Suppose we forced each truck, electron, to carry energy through

two light bulbs as pictured on the right.

The energy is now shared amongst the light bulbs. Each light bulb gets a share of the energy. At the end of its journey each truck has no more energy

Notice how trucks entering the first light bulb carry 6 bags of energy while the trucks leaving the light bulb have only three.

There is a difference of 3 bags of energy. We say there is a potential difference of 3 volts across the terminals of the light bulb.

What is the potential difference across the terminals of the second light bulb? 3 volts

What is the potential difference across the terminals of the battery?

6 volts

So in series circuitsVt = V 1 + V 2

Is the number of electrons leaving the battery the same as the number of electrons passing through each light bulb? YES

So in series circuits It = I1 = I2

Potential difference in series and parallel circuits

The animation on the right shows another way of connecting two light bulbs.

Parallel circuit

What is the potential difference across the terminals of each light bulb?

The potential difference is 6 volts

What is the potential difference across the terminals of the battery?

The potential difference is6 volts

So in the parallel circuitVt = V 1 = V 2

Current in series and parallel circuits

now look at the animation on the right again.

Is the number of electrons leaving the battery the same as the number of electrons passing through each light bulb?

The answer is no, as the number of electron that leaving the battery is bigger than the number of electrons passing through each light bulb.

Or we can say that:

the number of electron that leaving the battery equals to

the number of electrons passing through the two light bulbs.

So, The Total current equals to the sum of the current that passing through the two bulbs.

So in the parallel circuitIt = I1 + I 2

Adding resistors to series and parallel circuits

In the previous part we examined what happens to the resistance and the voltage in the series, and parallel circuits, and we detected the values of the total electric current (It) and the total voltage (Vt) in both of them. Now we will figure out how adding resistors to a circuit in series, and parallel changes the total resistance of the circuit (Rt).

YouTube Video

You can check how adding resistors affect the total resistance

in series and parallel circuit from this link.

So from the previous information about the total current, total voltage, and Ohm's law

we will detect the value of :

1- The total resistance in series circuits:

In the above series circuit:

The value of the electric current is constant through the resistors.

from Ohm's law:

The total potential difference is (Vt) , the voltage on resistor R1is V1 - on R2 is V2 -- on R3is V3

V1= IR1 V2 = IR2 V3 = IR3

Vt = V1 + V2 + V3

Vt = IR1 + IR2 + IR3

Vt = I ( R1 +R2 +R3 )

Vt = I Rt

So the total resistance in the series circuit is: Rt = R1 +R2 +R3

So when you add resistors in series, the total resistance is increased.

2- The total resistance in parallel circuits:

In the above Parallel circuit:

The value of the voltage on each one of the resistors is constant (Vt).

.

. the total resistance in the parallel circuit is

.

So when you add resistors in parallel, the total resistance is decreased

You can calculate the resistance, current, and voltage in parallel and series circuit from this game:

The relationship between the length and diameter of a wire and its resistance

Watch this videos at investigate how does the length and diameter of a wire and its resistance

YouTube Video

Imagine that there are trucks loaded with goods, and we want it to reach the store as soon as possible, Which is faster? Traveling through a long. or a short road?

And if we have many of these trucks, and we want all of them to reach the store as soon as possible, Which is better to travel through a narrow road or a wider one?

The same thing for electricity as

(trucks are the electric charges - goods are electric energy-road is the wire-the store is the bulb )

So using a thinner (with a small diameters) wire
makes the bulb dimmer, that’s because when the wire is thick it will
let more electricity pass to the bulb.

Lengthening
the wires in a circuit will reduce the electrical energy, as it has further to travel. The
extra distance will make the bulb dimmer.

YouTube Video

So the electrical resistance(R) is directly proportional with the length of the wire (l)

R ∝l

the electrical resistance(R) is inversely proportional with the cross sectional area (or diameter) of the wire (A)

R ∝ (1/A)

so R ∝ (l/A)

ρ

resistivity [Ωm] . Resistivity is material property (each material has its own resistivity). This means that resistivity only applies to a given object. It describes how well a material resists an electric current.

Check this link to see how does the length, and cross sectional area affect the resistance in a wire,

this link combines the actual experiment with the algebraic relationship.

If we measured the resistance of the wire (using R = V/I) for at least five lengths of wire with the same diameter (the same cross sectional area). We will use these results to plot a graph of length against resistance.

This investigation can then be repeated using wires of different diameters (with the same length).

Watch this video to help you to plot the data in a line graph.

YouTube Video

Problems

1. A pure copper wire has a radius of 0.5mm, a resistance of 1 MΩ, and is 4680 km long. What is the resistivity of copper?

2-Determine the overall resistance of a 100-meter length of (0.163 cm diameter) wire made of the following materials

According to these values of resistivty, What is the best metal to conduct electricity?

a.

copper

(resistivity = 1.67x10-8Ω•m)

b. silver (resistivity = 1.59x10-8Ω•m)

c. aluminum (resistivity = 2.65x10-8Ω•m)

d. iron (resistivity = 9.71x10-8Ω•m)

Answer :

a. 0.800 Ω

b. 0.762 Ω c. 1.27 Ω d. 4.65 Ω

.

l

1.

A wire 12m long has a resistance of 1.5 Ω. What will be the resistance of 16m of the same wire?

a) 1.6Ω

b) 2 Ω

c) 12Ω

d) 1.4Ω

2.

A sample of copper wire of 0.2mm radius has a length of 5m. If the resistivity of the copper is 1.7 x 10^{-8}, what would be the resistance of the wire?

a) 60.49Ω

b) 42.5Ω

c) 4.25Ω

d) 0.676Ω

3.

What is the resistance of 100m of copper wire, having a diameter of 1.024mm?

a) 2.06Ω

b) 5.15Ω

c) .515Ω

d) 3.7Ω

4.

The resistance of a 50m length of wire is found to be 0.83 ohms. Its diameter is 1.15mm. What is its resistivity and from what metal is it most probably made?

a) 2.7 x 10^{-8} aluminium

b) 10.5 x 10^{-8} iron

c) 17.24 x 10^{-9} aluminium

d) 1.72 x 10^{-8} copper

5.

What is the length of a copper wire having a radius of 0.25mm if its resistance is found to be 0.2 Ω?

a) 2m

b) 2.31m

c) 16.96m

d) 1.7m

Now to revise all what we have covered, please watch this video

YouTube Video

Check your Learning:

Complete the tables of current and voltage results

for all the circuits in the word file attached below: