Electricity Part 2

Question 1

What are the current and potential difference rules for series circuits?

Answer 1

Rule 1 : In a series circuit, current is the same all the way round the circuit.
Rule 2: In a series circuit, the potential difference is shared between all the components in the circuit.

Question 2

What are the current and potential difference rules for parallel circuits?

Answer 2

Rule 1: In a parallel circuit, the current splits up to go down different routes. The total current through the cell, is the sum of all the seperate currents.
Rule 2: In a parallel circuit, the potential difference is the same across each route.

Question 3

The potential difference across each component (in a loop) is dependant on what?

Answer 3

The potential difference is shared between the component in the circuit in ratios of their resistances. The higher the resistance, the greater the share of voltage (pd).

Question 4

The current across each loop in a parallel circuit is dependant on what?

Answer 4

In a parallel circuit, the current through a route depends on the resistance of each route available. The higher the resistance of a route, the lower the current that will flow along it.

Question 5

Why is the current through each component (or anywhere) in a loop, the same?

Answer 5

This is because the rate of flow of charge through each component or any point in the loop is the same - charge is conserved (Kirchoff’s 1st Law)

Question 6

Given that each component has a different resistance, why is the voltage through each component in series, different?

Answer 6

Different resistances mean that different amounts of work is done per coloumb of charge across the different components. This means the greater the resistance across a component, the more work done per coloumb of charge across that component.

Question 7

What is a potential drop?

Answer 7

If a charge carrier loses energy, the potential difference is described as a potential drop (e.g. at a component)

Question 8

What is a potential gain?

Answer 8

If a charge carrier gains energy, the potential difference is potential gain (e.g. a battery/cell terminal - the potential gain is equal to the pd across the battery or source).

Question 9

What are the resistance rules in series or parallel circuits?

*VIEW DERIVATIONS IN NOTES

Answer 9

Series: For two or more resistors in series, the total resistance is equal to the sum of all the individual resistances : Rₜₒₜₐₗ = R₁ + R₂ + R₃ + …
Parallel: For two or more resistors in parallel, the total resistance R is given by: 1/Rₜₒₜₐₗ = 1/R₁ + 1/R₂ + 1/R₃ + …

Question 10

What is the heating effect of an electric current in any component due to ?

Answer 10

The heating effect of an electric current in any component is due to the resistance of that component.

Question 11

The greater the resistance of the component…

Answer 11

…the greater the frequency of collisions between the charge carriers collide and the fixed positive metal ions of the conducting material. This causes the electrons to do more work.

Question 12

For a component of resistance R, when current I passes thriough it, the pd across it is equal to 1) _______. Therefore the power supplied to the component is 2)_____

Answer 12

1) V = IR
2) P = IV = I² x R = (v²/ R)

P = Power (W - watts)
I = Current (A - amperes)
V = Potential Difference (V - volts)
R = Resistance (Ω - ohms)

Question 13

If the component is at a constant temperature, what does this tell us about the rate of heat transfer to the surroundings? Equation?

Answer 13

Rate of heat transfer to the surroundings = rate that energy is transferred (power) to the component (as thermal energy) . Therefore:
I² x R = rate of heat transfer

Question 14

If the component does heat up what does the temeprature rise depend on?

Answer 14

1) The rate at which thermal energy is transferred to it (I² x R)
2) The rate at which thermal energy is transferred to the surroundings

Question 15

How can you calculate the energy transferred to the object by the electric current in time t?

Answer 15

Pt = E
P =  I² x R or v²/ R or IV
Pt = (I² x R or v²/ R or IV )x t = E

Question 16

Define ‘internal resistance’

*Look how they are drawn in a circuit diagram in notes

Answer 16

Internal resistance of a source is the loss of potential difference per unit current in the source when current passes through the source.

Question 17

Define ‘electromotive forve’

Answer 17

Symbol : ε

The electrical energy produced per unit charge passing through the source.

Question 18

If electrical energy, E, is given to a charge Q in the source, what is the equation to calculate emf?

Answer 18

ε = E/Q (Variation on V = W/Q)

ε = e.m.f ( V - volts)
E = Electrical energy (J - joules)
Q = Charge (C - coulombs)

Question 19

What is ‘terminal pd’? compare to emf.

Answer 19

• The terminal pd is the pd across the terminals of a power supply.
• When current is not flowing through the source, emf = terminal pd.
• When current is flowing through the source, the terminal pd is less than the end.
• The difference is due to the internal resistance of the source that results in ‘lost volts’.

Question 20

What is internal resistance due to?

Answer 20

• The internal resistance of a source is due to opposition to the flow of charge through the source (electrons colliding with ions in the battery).
• This causes electrical energy produced by the source to be dissipated inside the source when the charge flows through it.

Question 21

When a cell of emf ε and internal resistance ‘r’ is connected to an external resistor of resistance ‘R’, all the current through the cell, passes through its internal and external resistor. So the two resistors are in series, which means total resistance of the circuit = ‘r + R’. Therefore what is the current through the cell and thus the cell emf?

Answer 21

1) Current through the cell

I = ε / ‘R + r’

ε = I (R+r) → IR + Ir → ε = IR + Ir

ε = Emf (V - volts)
I = Current (A - amperes)
R = external resistance
r = internal resistance
Ir = lost or wasted pd due to internal resistance
IR = terminal pd

Question 22

What is the equation to calculate the :

1) Power supplied by the cell
2) Power wasted by the cell due to Ir
3) Power supplied to R

Answer 22

Iε = I²r + I²R (multiplying each term from the equation above by the cell current)

Iε  = power supplied by the cell (W - watts)
I²r = power wasted by the cell due to internal resistance
I²R = power supplied to R

Question 23

How does the value of the power delivered to R (the load) varies with the value of R?

Answer 23

• On a graph of power against resistance of the load, the peak of the curve is when R (the resistance of the load) = r (resistance of the internal resistor)
• This means that the maximum power is delivered to the load when the load’s resistance is equal to the internal resistance of the source.
• The power delivered to the load increases as the value of R (resistance of the load) increases up to the value of internal resistance, however when the value of R increases beyond that of r, the internal resistance, the power delivered, decreases.

Question 24

Outline an experiment that you can do to calculate the internal resistance

Answer 24

1) Connecting a voltmeter across the cell terminals, allows you to see how terminal pd changes with different values of current.
2) The current is changed by adjusting the variable resistor. The lamp (or fixed resistor) limits the maximum current that can pass through the cell. An ammeter is used to measure the cell current.
3) The measurements of terminal pd are plotted against current to give a straight line graph. The gradient of this graph will give you the negative of the internal resistance (terminal pd/current). Multiply this by -1 to give the internal resistance of the cell.

*Look at the circuit diagram and graph!!

Question 25

What happens to the terminal pd as the current increases?

Answer 25

As the current increases, the amount of pd that is lost due to internal resistance increase, and so the terminal pd (emf - wasted pd) decreases.

Question 26

Why is the emf equal to terminal pd when the current is zero?

Answer 26

Emf equals terminal pd when the current is zero, because no pd is lost or wasted due to internal resistance.

Question 27

The graph produced by plotting terminal pd against current produces what kind of graph line and why?

Answer 27

• It produces a straight graph line with a negative gradient.
• ^^Produces this because ε = IR + Ir rearranged to get → IR = -Ir + ε

Ir = V (Terminal P.D.):
V = - Ir + ε  → Same format at y = mx + c:
*m = -r (Gradient → internal resistance)
*c = ε (Y-intercept → emf)
*x = I (X-values → current)
*y = V (Y-values → terminal pd)

Question 28

As long as you know the 1)________ for two different values of 2)_________, you can calcualte the internal resistance and emf of the cell.

Answer 28

1) Terminal pd
2) Current

r = V₂ - V₁ / I₂ - I₁

Question 29

How to calculate the current through a cell?

Answer 29

Cell current = cell emf / total resistance of the circuit.

Question 30

If the cells (in series) are connected in the same direction in the circuit, the net/overall emf is…

Answer 30

…the SUM of the individual emfs

Question 31

If the cells (in series) are connected in the opposite directions to each other, the net emf is…

Answer 31

…the difference between the emfs in each direction

Question 32

The total internal resistance (of each cell in series) is the…

Answer 32

…the SUM of the individual internal resistances.
[This is because, the cells, and therefore the internal resistors, ar in series].

[Sees no direction like emf]

Question 33

For a circuit, with n IDENTICAL cells in parallel, the current through each cell is?

Answer 33

I/n where I is the total current supplied by the cells.

*Identical should tell you that each cell has the same internal resistance

Question 34

1) If I/n is the current through each of the n identical cells, what is the lost pd in each cell?
2) Hence what is the terminal pd through each cell?

Answer 34

1) I/n x r = Ir/n, where r is the internal resistance of each cell.
2) V = ε - Ir/n

Question 35

In a circuit with one or more diodes…?

Answer 35

1) A pd of 0.6V (threshold voltage) exists across the diode that is forward biased and it passes a current
2) A diode that is reverse biased has infinite resistance so does not pass a current

Question 36

What is a potential divider?

Answer 36

• A potential divider consists of two or more resistors in series with each other and a fixed source of potential difference.
• Used to ​produce a required fraction of the source potential difference

Question 37

By making a suitable choice of components a potential divider can be used:

*State suitable components

Answer 37

1) To supply a pd which is fixed at a value between zero and the source pd. (2 fixed resistors)
2) To supply a variable pd (potentiometer or variable resistor + e.g. thing that needs pd varied across it, e.g. resistor, diode, filameent lamp)
3) To supply a pd that varies with a physical condition such as temperature or pressure. (variable resistor/fixed resistor + thermistor/LDR)

Question 38

2) How is the pd shared across the two e.g. resistors?

Answer 38

V₁ / V₂ = R₁ / R₂

This tells us that ratio of pds across the resistors is equal to the ratio of resistances across the resistors.

Therefore to calculate the potential difference across, e.g. R₁
Vₒᵤₜₚᵤₜ/V₁ = Vᵢₙₚᵤₜ x R₁/R₁ + R₂

Question 39

How can a potential divider used to supply a variable pd?

Answer 39

The source pd is connected to a fixed length of uniform resistance wire. A sliding contact can then be moved along the wire, varying a pd across a specific component (this is using a potentiometer). The track may be linear of circular.
2) Instead of a potentiometer, a variable pd can be produced by changing resistance of the variable resistor. Increasing the resistance of it will increase the potential difference across it.

Question 40

How is a potential divider used in controlling the volume of a loudspeaker or the brightness of a light bulb?

Answer 40

• To change the volume of a loudspeaker:
  1) audio signal pd is supplied to the potential divider in place of a cell/battery.
  2) The variable output pd from the potential divider is supplied to the speaker.
  3) Increasing the resistance across the device in parallel with the loudspeaker, increases the pd across it and thus the pd supplied to the loudspeaker which increases the volume.
• The brightness of a light bulb can be altered between zero and normal brightness:
  1) A variable potential divider (with a sliding contact) can be used to do this.
  2) The variable output pd is supplied to a light bulb. Increasing the resistance across the potential divider, increases the output pd across the lamp, so brighter lamp.

Question 41

Draw the ways in which a heater can be used in a potential divider circuit, to act as a heater and air conditioner. (Replace with LDR, for light sensors etc)

Answer 41

DRAW ON PIECE OF PAPER - Figure 1 and 2 in notes, last page.

Question 42

Describe how each potential divider circuit with a thermistor works as a heater/ air conditioner.

Answer 42

HEATER = As temperature of thermistor decreases, the resistance across it increases and thus potential difference across it also increases. As a result, the heater connected in parallel with the thermistor, recieves a higher output pd, - acts more powerfully (releases heat) to reheat the area.
AIR CONDITIONER = If the position of the variable resistor and thermistor was switched and an air conditioner was put in parallel, in place of the heater, the opposite effect would be observed. As the temperature of the thermistor increases, the resistance across it decreases, and thus the potential difference across it also decreases. This means the variable resistor recieves a greater share of the pd (don’t say the resistance increases, because it doesn’t - its only the ratio that changes) and so the output pd supplied to the air conditioner in parallel with it would be greater - acts more powerfully to cool down the area

Question 43

Describe how each potential divider circuit works when the light intensity increases if an LDR replaces its position in figure 1 and figure 2,

Answer 43

Figure 1 : As light intensity on the LDR increases, it resistance decreases and the pd across it decreases. The output pd supplied to the component conencted in parallel with the LDR(e.g. bulb) decreases (and so the bulb will dim).
If the LDR in figure 1 was completely covered - i.e. light intensity at a minimum, its resistance would be at its maximum value, the output pd would also be at the maximum value (and thus bulb would shine brightly)
Figure 2 : If the position of the two resistors was reversed and the variable resistor was placed in parallel with the component (e.g. bulb) the opposite effect would be observed. As the light intensity increases, the resistance across LDR would drop and a greater proportion of the cell’s pd would be used up over the resistor; and supplied as output pd to the component in parallel (if a bulb would be brighter when the incident light is brighter)

Question 44

State Kirchoff’s 1st Law

Answer 44

“The total current flowing into a junction is equal to the total current flowing out of that junction​.” [This shows that no charge is lost at any point in the circuit/charge is conserved]

Question 45

State Kirchoff’s 2nd Law

Answer 45

“For any complete loop of a circuit, the sum of emfs around that loop is equal to the sum of potential drops around that loop” [This shows that no energy is lost at any point in that circuit/ energy is conserved]

Question 46

What is a difference between potential difference and emf?

In an exam, first state their definitions (which are different) and then state this extra difference.

Answer 46

• Potential difference is used to describe when work is doine by charge carriers (lose energy as they pass through a component) whereas emf is used to describe when work is done on the charge carriers (gain energy as they pass through the source).

Question 47

If we apply V = IR to internal resistance we can see that:
- lost volts = Ir, where r is the internal resistance

If r is fixed, what does thsi tell us about the relationship between the ‘lost volts’ and the current?
What does this then tell us about the terminal pd in the circuit?

Answer 47

If r is fixed, this means that the current through the source is directly proportional to the lost volts. This means increase in current = increase of the lost volts = decrease in terminal pd, given that the emf remains constant.

Question 48

To produce a higher current or higher emf, how can we position the cells in a circuit?

Answer 48

Depending on the desired effect, we can connect them in series or parallel:

1) Connecting cells in series increase the available emf, but also increases the internal resistance. This limits current that the combination can produce.
2) The same two cells connected in parallel produce the same emf as one cell, but have a much smaller internal resistance, so provide a greater current.

Question 49

Advantages of using a potentiometer instead of a potential divider?

Answer 49

• Can be made very compact
• Uses fewer components
• Can easily be made into a rotary dial
• Allows the full range of output potential difference from 0V to Vᵢₙₚᵤₜ