IB Physics SL B.5 Current & Circuits

[Maximum number: 8]

This question is in two parts. Part 1 is about a thermistor circuit. Part 2 is about vibrations and waves.
Part 1 Thermistor circuit
The circuit shows a negative temperature coefficient (NTC) thermistor X and a $100 \mathrm{k} \Omega$ fixed resistor R connected across a battery.

The battery has an electromotive force (emf) of 12.0 V and negligible internal resistance.

Part 2 Vibrations and waves

The cone and dust cap D of a loudspeaker L vibrates with a frequency of 1.25 kHz with simple harmonic motion (SHM).

(a)

Define electromotive force (emf).

[ 1 ]

(b)

The graph below shows the variation with temperature T of the resistance $R_{\mathrm{x}}$ of the thermistor.

[ 7 ]

(i)

Determine the temperature of X when the potential difference across R is 4.5 V .

[ 4 ]

$V_X=7.5\,\mathrm{V}$ $I=\frac{4.5\,\mathrm{V}}{100\times10^3\,\Omega}=4.5\times10^{-5}\,\mathrm{A}$ $R_X=\frac{7.5\,\mathrm{V}}{4.5\times10^{-5}\,\mathrm{A}}=1.67\times10^5\,\Omega$ $T=-37\text{ to }-38\,^{\circ}\mathrm{C}$

Alternatively, use the potential-divider relation V_X/V_R=R_X/R_R or the equivalent resistance calculation.

(ii)

State the range of temperatures for which the change in the resistance of the thermistor is most sensitive to changes in temperature.

[ 1 ]

50 to (up to) $-30^{\circ} \mathrm{C} /$ at low temperatures;

(iii)

State and explain the effect of a decrease in temperature on the ratio

\frac{\text { voltage across } \mathrm{X}}{\text { voltage across } \mathrm{R}}

[ 2 ]

as the temperature decreases $R_{X}$ increases;
same current through R and X so the ratio increases or $V_{X}$ increases and $V_{R}$ decreases so the ratio increases;

[Maximum number: 10]

(a)

A cell is connected to an ideal voltmeter, a switch S and a resistor R. The resistance of R is $4.0 \Omega$ .

When S is open the reading on the voltmeter is 12 V . When S is closed the voltmeter reads 8.0 V .

[ 3 ]

(i)

State the emf of the cell.

[ 1 ]

$12 \mathrm{~V}$

(ii)

Deduce the internal resistance of the cell.

[ 2 ]

I=2.0 A OR 12=I(r+4) OR 4=I r OR 8=4 I

«Correct working to get » r=2.0 « $\Omega$ »

Marking guidance:

Allow ECF from (b)(i)

(b)

The voltmeter is used in another circuit that contains two secondary cells.

Cell A has an emf of 10 V and an internal resistance of $1.0 \Omega$ . Cell B has an emf of 4.0 V and an internal resistance of $2.0 \Omega$ .

Calculate the reading on the voltmeter.

[ 3 ]

(c)

A fully charged cell of emf 6.0 V delivers a constant current of 5.0 A for a time of 0.25 hour until it is completely discharged.

The cell is then re-charged by a rectangular solar panel of dimensions $0.40 \mathrm{~m} \times 0.15 \mathrm{~m}$ at a place where the maximum intensity of sunlight is $380 \mathrm{Wm}^{-2}$ .

The overall efficiency of the re-charging process is 18 %.
Calculate the minimum time required to re-charge the cell fully.

[ 3 ]

ALTERNATIVE 1

«energy output of the panel = » VIt OR $6 \times 5 \times 0.25 \times 3600$ OR 27000 «J»

«available power = » $380 \times 0.4 \times 0.15 \times 0.18$ OR 4.1 «W»

$t=\frac{27000}{4.1}\approx6600\,\mathrm{s}$

ALTERNATIVE 2

«energy needed from Sun = » $\frac{V I t}{\text { eff }}$ OR $\frac{6 \times 5 \times 0.25 \times 3600}{0.18}$ OR 150000 «J»

« incident power=» $380 \times 0.4 \times 0.15$ OR 22.8 «W»

$t=\frac{150000}{22.8}\approx6600\,\mathrm{s}$

Marking guidance:

Allow ECF for MP3

Accept final answer in minutes (110) or hours (1.8).

(d)

Outline why research into solar cell technology is important to society.

[ 1 ]

[Maximum number: 6]

The graph shows how current I varies with potential difference V for a resistor R and a non-ohmic component T.

(a)

State how the resistance of T varies with the current going through T .

[ 1 ]

$R_{\mathrm{T}}$ decreases with increasing I

OR

$R_{\mathrm{T}}$ and I are negatively correlated

Must see reference to direction of change of current in first alternative.

Marking guidance:

Do not allow "inverse proportionality".

May be worth noting any marks on graph relating to 7bii.

(b)

Deduce, without a numerical calculation, whether R or T has the greater resistance at $I=0.40 \mathrm{~A}$ .

[ 2 ]

$\text { at } 0.4 \mathrm{~A}: V_{\mathrm{R}}>V_{\mathrm{T}} \text { or } V_{R}=5.6 \mathrm{~V} \text { and } V_{\mathrm{T}}=5.3 \mathrm{~V}$

so $\mathrm{R}_{\mathrm{R}}>\mathrm{R}_{\mathrm{T}}$ because $V=I R / V \propto \mathrm{R}$ «and I same for both»

Marking guidance:

Award [0] for a bald correct answer without deduction or with incorrect reasoning.

Ignore any references to graph gradients.

Both elements must be present for MP2 to be awarded.

(c)

Components R and T are placed in a circuit. Both meters are ideal.

Slider Z of the potentiometer is moved from Y to X .

[ 3 ]

(i)

State what happens to the magnitude of the current in the ammeter.

[ 1 ]

(ii)

Estimate, with an explanation, the voltmeter reading when the ammeter reads 0.20 A .

[ 2 ]

realization that V is the same for R and T

OR

identifies that currents are 0.14 A and 0.06 A

$V=2 \mathrm{~V}$ OR 2.0 V

Marking guidance:

Award [0] if pds 2.8 V and 3.7 V or 1.4 V and 2.6 V are used in any way. Otherwise award [1 max] for a bald correct answer. Explanation expected.

[Maximum number: 1]

This question is in two parts. Part 1 is about a thermistor circuit. Part 2 is about vibrations and waves.
Part 1 Thermistor circuit
The circuit shows a negative temperature coefficient (NTC) thermistor X and a $100 \mathrm{k} \Omega$ fixed resistor R connected across a battery.

The battery has an electromotive force (emf) of 12.0 V and negligible internal resistance.

Part 2 Vibrations and waves

The cone and dust cap D of a loudspeaker L vibrates with a frequency of 1.25 kHz with simple harmonic motion (SHM).

Define electromotive force (emf).

[Maximum number: 1]

Outline why research into solar cell technology is important to society.

[Maximum number: 2]

The graph shows how current I varies with potential difference V for a resistor R and a non-ohmic component T.

Deduce, without a numerical calculation, whether R or T has the greater resistance at $I=0.40 \mathrm{~A}$ .

$\text { at } 0.4 \mathrm{~A}: V_{\mathrm{R}}>V_{\mathrm{T}} \text { or } V_{R}=5.6 \mathrm{~V} \text { and } V_{\mathrm{T}}=5.3 \mathrm{~V}$

so $\mathrm{R}_{\mathrm{R}}>\mathrm{R}_{\mathrm{T}}$ because $V=I R / V \propto \mathrm{R}$ «and I same for both»

Marking guidance:

Award [0] for a bald correct answer without deduction or with incorrect reasoning.

Ignore any references to graph gradients.

Both elements must be present for MP2 to be awarded.

[Maximum number: 1]

Which of the following is a statement of Ohm's law?

A

The resistance of a conductor is constant.

B

The current in a conductor is inversely proportional to the potential difference across the conductor provided the temperature is constant.

C

The resistance of a conductor is constant provided that the temperature is constant.

D

The current in a conductor is proportional to the potential difference across it.

[Maximum number: 2]

A cable consisting of many copper wires is used to transfer electrical energy from a generator to an electrical load. The copper wires are protected by an insulator.

The cable consists of 32 copper wires each of length 35 km . Each wire has a resistance of $64 \Omega$ . The resistivity of copper is $1.7 \times 10^{-8} \Omega \mathrm{~m}$ .

There is a current of 730 A in the cable. Show that the power loss in 1 m of the cable is about 30 W .

$R=2\,\Omega$ $P=I^2R=(730\,\mathrm{A})^2\times2\,\Omega=1.07\times10^6\,\mathrm{W}=1.07\,\mathrm{MW}$ $\text{power loss per metre}=\frac{1.07\times10^6\,\mathrm{W}}{35\times10^3\,\mathrm{m}}=30.6\,\mathrm{W\,m^{-1}}$

Allow [2] for a solution using the resistivity and the answer to (b)(i), giving a resistance per unit length of $5.7\times10^{-5}\,\Omega\,\mathrm{m^{-1}}$ .

[Maximum number: 3]

A fully charged cell of emf 6.0 V delivers a constant current of 5.0 A for a time of 0.25 hour until it is completely discharged.

The cell is then re-charged by a rectangular solar panel of dimensions $0.40 \mathrm{~m} \times 0.15 \mathrm{~m}$ at a place where the maximum intensity of sunlight is $380 \mathrm{Wm}^{-2}$ .

The overall efficiency of the re-charging process is 18 %.
Calculate the minimum time required to re-charge the cell fully.

ALTERNATIVE 1

«energy output of the panel = » VIt OR $6 \times 5 \times 0.25 \times 3600$ OR 27000 «J»

«available power = » $380 \times 0.4 \times 0.15 \times 0.18$ OR 4.1 «W»

$t=\frac{27000}{4.1}\approx6600\,\mathrm{s}$

ALTERNATIVE 2

«energy needed from Sun = » $\frac{V I t}{\text { eff }}$ OR $\frac{6 \times 5 \times 0.25 \times 3600}{0.18}$ OR 150000 «J»

« incident power=» $380 \times 0.4 \times 0.15$ OR 22.8 «W»

$t=\frac{150000}{22.8}\approx6600\,\mathrm{s}$

Marking guidance:

Allow ECF for MP3

Accept final answer in minutes (110) or hours (1.8).

[Maximum number: 1]

With reference to internal energy conversion and ability to be recharged, what are the characteristics of a primary cell?

Internal energy conversion

Ability to be recharged

chemical to electrical

rechargeable

chemical to electrical

not rechargeable

electrical to chemical

rechargeable

electrical to chemical

not rechargeable

[Maximum number: 1]

In an experiment a student constructs the circuit shown in the diagram. The ammeter and the voltmeter are assumed to be ideal.

A connecting wire in the circuit has a radius of 1.2 mm and the current in it is 3.5 A . The number of electrons per unit volume of the wire is $2.4 \times 10^{28} \mathrm{~m}^{-3}$ . Show that the drift speed of the electrons in the wire is $2.0 \times 10^{-4} \mathrm{~ms}^{-1}$ .

$3.5=2.4 \times 10^{28} \times \pi\left(1.2 \times 10^{-3}\right)^{2} \times 1.6 \times 10^{-19} \times v\left\|\Rightarrow v=2.0 \times 10^{-4} \mathrm{~ms}^{-1}\right\| v$

[Maximum number: 2]

A cable consisting of many copper wires is used to transfer electrical energy from a generator to an electrical load. The copper wires are protected by an insulator.

The cable consists of 32 copper wires each of length 35 km . Each wire has a resistance of $64 \Omega$ . The resistivity of copper is $1.7 \times 10^{-8} \Omega \mathrm{~m}$ .

Calculate the radius of each wire.

A=\frac{(1.7\times10^{-3}\,\Omega\,\mathrm{m})(35\times10^3\,\mathrm{m})}{64\,\Omega}=9.3\times10^{-6}\,\mathrm{m^2}

[Maximum number: 4]

This question is in two parts. Part 1 is about a thermistor circuit. Part 2 is about vibrations and waves.
Part 1 Thermistor circuit
The circuit shows a negative temperature coefficient (NTC) thermistor X and a $100 \mathrm{k} \Omega$ fixed resistor R connected across a battery.

The battery has an electromotive force (emf) of 12.0 V and negligible internal resistance.

Part 2 Vibrations and waves

The cone and dust cap D of a loudspeaker L vibrates with a frequency of 1.25 kHz with simple harmonic motion (SHM).

The graph below shows the variation with temperature T of the resistance $R_{\mathrm{x}}$ of the thermistor.

Determine the temperature of X when the potential difference across R is 4.5 V .

$V_X=7.5\,\mathrm{V}$ $I=\frac{4.5\,\mathrm{V}}{100\times10^3\,\Omega}=4.5\times10^{-5}\,\mathrm{A}$ $R_X=\frac{7.5\,\mathrm{V}}{4.5\times10^{-5}\,\mathrm{A}}=1.67\times10^5\,\Omega$ $T=-37\text{ to }-38\,^{\circ}\mathrm{C}$

Alternatively, use the potential-divider relation V_X/V_R=R_X/R_R or the equivalent resistance calculation.

IB Physics SL B 5 Current and Circuits Questions

Exam points

Covered syllabus points

Covered syllabus points

Question 1

Question (a)

Question (b)

Question (i)

Question (ii)

Question (iii)

Question 2

Question (a)

Question (i)

Question (ii)

Question (b)

Question (c)

Question (d)

Question 3

Question (a)

Question (b)

Question (c)

Question (i)

Question (ii)

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13