ELECTRICITY

Question 1

Define Potential difference

Answer 1

Potential difference is the energy transferred (or work done) between two points in a circuit per unit charge (V = W/Q)

Question 2

Define emf

Answer 2

emf is the work done per unit charge by the power supply or cell, converting
energy into electrical potential energy of the charges

Question 3

Define current

Answer 3

the rate of flow of charge (I = ∆Q/∆t)

Question 4

Define resistance

Answer 4

the ratio of pd to current (R=V/I)

Question 5

Define power

Answer 5

the rate of transfer of energy (P=E/t)

Question 6

Name and define each of the terms in the equation I = nqvA

Answer 6

1. I is current – rate of flow of charge.
2. n is the number density of free charge carriers – the number of charge carriers per unit volume.
3. q is the charge on an electron ( - 1.6x10 -19 C.)
4. v is the drift velocity of electrons – the mean speed of electrons due to an applied potential difference.
5. A is the cross sectional area of the material carrying current.

Question 7

State Kirchoff’s potential difference law:

Answer 7

The sum of the potential difference drops is equal
to the sum of the emfs around a closed loop within a circuit – this is due to conservation of
energy

Question 8

State Kirchoff’s current law:

Answer 8

The sum of the currents into a junction is equal to the sum of
the currents out of the junction – this is due to conservation of charge

Question 9

Describe how to perform an experiment to determine the EMF and internal resistance of a cell:

Answer 9

1. set up a series circuit with an ammeter in series with a variable resistor and cell.
2. A voltmeter should be connected in parallel with the cell to measure the pd across it.
3. Measure current and potential difference.
4. Change the current by changing the resistance of the variable resistor and measure the current and potential difference again for a range of currents.
5. Plot current on x axis, terminal potential difference on the y
   axis.
6. V = -r I + EMF
7. y = mx + c
8. Gradient of line of best fit graph = -r, y-intercept = EMF. To determine the internal resistance multiply the gradient by -1

Question 10

Define lost volts

Answer 10

The energy per unit charge transferred to the internal resistance of a cell (Ir)

Question 11

Define resistivity

Answer 11

The resistance of a unit cube

Question 12

Describe what resistivity depends on:

Answer 12

Resistivity is independent of dimensions of the wire –

it is unique to an individual material only and depends on temperature.

Question 13

Describe how the resistivity of a thermistor changes with temperature:

Answer 13

Resistivity of a thermistor decreases with temperature (as number density of charge carriers, n increases)

Question 14

Describe how the resistivity of a metal filament changes with temperature:

Answer 14

Resistivity of a metal filament increases with temperature (as drift velocity, v decreases due to more frequent collisions between electrons and ions)

Question 15

Describe how to perform an experiment to determine the resistivity of a wire:

Answer 15

1. Measure the resistance of a wire using an ohmmeter for different values of the length of the wire.
2. Measure diameter of wire using micrometer (in three places along the wire, taking an average).
3. Calculate the cross sectional area of the wire using A = π (d/2) 2
4. Plot a graph of resistance against length, the gradient of the line of best fit will be the resistivity / cross sectional area.
5. Multiply the gradient by cross sectional area to get the resistivity.

Question 16

Explain the IV relationship for a filament bulb

Answer 16

1. Initially, current is directly proportional to potential difference (the filament obeys Ohm’s law)
2. As current increases, the electrons collide more frequently with lattice resistor ions, transferring more energy to them per second OR as pd increases, electrons gain more energy and transfer more energy to lattice resistor ions.
3. This causes the resistor ions to oscillate with greater amplitudes so the temperature of the resistor increases.
4. This causes more frequent collisions between the electrons and the lattice ions.
5. This results in a reduced drift velocity.
6. This causes the current, I=nqvA to increase by a smaller factor than the potential difference (as n, q, A all fixed)
7. So, as resistance is the ratio of pd to current, if pd increases more than current does, the ratio R=V/I increases.

Question 17

Explain the resistance behaviour of a thermistor as temperature increases (due to external factors)

Answer 17

1. As temperature increases, the electrons in the thermistor are provided with more
   energy.
2. This means they can be promoted from the valence band of the semiconductor to
   the conduction band.
3. As they are free and able to conduct, the number density of free charge carriers, n
   increases.
4. This results in a higher current, as I=nqvA and q, v, A are all fixed.
5. As resistance is the ratio of pd to current, R=V/I, as current can be higher for the
   same potential difference at this higher temperature, the resistance decreases.

Question 18

Explain the IV relationship for a thermistor (here, the temperature increases due to the
current flowing through the thermistor so we must explain this step first)

Answer 18

1. As current increases, the electrons collide more frequently with lattice resistor ions,
   transferring more energy to them per second OR as pd increases, electrons gain
   more energy and transfer more energy to lattice resistor ions.
2. This causes the resistor ions oscillate with greater amplitudes so the temperature of
   the resistor increases.
3. But, as temperature increases, electrons can gain energy and so more electrons are
   promoted into the conduction band of the thermistor, so n, the number density of free
   charge carriers increases.
4. Thus, current, I= nqvA will increase by a greater factor than the potential difference.
5. So, as resistance is the ratio of pd to current, if current increases by a greater factor
   than pd does, the ratio R=V/I decreases.

Question 19

Explain the resistance, light intensity behaviour for an LDR

Answer 19

1. As light intensity increases, electrons can gain energy from the incident photons
2. Electrons are promoted into the conduction band of the LDR, so n, the number density of free charge carriers increases.
3. Current increases
4. So, as resistance is the ratio of pd to current, if pd remains constant as current increases, resistance = V/I decreases

Question 20

Derive the formula for resistors in series

Answer 20

(1): Itotal = I1 = I2 = I3 (charge conservation)
(2): Vtotal = V1 + V2 + V3 (energy conservation)
V= IR, so, substituting into (2)
ItotalRtotal = I1R1 + I2R2 + I3R3
As current is the same from (1)
Rtotal = R1 + R2 + R3

Question 21

Derive the formula for resistors in parallel

Answer 21

(1) Itotal = I1 + I2 + I3 (charge conservation)
(2): Vtotal = V1 = V2 = V3 (energy conservation)
I=V/R, so, substituting into (1)
Vtotal/Rtotal = V1/R1 + V2/R2 + V3/R3
As potential difference is the same from (2)
1/Rtotal = 1/R1 + 1/R2 + 1/R3

Question 22

Describe what is meant by a potential divider circuit

Answer 22

A potential divider circuit is a simple circuit involving a fixed supply potential difference
that is shared between two resistors.