IB Physics HL B.5 Current & Circuits

[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 ]

$12 \mathrm{~V}$

(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 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 ]

[Maximum number: 1]

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 .

State the emf of the cell.

$12 \mathrm{~V}$

[Maximum number: 1]

This question is in two parts

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 $1.44 \times 10^{-18} \mathrm{~J}$ .

Define resistance of a resistor.

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

[Maximum number: 1]

The graph shows how current I varies with potential difference V across a component X .

Outline why component X is considered non-ohmic.

[Maximum number: 1]

This question is in two parts

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 $1.44 \times 10^{-18} \mathrm{~J}$ .

Calculate the total power supplied by the cell.

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

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).

[Maximum number: 1]

Two secondary cells require recharging. Which circuit will recharge both cells?

A

Figure for Question 9 — IB Physics HL — Option A

B

Figure for Question 9 — IB Physics HL — Option B

C

Figure for Question 9 — IB Physics HL — Option C

D

Figure for Question 9 — IB Physics HL — Option D

[Maximum number: 1]

A1.0 mF capacitor is connected in series to a $2.3 \mathrm{k} \Omega$ resistor and a cell of negligible internal resistance.

The capacitor is initially uncharged. At time t=0 the switch is moved to position A. The initial current in the resistor is 3.9 mA .

Draw, on the circuit diagram above, an arrow showing the direction of the conventional current in the resistor for $t>5.0 \mathrm{~s}$ .

[Maximum number: 3]

A cable consisting of many copper wires is used to transfer electrical energy from an alternating current (ac) generator to an electrical load. The copper wires are protected by an insulator.

The copper wires and insulator are both exposed to an electric field. Discuss, with reference to charge carriers, why there is a significant electric current only in the copper wires.

[Maximum number: 2]

A cable consisting of many copper wires is used to transfer electrical energy from an alternating current (ac) 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 cable is connected to the ac generator which has an output power of 110 MW when the peak potential difference is 150 kV . The resistivity of copper is $1.7 \times 10^{-8} \Omega \mathrm{~m}$ .
output power =110 MW

Calculate the radius of each wire.

$\begin{aligned} \text{area} &= \frac{1.7\times10^{-8}\times35\times10^{3}}{64}=9.3\times10^{-6}\ \mathrm{m^2}\\ \text{radius} &= \sqrt{\frac{9.3\times10^{-6}}{\pi}}=0.00172\ \mathrm{m} \end{aligned}$

[Maximum number: 3]

A girl rides a bicycle that is powered by an electric motor. A battery transfers energy to the electric motor. The emf of the battery is 16 V and it can deliver a charge of 43 kC when discharging completely from a full charge.

(a)

The battery is made from an arrangement of 10 identical cells as shown.

Calculate

[ 3 ]

(i)

the emf of one cell.

[ 1 ]

$\frac{16}{5}=3.2$ « V »

(ii)

the internal resistance of one cell.

[ 2 ]

ALTERNATIVE 1:

2.5 r=0.62

r=0.25 «Ω»

ALTERNATIVE 2:

$\frac{0.62}{5}=0.124$ « $\Omega$ »

r=2(0.124)=0.248 « $\Omega$ »

Marking guidance:

Allow ECF from (d).

[Maximum number: 4]

This question is in two parts

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 $1.44 \times 10^{-18} \mathrm{~J}$ .

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

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 $=\frac{3.00}{5.00}=0.600 \mathrm{~A}$ ;
$R=\frac{9.00}{0.600}=15.0 \Omega$ ;

[Maximum number: 3]

This question is about an operational amplifier (op-amp).

An op-amp is used as a comparator in a circuit, as shown.

The circuit is used to monitor the temperature in a greenhouse. The resistance of a thermistor decreases as its temperature increases.

(a)

Explain how the potential at the inverting input changes as the temperature decreases.

[ 2 ]

(b)

Describe the purpose of component X in the circuit.
When the potential of the non-inverting input is set at 0 V , the heater switches on at a temperature of around $10^{\circ} \mathrm{C}$ . The ideal temperature to grow plants in the greenhouse is $18^{\circ} \mathrm{C}$ or greater. At $18^{\circ} \mathrm{C}$ the resistance of the thermistor is $193 \Omega$ .

[ 1 ]

IB Physics HL B 5 Current and Circuits Questions

Exam points

Covered syllabus points

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)

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question (a)

Question (i)

Question (ii)

Question 14

Question 15

Question (a)

Question (b)