Physics 9702/23 — May/June 2019

Cambridge AS Level · AS Level Structured Questions · worked solutions for every part, with the mark scheme

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Topics Electricity · Physical Quantities and Units · Work, Energy and Power · Kinematics · Dynamics · Forces, Density and Pressure · +5 more

Q1ElectricityPhysical Quantities and UnitsFree sample

Answer all the questions in the spaces provided.

(a)

(i)

Define resistance.

DifficultyEasy

Worked solution

Answer

Resistance is the ratio of potential difference to current:

R = \frac{V}{I}

Final answer

Resistance is the ratio of potential difference to current, R = V/I.

Detailed explanation

Background Concept

Resistance describes how strongly a component opposes the flow of electric current. For many circuit questions at AS level, it is defined using the relationship between potential difference $V$ across a component and the current $I$ through it.

Understanding the Question

You are asked to define resistance, so you should give the standard statement (often with the defining equation) using the correct quantities.

Approach

Use the defining equation for resistance:

R = \frac{V}{I}

Then state it in words: potential difference per unit current.

Step-by-Step Reasoning

Potential difference $V$ is measured in volts (V) and current $I$ in amperes (A).
Resistance $R$ is defined as the ratio $V/I$ .
Therefore, in words: resistance is the potential difference across a component per unit current through it.

Key Takeaways

Resistance is defined by the ratio $R = V/I$ .
A good definition uses both the words and the equation.

Common Mistakes

Writing $R = I/V$ (inverting the ratio).
Giving Ohm’s law wording (" $V$ is proportional to $I$ ") instead of the actual definition.

Things to Be Careful About

Use the correct symbols ( $V$ , $I$ , $R$ ) and ensure the ratio is in the correct order ( $V$ divided by $I$ ).

Techniques used

state the definition of resistance as a ratiouse correct symbols and units for potential difference and current

(ii)

A potential difference of $0.60\ \text{V}$ is applied across a resistor of resistance $4.0\ \text{G}\Omega$ .

Calculate the current, in $\text{pA}$ , in the resistor.

current = ______ $\text{pA}$

DifficultyMedium-Easy

Worked solution

Working

I = \frac{V}{R} = \frac{0.60}{4.0 \times 10^{9}} = 1.5 \times 10^{-10}\ \text{A}

I = 1.5 \times 10^{-10} \times 10^{12} = 1.5 \times 10^{2}\ \text{pA} = 150\ \text{pA}

Answer

$150\ \text{pA}$

Final answer

150 pA

Detailed explanation

Background Concept

Ohm’s law links potential difference $V$ , current $I$ , and resistance $R$ :

V = IR

So, if you know $V$ and $R$ , you can find the current using:

I = \frac{V}{R}

Unit prefixes:

$\text{G}\Omega = 10^{9}\ \Omega$
$\text{pA} = 10^{-12}\ \text{A}$

Understanding the Question

You are given:

$V = 0.60\ \text{V}$
$R = 4.0\ \text{G}\Omega = 4.0 \times 10^{9}\ \Omega$

You must calculate the current and express it in picoamperes (pA).

Approach

Convert $R$ from $\text{G}\Omega$ to $\Omega$ .
Use $I = V/R$ to find $I$ in amperes.
Convert amperes to picoamperes.

Step-by-Step Reasoning

Convert resistance:

4.0\ \text{G}\Omega = 4.0 \times 10^{9}\ \Omega

Calculate current:

I = \frac{V}{R} = \frac{0.60}{4.0 \times 10^{9}} = 1.5 \times 10^{-10}\ \text{A}

Convert to pA (multiply by $10^{12}$ because $1\ \text{pA} = 10^{-12}\ \text{A}$ ):

I = 1.5 \times 10^{-10} \times 10^{12} = 1.5 \times 10^{2}\ \text{pA} = 150\ \text{pA}

Key Takeaways

Use $I = V/R$ for a resistor.
Convert prefixes carefully: $\text{G} = 10^{9}$ , $\text{p} = 10^{-12}$ .

Common Mistakes

Forgetting to convert $4.0\ \text{G}\Omega$ into $\Omega$ .
Converting to pA the wrong way (dividing by $10^{12}$ instead of multiplying).
Dropping powers of ten or writing an answer with no unit.

Things to Be Careful About

Significant figures: $0.60$ and $4.0$ are 2 s.f., so $150\ \text{pA}$ (2 s.f.) is appropriate.
Keep track of units at every step to avoid conversion errors.

Techniques used

apply Ohm's law in the form I = V/Rconvert prefixes and express quantities in powers of tenconvert amperes to picoamperes

(b)

The energy $E$ transferred when charge $Q$ moves through an electrical component is given by the equation

E = QV

where $V$ is the potential difference across the component.

Use the equation to determine the SI base units of potential difference.

SI base units ______

DifficultyMedium

Worked solution

Working

From $E = QV$ ,

V = \frac{E}{Q}

[ E ] = \text{J} = \text{kg m}^{2}\text{ s}^{-2}, \qquad [ Q ] = \text{C} = \text{A s}

[ V ] = \frac{\text{kg m}^{2}\text{ s}^{-2}}{\text{A s}} = \text{kg m}^{2}\text{ s}^{-3}\text{ A}^{-1}

Answer

$\text{kg m}^{2}\text{ s}^{-3}\text{ A}^{-1}$

Final answer

kg m^2 s^-3 A^-1

Detailed explanation

Background Concept

A derived unit can be found by rearranging a physics equation and replacing each quantity with its units.

Here,

E = QV

where:

$E$ is energy transferred (unit: joule, J)
$Q$ is charge (unit: coulomb, C)
$V$ is potential difference (unit: volt, V)

In SI base units:

1\ \text{J} = 1\ \text{kg m}^{2}\text{ s}^{-2}

and

1\ \text{C} = 1\ \text{A s}

Understanding the Question

You must use $E = QV$ to find the SI base units of potential difference $V$ . That means your final unit must be written using only the base units: $\text{kg}$ , $\text{m}$ , $\text{s}$ , $\text{A}$ (and not J or C).

Approach

Rearrange to make $V$ the subject: $V = E/Q$ .
Replace $E$ with joules and then with base units.
Replace $Q$ with coulombs and then with base units.
Simplify.

Step-by-Step Reasoning

Start from:

E = QV

Rearrange:

V = \frac{E}{Q}

Now substitute units:

Energy $E$ in joules: $\text{J}$
Charge $Q$ in coulombs: $\text{C}$

So:

[ V ] = \frac{\text{J}}{\text{C}}

Convert to base units:

\text{J} = \text{kg m}^{2}\text{ s}^{-2}

and

\text{C} = \text{A s}

Therefore:

[ V ] = \frac{\text{kg m}^{2}\text{ s}^{-2}}{\text{A s}} = \text{kg m}^{2}\text{ s}^{-3}\text{ A}^{-1}

Key Takeaways

To find base units, rearrange the equation and substitute SI units.
Remember: $\text{J} = \text{kg m}^{2}\text{ s}^{-2}$ and $\text{C} = \text{A s}$ .
Potential difference has base units $\text{kg m}^{2}\text{ s}^{-3}\text{ A}^{-1}$ .

Common Mistakes

Leaving the answer as $\text{J C}^{-1}$ (not in base units).
Using $\text{C} = \text{A}\,\text{s}^{-1}$ (wrong; charge is current \u00d7 time).
Losing a power of $s$ when dividing by $\text{A s}$ .

Things to Be Careful About

The final expression must contain only base units, not derived names like J, C, V.
When dividing by $\text{A s}$ , subtract indices correctly: $s^{-2}/s^{1} = s^{-3}$ .

Techniques used

rearrange an equation to isolate the required quantitysubstitute SI units for each physical quantityexpress derived units in SI base unitsuse dimensional analysis to simplify units

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