IB Physics HL B 5 Current and Circuits Questions

Practise IB Physics HL B.5 by solving resistor networks, reading V–I curves, analysing real-cell voltage under load and tracing power through AC circuits.

Exam points

Trace conventional current on diagrams and apply junction/loop rules to solve series–parallel circuits.
Interpret V–I curves to distinguish ohmic devices and apply R = ρL/A to wires.
Calculate electrical power, energy transfer and cable loss; use RMS values for AC quantities.
Describe cell emf as energy per charge and distinguish photovoltaic output from solar heating.
Use ε = I(R+r) to relate emf, terminal pd, internal resistance and load changes.
Relate carrier density and drift speed to current and charge flow in metal wires.
Interpret diode rectifier waveforms and predict how a capacitor smooths the output.
Analyse thermistor comparator circuits to predict potential changes and switching temperature.

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: 7]

(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 ]

$12 \vee$

(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 any valid method.

Allow ECF from (b)(i)

(b)

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»
$t=\frac{27000}{4.1}=6600\ \mathrm{s}$

ALTERNATIVE 2

«energy needed from Sun = » $\frac{V I t}{e f f}$ 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»
$t=\frac{150000}{22.8}=6600\ \mathrm{s}$

Marking guidance:

Allow ECF for MP3

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

(c)

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

[ 1 ]

[Maximum number: 3]

This question is in two parts. Part 1 is about solar power and climate models. Part 2 is about digital storage of data.

(a)

Distinguish, in terms of the energy changes involved, between a solar heating panel and a photovoltaic cell.

[ 2 ]

(b)

State an appropriate domestic use for a

[ 1 ]

(i)

photovoltaic cell.

[ 1 ]

[Maximum number: 6]

This question is in two parts

(a)

The plates in (c) are replaced by a cell that has an emf of 12.0 V and internal resistance $5.00 \Omega$ . A resistor of resistance R is connected in series with the cell. The energy transferred by the cell to an electron as it moves through the resistor is $1.44 \times 10^{-18} \mathrm{~J}$ .

[ 6 ]

(i)

Define resistance of a resistor.

[ 1 ]

ratio potential difference/voltage (across resistor) to current (in resistor) / $\frac{V}{I}$
with symbols defined;

(ii)

Show that the value of R is $15.0 \Omega$ .

[ 4 ]

pd across $R=\frac{1.44 \times 10^{-18}}{1.6 \times 10^{-19}}=9.00 \mathrm{~V}$ ;
pd across internal resistance $=12.0-9.00=3.00 \mathrm{~V}$ ;
current in circuit $=\frac{3.00}{5.00}=0.600 \mathrm{~A}$ ;
$R=\frac{9.00}{0.600}=15.0 \Omega$ ;

(iii)

Calculate the total power supplied by the cell.

[ 1 ]

All question bank results loaded

Covered syllabus points

Question 1

Question (a)

Question (i)

Question (ii)

Question (b)

Question (c)

Question 2

Question (a)

Question (b)

Question (i)

Question 3

Question (a)

Question (i)

Question (ii)

Question (iii)