** **

Suppose you had the following circuit. If you started with a low voltage and gradually increased it (by changing the battery or turning the button on a power source), what would happen to the current?

*Temporary digression: Don’t be confused by the terms current and voltage.
If we referred to a river current, we would be talking about the amount of
water that went by every second. In electricity, current is the amount of charge
measured in Coulombs that goes by every second. 1A = 1C/s.(A Coulomb is
the amount of charge carried by 6.25 X 10^{18} electrons). *

When electrons move out of a battery they eventually return to its positive terminal. Within the battery there are chemicals that are eager to release electrons[ (-) side] and others that are willing to take them back [(+) side] There is a potential energy difference between the area where the electrons are crowded and the uncrowded area, just like there is an energy difference between the top and bottom of a ski hill. Voltage is a measure of that potential energy difference in joules per coulomb of charge. 1 V = 1 J/C.

The current would keep increasing in proportion to the voltage, as shown by the data table below.

Voltage, V ( |
0.0 |
1.5 |
2.0 |
2.5 |
3.0 |

Current Intensity, I ( |
0.0 |
2.0 |
2.7 |
3.3 |
4.0 |

The slope on a *current (y) versus voltage(x)* graph is
known as conductance, G, which is measured in Siemens (S)

In this
case the slope = = G = conductance of
the circuit which is approximately the conductance of the light bulb.

Another electrical property,
more commonly used than conductance is **resistance**,
which is measured in **ohms(****W)**.
Resistance is a measurement of how difficult it is for electrons to get through
a substance. Resistance converts

electrical energy into heat.

R = 1/G
= 1/1.33 = 0.75W.= resistance of the circuit, which is approximately the resistance
of the light bulb.

Based on this, what is the relationship between V, I and R ?

Since G = I/R and since R = 1/G, it follows that R = V/I. Cross multiplying we get:

**Ohm’s Law:** V
= IR

1.
A resistance of 10** W** is placed across a 9 V battery. What
current flows through the battery?

Using Ohm’s Law,

V = IR

9 = I (10)

I = 0.9 A

2. a. A resistor has a conductance of 0.100 **S**.
What is its resistance?

R = 1/G = 1/0.100 = 10 **W**

b. What voltage is needed to cause a 500 mA current to flow through the circuit?

500 mA =
0.500 A

V = IR

V = 0.500(10) = 5 V

3. What is the overall resistance of a CD portable player if it is operated by a 3 V battery and 0.75 A flow through its circuitry?

V = IR

3 = 0.75(R)

R = 4.0 ** W.**