IB Physics SL B 5 Current and Circuits Questions

Practise IB Physics SL B.5 by solving resistor networks, calculating charge and power, analysing real-cell voltage and predicting thermistor or LDR divider outputs.

Exam points

Use circuit diagrams, junction/loop laws and series/parallel rules to find currents and voltages.
Interpret V–I graphs to test Ohm’s law, then use R = ρL/A to calculate wire resistance.
Calculate electrical power and energy in cells, appliances and cables, including transmission losses.
Compare photovoltaic generation with other power sources and calculate array output or area.
Use ε = I(R+r) to relate emf, terminal pd, internal resistance and lost volts under load.
Use I = Δq/Δt to calculate charge flow and explain current through carriers in conductors.
Analyse thermistor and LDR potential dividers to predict output-voltage changes.

Covered syllabus points

B.5.1—Cells provide a source of emf B.5.2—Electrical energy sources B.5.3—Resistance B.5.4—Ohmic behaviour B.5.5—Electrical power B.5.6—Potential difference and electrical energy transfer B.5.7—Electric cells B.5.8—Circuit diagrams and conventional current B.5.9—Direct current and charge flow B.5.10—Direct and alternating current boundary B.5.11—Conservation in series and parallel circuits B.5.12—Resistance and resistivity B.5.13—Series and parallel circuits B.5.14—Emf and internal resistance B.5.15—Variable resistance

[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.

All question bank results loaded

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)