]

Ans: A light bulb can be lit with a battery if electrical contact is made between one end (or terminal) of the battery and the base of the bulb and the other end (or terminal) of the battery and the other contact of the bulb. A complete circuit must be formed. If the contact can be make without separate wires other end of the battery.

• Medium:
A current of 3 amps flows through a resistor with a voltage of 24 volts. What is the resistance?

Ans: A straightforward application of Ohm's law R=V/I results in 8 ohm (24V/3a).

• Hard:
Two 1.5 V batteries in a flashlight with resistance of 2 ohm were left running for 3 hours. How much power would be left?

Ans: Assuming that the 2 batteries are in series to provide 3.0 V to the bulb of 2 ohm, we find that 1.5 A current would be flowing (kind of large, but numerically correct). Referring to our table of battery information, we find that a an alkaline D cell can provide 8000 mA-hr at a current of 100 mA. (Since the current here is much larger, this value of capacity is unlikely to be very accurate.) So with 8 A-hr available, the battery would discharge its energy in just 8/3 hr (160 min) at 3 A, an aggravatingly short time. So final answer is that no energy would be left -- battery would already be "dead."

An alternative answer: Just for interest, let's see what happens if the 2 batteries are used in parallel. Now only 1.5 V is applied to the lamp, resulting in a current lowered by half from before -- now 0.75 A. Also, since the identical batteries are in parallel, they each "push" half of this current through the lamp. So now each battery is driving about 0.4 A (much closer to the value of 100 mA in the table). Each battery will discharge in (8000 mA-hr/400 mA) = 20 hr -- a more reasonable lifetime. So final answer is battery has about 17 hr "worth of energy" left.

Ohms
• Easy:
How do you lower current associated with power distribution?

Ans: If the current is ac, a transformer may be used to boost the voltage, and subsequently lower the current. (The product of current and voltage, namely the power, is the same on the input and output sides of an ideal transformer.)

• Medium:
Why does ac come out of the wall and not dc?

Ans: The convenience of converting to high voltage for power distribution and back to lower voltage for safe use has led to the almost universal adoption of ac as the means for transporting electrical energy from power plants to the home.

• Hard:
Explain the difference between voltage and voltage difference.

Ans: Short answer is that these is none. When voltage is referred to in this course, I always refer to the voltage across two points, as in 3 V across a 6 ohm reistor. The voltage across the object is the voltage difference between those two points. Many electrical measurements assume that one reference point is at ground (reference point taken to be 0 V) -- thus the voltage reading at a point in a circuit is the voltage difference between ground and that point. The author of the handout is making a careful distinction when the term voltage difference is used.

Whatever
• Easy:
What is the ultimate goal of this class?

Ans: It depends on your perspective. My goal for this class is to create a comfortable yet challenging, enjoyable yet productive environment for learning the science concepts behind high technology. I hope to increase your science and computer literacy, to enable you to make sound judgements regarding scientific issues facing you in the future, to help you learn how to learn, and to exercise and expand your ability to think critically.

• Medium:
Using the circuit below, what is the current through the 20 ohm resistance?
[Drawing of 20 and 25 ohm resistors in series with 1.5 V battery.]

Ans: Since the two resistors are in series, the same current will be flowing in each resistor. For a series circuit, the total "effective" resistance of the circuit is the sum of the two resistances: 20 ohm + 25 ohm = 45 ohm. From Ohm's law, the current is I=V/R=(1.5 V)/(45 ohm)=0.033 A or 33 mA.

• Hard:
How do you determine the voltage difference of an 11 ohm resistor?
[Drawing of an 11 ohm resistor connected to a 6 V battery with 1 ohm internal resistance.]

Ans: Again, since the two resistors are in series, the effective resistance is 12 ohm (=1 ohm + 11 ohm). The current in the circuit is then I=V/R=(6 V)/(12 ohm) = 0.5 A from Ohm's law. Finally, we may now find the voltage across either of the batteries by using Ohm's law, the value of current, and the associated resistance: V=IR=(0.5 A)(11 ohm) = 5.5 V.

Voltage
• Easy:
Which resistor has the most current flowing throught it? R1>R2>R3
[Drawing showing three resistors, R1, R2, and R3, connected in parallel to battery.]

Ans: Connected in parallel, each resistor will have the same voltage across it. From Ohm's law, I=V/R, for a given voltage, a larger resistance will have a smaller current --> I3 > I2 > I1.

• Medium:
Water flows from high to low pressure [was low to high in original statement]; if you want to reverse this process, how would you do it?

Ans: A water pump in a fluid system palys the analog to a battery in an electrical circuit. The pump is capable of moving the fluid from a low pressure to a high pressure; a battery moves charge (moving charge constitutes the current) to a point having a higher "voltage." Without a source of energy to restore it, current flows from high to low voltage.

• Hard:
When going from low resistance to high resistance, how does the bottleneck theory play a role?

Ans: Attempting to visualize a region of high resistance as a "bottleneck" has some merit. Since the resistance of a cylinder of uniform material depends inversely on its cross-sectional area, the resistance becomes larger where the area is smaller -- hence the "bottleneck" to flow.

Threesome
• Explain all the similarities and differences between series and parallel circuits.

Ans: Let's consider the case of just two resistors.

In series, the two resistors will have the same current through each. In parallel, they will have the same voltage across each.

In series, the effective resistance is the sum of resistances; thus the effective resistance is always larger than either one. In parallel, the reciprocal of the effective resistance is the sum of the reciprocal resistances. It turns out that the effective resistance is always smaller than either one.

In series, the larger resistor will have a proportionally larger voltage across it. In parallel, the smaller resistor will have a larger current flowing through it.

In series, the two resistors have only one end tied together, with nothing else attached at that point. In parallel, the two resistors have both ends tied together, with nothing in between.

In parallel, each resistor has access to the same voltage (good for household wiring). In series, if one of the resistors "breaks" (as in a lamp "burning out"), no current flows through any resistor (bad for useful wiring).

Perhaps you can think of more...

• Why is the resistance of a cylinder proportional to the length, but inversely proportional to the cross-sectional area?

Ans: One answer would be "That's the way it is!" After all, we did the Play-Doh experiment in class and easily saw a direct proportionality between resistance and length. The inverse proportionality between resistance and cross-sectional area came after a great deal of sample fabrication and analysis.

The physical origin of resistance is a microscopic scattering of charge carriers as they attempt to traverse the material. The longer the distance that the charge carriers must migrate, the more scattering that takes place, and the larger the overall resistance. Think of putting a bunch of identical resistors in series; the more resistors, the greater the effective resistance.

The lowering of resistance by increasing area may be thought of as a bunch of resistors in parallel. The more resistors in parallel, the lower the effective resistance, given that there are more possible paths.

• Why is ac current preferable to dc current in reference to household wiring?

Ans: The ease of transforming ac voltages was mentioned above. In addition, motors of simpler design are possible for ac. In particular, three-phase motors offer numerous advantages not possible with dc motors (increase starting torque, more uniform torque).

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Comments, suggestions, or requests to ghw@udel.edu.