Model Answers - electric circuits - complete

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

Derive the equation linking current with the number of electrons N flowing past a point in a time ∆t with the current

Answer 4

The charge flow ∆Q = Ne where e is the magnitude of the charge on each electron, e=1.6x10-19 C
I = Ne / ∆t

Question 5

Explain how to calculate the number of electrons in a certain amount of charge

Answer 5

Number of electrons = total charge / charge on one electron
o N = Q/ 1.6 x10 ^-19

Question 6

Define resistance

Answer 6

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

Question 7

Derive the formula for the ratio of currents down parallel branches of a circuit

Answer 7

The pd across each branch is the same V1 = V 2
o V = IR so
o I1R1 = I2R2
o So I1 / I2 = R2 / R1
o Or, the ratio of the currents is the reciprocal of the ratio of the resistances
o Eg. If resistor 1 is 100 times greater in resistance than resistor 2, it will receive 100 times less current than resistor 1

Question 8

Derive the formula for resistors in series

Answer 8

Vtotal = V1 + V2 + V3 (due to energy conservation)
o V= IR so
o ItotalRtotal = I1R1 + I2R2 + I3R3
o Itotal = I1 = I2 = I3 = I (charge conservation)
o lRtotal = IR1 + IR2 + IR3
o Rtotal = R1 + R2 + R3

Question 9

Derive the formula for resistors in parallel

Answer 9

o Itotal = I1 + I2 + I3 (due to charge conservation)
o I= V/R so
o Vtotal / Rtotal = V1/R1 + V2/R2 + V3/R3
o Vtotal = V1 = V2 = V3 = V (due to energy conservation)
o V/ Rtotal = V/R1 + V/R2 + V/R3
o 1/Rtotal = 1/R1 + 1/R2 + 1/R3

Question 10

Answer 10

Question 11

State Kirchoff’s potential difference law

Answer 11

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

Question 12

State Kirchoff’s current law

Answer 12

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

Question 13

Answer 13

To determine the resistance of the resistor, the student would use R = V/I where V is the pd across the resistor and I is the current through the resistor
o An ideal ammeter has zero resistance
o If the ammeter had non-zero resistance, then there would be potential difference across it
o Therefore the potential difference measured by the voltmeter would be the sum of the pd across the ammeter and the pd across the resistor. This would be problematic as the resistance calculation requires only the pd across the resistor (the voltmeter position would have to change to be only around the resistor).
o An ideal voltmeter has infinite resistance
o If the voltmeter had non-infinite resistance then it would draw some current and reduce the total resistance of the circuit, increasing the overall current.
o However, this would not affect the resistance determination, as the ammeter would still be measuring the genuine current through the resistor
o So student B is correct

Question 14

Answer 14

To determine the resistance of the resistor, the student would use R = V/I where V is the pd across the resistor and I is the current through the resistor
o An ideal voltmeter has infinite resistance
o If the voltmeter had non-infinite resistance then it would draw some current (and increase the total current in the circuit as the total resistance of the circuit would decrease)
o This would mean that the ammeter reading would give the sum of the current through the resistor and the current through the voltmeter
o This would mean the resistance calculation would not be possible as this should only use the current through the resistor, which is not measured here
o If the ammeter had non-zero resistance, then the total resistance of the circuit would increase and there would be potential difference across it
o Neither of those consequences are problematic though, as the voltmeter still only reads the pd across the resistor, and the current reading is still the same current passing through the resistor
o So the ammeter can have a non-zero resistance and student B is correct.

Question 15

Define lost volts

Answer 15

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

Question 16

Define terminal potential difference

Answer 16

V = EMF – Ir where r is the internal resistance of the cell and I is the current through the cell

Question 17

Explain why connecting an ideal voltmeter directly around a cell allows us to measure the EMF directly

Answer 17

The ideal voltmeter has infinite resistance
o So the current is approximately zero
o So as terminal pd = EMF – lost volts,
o V = E – Ir
o The lost volts is zero
o And terminal pd is equal to EMF

Question 18

Answer 18

Student A is correct
o Because the wires we assume to have zero resistance
o Therefore there is no potential difference dropped across the wires
o So, as the sum of the EMF is equal to the sum of potential difference drops
o Both voltmeters read the terminal potential difference, V = EMF- Ir

Question 19

Explain what happens to the reading on the voltmeter as the resistance of the variable resistor is decreased

Answer 19

The voltmeter reads the terminal potential difference, V
o V = EMF – Ir
o As the resistance of the variable resistor is decreased, the total resistance in the circuit decreases
o Meaning the current, I = EMF / (Rtotal) will increase
o So the lost volts, Ir will increase
o And as the EMF is fixed
o Terminal pd will decrease

Question 20

Explain why a battery will get hot when the current passing through it is high

Answer 20

If the current is high, the lost volts, Ir will be high
o So there will be a large amount of energy per unit charge transferred to the internal resistance of the cell
o This energy is transferred to the resistor ions
o Causing them to vibrate with higher amplitudes, increasing the temperature

Question 21

Draw the circuit diagram required to determine the emf and internal resistance of a cell:
 

Answer 21

Question 22

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

Answer 22

Question 23

Answer 23

The x intercept is when the terminal potential difference is zero
o This means that all of the EMF is being transferred to the internal resistance of the cell as ‘lost volts’
o no energy is being transferred to the load resistance and the current cannot continue to increase
o the y-intercept is when the EMF is equal to the terminal potential difference
o this is when the current is zero and therefore no energy is being transferred to the internal resistor
o lost volts, Ir is zero

Question 24

Answer 24

When the extra light bulb is connected in parallel, the total resistance of the circuit decreases
o This means the total current in the circuit increases, as Itotal = EMF / Rtotal and the EMF is fixed
o So the lost volts, Ir, increases
o This means the terminal pd across AB decreases, as Vterminal = EMF - Ir

Question 25

What can we say about the brightness of each light bulb after an extra light bulb is connected to the circuit (assuming the brightness of the bulb increases with the energy transferred)?

Answer 25

The power across AB is given by P=V2/R where V is the pd across AB and R is the resistance of the bulb between A and B
o As the terminal pd decreases as extra bulbs are added (because lost volts increases due to the increase in current caused by the decrease in total resistance), the power across AB will decrease
o And therefore the brightness will decrease as extra bulbs are added

Question 26

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

Answer 26

I is current – rate of flow of charge

o n is the number density of free charge carriers – the number of charge carriers per unit volume (n = N/V)

o q is the charge on an electron - - 1.6x10-19C

o v is the drift velocity of electrons – the mean speed of electrons due to an applied potential difference

o A is the cross sectional area of the material carrying current

Question 27

30. Derive I=nqvA

Answer 27

There are N electrons moving with a drift velocity v along a wire of length l and cross sectional area A

o The current I = ∆Q/∆t ———— 1)

o ∆Q is total charge flowing passed a point in a time ∆t

o ∆Q = N q where q is the charge on the electron

o v = l/∆t so ∆t = l/v

o Substituting into 1)

o I = ∆Q/∆t = N q v / l —————————— 2)

o There is a number density n of electrons n = N/V, so N = Vn = Al n

o Substituting into 2)

o I = N q v / l

o = A l n q v / l

o = nqvA

Question 28

Draw a diagram of a circuit you could use to determine the resistance of a wire

Answer 28

or simply an ohmmeter connected to a resistor in series (with nothing else in the circuit)
(note that there is no mechanism in either diagram to change the readings: these would take single readings to determine the resistance once. If you wanted to plot a graph, for the second diagram you could change the current by adding a variable resistor and changing its resistance and find corresponding values of the potential difference)

Question 29

32. Explain how to determine the resistance of a component from a V, I graph

Answer 29

o Choose a point on the IV graph and identify the current and pd values
o R = V/I
o Note that this is NOT the gradient of the line, as R is not ΔV/ΔI

Question 30

33. Draw the IV graphs and the VI graphs for an ohmic conductor of high resistance (a), an ohmic conductor of low resistance (b), a filament lamp, a thermistor and a diode

Answer 30

Question 31

34. Explain the IV relationship for a filament bulb

Answer 31

o 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
o This causes the resistor ions to oscillate with greater amplitudes so the temperature of the resistor increases
o This causes more frequent collisions between the electrons and the lattice ions, limiting the drift velocity
o This causes the current, I=nqvA to increase by a smaller factor than the potential difference (as n, q, A all fixed)
o So, as resistance is the ratio of pd to current, if pd increases by a greater factor than current does, the ratio R=V/I increases

Question 32

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

Answer 32

o As temperature increases, the electrons in the thermistor are provided with more energy
o This means they can be promoted from the valence band of the semiconductor to the conduction band
o As they are free and able to conduct, the number density of free charge carriers, n increases
o Thus, current, I= nqvA will increase by a greater factor than the potential difference.
o 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 33

36. 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 33

o As current increases, the electrons collide more frequently with lattice resistor ions, transferring more energy overall to them per second OR as pd increases, electrons gain more energy and transfer more energy to lattice resistor ions
o This causes the resistor ions oscillate with greater amplitudes so the temperature of the resistor increases
o 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.
o Thus, current, I= nqvA will increase by a greater factor than the potential difference.
o 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 34

37. Explain the resistance, light intensity behaviour for an LDR

Answer 34

Long answer:
o A higher intensity means a greater number of photons incident on the LDR per second, per m2
o So as intensity increases, more photons incident per second means more electrons gain energy from the incident photons per second (note that each electron still gains the same amount of energy from each photon, as photon energy is proportional to frequency not intensity)
o More electrons are promoted into the conduction band of the LDR per second, so n, the number density of free charge carriers increases.
o Current increases
o So, as resistance is the ratio of pd to current, if pd remains constant as current increases, resistance = V/I decreases
Short answer:
o Higher intensity of EM radiation means more electrons gain energy from photons
o Increasing the number density of free charge carriers
o Increasing current, reducing resistance

Question 35

38. Explain the shape of the IV graph shown below

Answer 35

o This is a diode
o It allows current to flow in one direction only
o The diode conducts when the pd is above the threshold voltage, around 0.7V
o For negative pds (in the reverse bias direction) the resistance is very high, but at a very large negative potential difference the diode will conduct in the reverse direction

Question 36

39. Derive the formula for potential dividers

Answer 36

Question 37

40. Describe what is meant by a potential divider circuit and draw a circuit diagram to illustrate this:

Answer 37

o a potential divider circuit is a simple circuit involving a fixed supply potential difference that is shared between two resistors.
o This can either be two separate resistors as shown in a) or a rheostat or potentiometer as shown in b).

Question 38

41. Describe what will happen to the voltmeter reading in circuit a if the resistance of the variable resistor increases:

Answer 38

o There is a fixed supply potential difference and the other resistor is of fixed resistance
o When the variable resistor’s resistance increases, the voltmeter reading increases
o This is because the potential difference is shared in the ratio of the resistances

Question 39

42. Describe how the voltmeter reading in circuit b changes as the slider is moved from the bottom of the resistor to the top

Answer 39

o At the bottom of the resistor, the voltmeter reading would be 0V
o At the top of the resistor, the voltmeter reading would be equal to the supply potential difference
o Potential difference is proportional to the length of the wire

Question 40

43. A student moves the slider along the potentiometer. Explain why potential difference is proportional to the length of wire

Answer 40

o resistance is proportional to length for a constant cross sectional area and resistivity, R=ρl/A
o The current is fixed as the total resistance in the circuit is unchanged as the slider is moved
o V is directly proportional to R as V=IR
o So V α R α l

Question 41

44. Explain which of these circuits would be better for plotting a graph of current against potential difference for plotting an IV graph for the component

Answer 41

o The right hand circuit would be better as it achieves a range of pds and currents from 0V to the EMF of the cell, whereas the left hand circuit has a minimum pd and current that is not close to zero.
o For the right hand circuit, the potential difference across the component can vary from 0V (when the slider is on the left) to the maximum pd of the power supply (when the slider is on the right)
o and the current through the component will be 0A when the slider is on the left and maximum when the slider is on the right
o for the left hand circuit, the component would always receive a share of the fixed total pd, and the current through it would never be zero
o minimum pd across the component in the left circuit would be V = EMF x R component/ R total, and minimum current would be EMF / Rtotal
o so you could not plot an IV graph that starts close to zero for the left hand circuit
(left hand circuit - data cannot be collected for small values of V, I)

Question 42

45. Define resistivity:

Answer 42

o the resistance of a unit cube

Question 43

46. Describe what resistivity depends on:

Answer 43

o Resistivity is independent of dimensions of the wire – it is unique to an individual material only and depends on temperature.

Question 44

47. Describe how the resistivity of a thermistor changes with temperature:

Answer 44

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

Question 45

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

Answer 45

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

Question 46

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

Answer 46

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