Chapter 18
The Giancoli onLine Tutor

Electric Currents

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I. Ohms Law

The basic equation here is Ohm's Law:

V=RI.

where V is the applied voltage and I is the current defined by

I=DQ/Dt.

The unit of current, the ampere (A) equal to 1 coulomb/second. R stands for resistance and has the unit ohm (W). Ohm's law applies to resistive elements like wire, resistors, light filaments and the like.

Interactive Example 1

In each row inthe following table you can select the values for two of three elements. Do so, and enter the third value; press the button to see your result. Try other values until you are sure that you have mastered the equation.

V R I

FEEDBACK:

The resistance of a wire is proportional to its length, l, and inversely proportional to its cross-sectional area, A. The proportionality constant, r, is called the resistivity and depends upon the wire's material .

R= rl/A

Resistance also has a small temperature dependence, depending on the Temperature Coefficient, a.

R=Ro(1+a(T-20 ^oC)

where R_o is the resistance at 30 ^oC.

Typically you will be given three of the four unknowns( R, r, l, A) and be asked to find the fourth. Variations might include:

Giving the diameter, d, of the wire.
- Use A=p(d/2)².
Asking for the diameter of the wire.
- Find A and use d=SQRT(A/p).
Asking for the wire's material.
- Find r and use Table 18-1 in the book.
Asking for the temperature of the wire.
- Find R and use T= T_o+(R-R_o)/aR_o. Use table 18-1 for a.

II. Electric Power

The electric Power, P, in a circuit can easily be calculated by using

P=VI,

where P is in watts (VI= [J/C][C/s]= J/s=watt).

If a typical circuit like a penlight powered by a 1.5 volt AA battery there are two elements the battery and the light bulb. The battery is the source of the electrical energy and is called the Seat (or Source) of EMF. The light bulb is the resistance element. Once the current is established the power drawn from the battery is given by VI and the power expended in the light bulb is also given by VI. Since the filament in the light bulb is a resistance element Ohm's Law (V=RI) applies to it and combining this with the above equation gives us two more formulae for the power in a resistance element.

P= RI²= V²/R
(Resistance Elements Only!).
Interactive Example 2
In the following complete the table just as in example 1. Repeat to you have mastered it.

P R I

FEEDBACK:

Be careful with units. Power is commonly given mW (10^-3 W), kW (10³W) and MW (10⁶W). Similarly you will come across mA (10^-6A), mA (10^-3A), mV (10^-3V), kV (10³V) and MV (10⁶V). Other units you will encounter are killowatt-hour (kWh) which is a unit of energy equal to 3.6x10⁶ J, Horsepower (hp) which equals 746 W and ampere-hour (Ah) which is a measure of charge equal to 3600 C.

Some things to note:

When power is transmitted over real wires which have resistance energy is lost in the resistance heating (RI²) of the wires. Since the power is given by VI, for a fixed power, the higher V is the smaller I and hence RI² will be.

If you are given the Ah rating of a battery you can find the energy stored in it by converting this to coulombs (3600C/Ah) and multiplying by the voltage.
Thus a 100 Ah, 12 V battery has 100 Ah*12 V=100 Ah*3600 C/Ah*12 J/C= 4.32x10⁶ J.

The efficiency of a device is defined as Power Out/Power In.

III. AC Current

AC Current and Voltage
So far we have discussed DC currents, currents in which charge flows in one direction only and by convention from "+" to "-". We are all aware that there is also Alternating Current, AC, the common source of electrical energy in our homes. In alternating circuits the voltage across a circuit element such as a toaster oscillates back and forth fron (+) -> (-) to (-) -> (+) and consequentially the current through the toaster oscillates at the same frequency, which in the US is 60 Hz. See the figure below (the numbers on the time axis are arbitrary). I_o and V_o are the peak values for the current and voltage, respectively. I_rms and V_rms are the root-mean-square values for these variables (more about these later).
What follows applies (completely) only to AC circuits with resistance elements only. The voltage can be taken to be

V=V_osin 2pft,

where f is the frequency and t is the time. From Ohm's Law then
I=I_osin 2pft,

where I_o= V_o/R.

AC Power
It can be shown that the average power is given by

P_ave= RI_o²/2,

as shown on the right (RI_o² is the Peak or Maximum Power).
The I_rms and V_rms (root-mean-square) current and voltage are found to be

I_rms = I_o/SQRT(2)
V_rms= V_o/SQRT(2)

Given this, another formula for average power is
P_ave= V_rmsI_rms

Still other equivalent formulae for average power are:

P_ave= V_oI_o/2

P_ave= RI_rms²

P_ave= V_o²/2R

P_ave= V_rms²/R

(and collecting the two from above)

P_ave= RI_o²/2

P_ave= V_rmsI_rms

In working problems keep in mind that

If power, current or voltage is mentioned without the modifier "Peak", it is assumed that average power or rms current or voltage are being given. If your house voltage is 120 V, that is the rms voltage; your peak voltage is 120*SQRT(2)=170 V.

For most problems two pieces of information will be given out of [P_ave, P_o, V_rms, V_o, I_rms, I_o, R] and one from this group will be the target. Use the formula which contains these three elements. Choose from the 6 formulae above and
I_o= V_o/R

P_o= V_oI_o

P_ave= P_o/2

I_rms = I_o/SQRT(2)

V_rms= V_o/SQRT(2)

Interactive Example 3

Enter the missing values in each row to complete the table below.

P_ave P_o V_rms V_o I_rms I_o R

© 1999 Carl Adler mailto:Carl@Image-ination.com