3.5 Electricity — printable coverage pack

3.5.1.1 · Basics of electricity

Electric current is the rate of flow of charge: $I=\dfrac{\Delta Q}{\Delta t}$ . Use amperes for current, coulombs for charge and seconds for time.
Potential difference is work done per unit charge: $V=\dfrac{W}{Q}$ . One volt is one joule per coulomb.
Resistance is defined by $R=\dfrac{V}{I}$ ; this definition applies even when the resistance is not constant.
A common error is to confuse charge with current. Charge is an amount transferred, whereas current states how quickly it is transferred.

Tier 1 · Easy

1. A charge of $3.2\,\text{C}$ passes a point in $0.80\,\text{s}$ . Calculate the current.[1 mark]
Answer
- $4.0\,\text{A}$
Method: Use $I=\dfrac{\Delta Q}{\Delta t}$ . Hence $I=\dfrac{3.2}{0.80}=4.0\,\text{A}$ .

Tier 2 · Standard

1. A component transfers $18\,\text{J}$ of energy when $3.0\,\text{C}$ passes through it. The current is $0.75\,\text{A}$ . Determine the potential difference and the resistance of the component.[3 marks]
Answer
- $6.0\,\text{V}$
- $8.0\,\Omega$
Method: First use $V=\dfrac{W}{Q}=\dfrac{18}{3.0}=6.0\,\text{V}$ . Then $R=\dfrac{V}{I}=\dfrac{6.0}{0.75}=8.0\,\Omega$ .

Tier 3 · Hard

1. A heater operates at $24\,\text{V}$ and carries a current of $2.0\,\text{A}$ for $150\,\text{s}$ . Determine the charge transferred, the energy transferred and the resistance of the heater.[5 marks]
Answer
- $3.0\times10^2\,\text{C}$
- $7.2\times10^3\,\text{J}$
- $12\,\Omega$
Method: The charge is $Q=It=2.0\times150=3.0\times10^2\,\text{C}$ . The energy is $W=VQ=24\times300=7.2\times10^3\,\text{J}$ . The resistance is $R=\dfrac{V}{I}=\dfrac{24}{2.0}=12\,\Omega$ .

3.5.1.2 · Current-voltage characteristics

An ohmic conductor obeys $I\propto V$ only while its physical conditions, especially temperature, remain constant.
A filament lamp heats as current rises, so its resistance increases and its current-voltage graph becomes less steep when current is plotted vertically.
A semiconductor diode conducts readily in the forward direction after its turn-on region but carries negligible reverse current at the potential differences considered here.
Check which quantity is on each graph axis before using a gradient: for an $I$ - $V$ graph the gradient is conductance, not resistance.
Unless a question states otherwise, treat an ammeter as having zero resistance and a voltmeter as having infinite resistance.

Tier 1 · Easy

1. State the condition under which a conductor obeys Ohm's law.[1 mark]
Answer
- The current is directly proportional to the potential difference when physical conditions, including temperature, are constant.
Method: Award the statement $I\propto V$ together with the requirement that physical conditions, particularly temperature, remain constant.

Tier 2 · Standard

1. Describe the current-voltage characteristic of a semiconductor diode for forward and reverse potential differences.[3 marks]
Answer
- There is negligible reverse current; the forward current is also small at first, then rises rapidly after the turn-on region.
Method: For reverse bias, state that the current is negligible. For small forward potential differences, the current remains small. Beyond the turn-on region, a small further increase in potential difference produces a large increase in current.

Tier 3 · Hard

1. A filament lamp and a fixed metal resistor carry the same small current. Explain why, as the potential difference across each component is increased, the lamp's current-voltage graph curves while the resistor's graph remains approximately straight.[5 marks]
Answer
- The lamp filament becomes hotter, increasing lattice vibrations and its resistance, so current rises less rapidly; the fixed resistor stays approximately at constant temperature and remains ohmic.
Method: Increasing the potential difference increases the current and therefore the power dissipated in the lamp. Its filament temperature rises. Greater lattice vibrations cause more frequent scattering of conduction electrons, so the filament resistance increases. Consequently the current rises by progressively smaller amounts for equal potential-difference increases, producing a curved graph. The fixed resistor is assumed to remain at approximately constant temperature, so its resistance is constant and $I\propto V$ gives a straight line through the origin.

3.5.1.3 · Resistivity

For a uniform conductor, $\rho=\dfrac{RA}{L}$ . Resistivity is a material property measured in $\Omega\,\text{m}$ .
For a circular wire, use $A=\pi d^2/4$ and convert the diameter to metres before squaring it.
The resistance of a metal increases with temperature, whereas an NTC (negative temperature coefficient) thermistor's resistance decreases as its temperature increases.
Below its material-dependent critical temperature, a superconductor has zero resistivity. This permits strong electromagnets and reduces power loss in transmission.
In the resistivity practical, measure several wire diameters with a micrometer and use a best-fit gradient or repeated readings; a common error is to use diameter in place of cross-sectional area.

Tier 1 · Easy

1. A wire of length $1.5\,\text{m}$ and cross-sectional area $0.20\,\text{mm}^2$ has resistance $4.2\,\Omega$ . Calculate its resistivity.[2 marks]
Answer
- $5.6\times10^{-7}\,\Omega\,\text{m}$
Method: Convert the area: $0.20\,\text{mm}^2=0.20\times10^{-6}\,\text{m}^2=2.0\times10^{-7}\,\text{m}^2$ . Then $\rho=\dfrac{RA}{L}=\dfrac{4.2\times2.0\times10^{-7}}{1.5}=5.6\times10^{-7}\,\Omega\,\text{m}$ .

Tier 2 · Standard

1. An NTC thermistor is used as a temperature sensor. Explain the microscopic change that causes its resistance to fall when its temperature rises.[3 marks]
Answer
- Heating releases more charge carriers in the semiconductor; the increased carrier density outweighs increased scattering, so resistance decreases.
Method: A higher temperature provides energy that releases more charge carriers in the semiconductor. This increases the number density of mobile carriers. Although collisions may also become more frequent, the carrier-density increase dominates, so conductivity rises and resistance falls.

Tier 3 · Hard

1. A $1.80\,\text{m}$ wire has diameter $0.400\,\text{mm}$ . A potential difference of $2.40\,\text{V}$ produces a current of $0.800\,\text{A}$ . Determine the wire's resistivity. Explain why replacing a transmission cable by a material operating below its superconducting critical temperature reduces energy loss.[5 marks]
Answer
- $2.09\times10^{-7}\,\Omega\,\text{m}$ ; below the critical temperature the resistivity and hence the resistive power loss are zero.
Method: The resistance is $R=\dfrac{V}{I}=\dfrac{2.40}{0.800}=3.00\,\Omega$ . The radius is $0.200\,\text{mm}=2.00\times10^{-4}\,\text{m}$ , so $A=\pi r^2=1.257\times10^{-7}\,\text{m}^2$ . Hence $\rho=\dfrac{RA}{L}=\dfrac{3.00\times1.257\times10^{-7}}{1.80}=2.09\times10^{-7}\,\Omega\,\text{m}$ . Below the critical temperature a superconductor has $\rho=0$ , so the cable resistance and the resistive loss $P=I^2R$ are zero.

3.5.1.4 · Circuits

Series resistances add: $R_{\text{T}}=R_1+R_2+\cdots$ . Parallel conductances add: $\dfrac{1}{R_{\text{T}}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\cdots$ .
Charge conservation gives equal current through series components and junction current balance; energy conservation gives the loop rule for potential differences.
Electrical power may be calculated using $P=IV=I^2R=\dfrac{V^2}{R}$ , and transferred energy using $E=IVt$ .
Cells in series have additive emfs. Identical cells in parallel have the same emf as one cell but can supply current with a smaller effective internal resistance.
A common error is to use the supply potential difference across a single component before reducing the network or applying the junction and loop rules.

Tier 1 · Easy

1. A $4.0\,\Omega$ resistor and a $6.0\,\Omega$ resistor are connected in series. State their total resistance.[1 mark]
Answer
- $10\,\Omega$
Method: For series resistors, $R_{\text{T}}=R_1+R_2=4.0+6.0=10.0\,\Omega$ .

Tier 2 · Standard

1. A $12\,\Omega$ resistor and a $6.0\,\Omega$ resistor are connected in parallel across an $18\,\text{V}$ supply. Determine the total resistance, the supply current and the total power.[3 marks]
Answer
- $4.0\,\Omega$
- $4.5\,\text{A}$
- $81\,\text{W}$
Method: $\dfrac{1}{R_{\text{T}}}=\dfrac{1}{12}+\dfrac{1}{6}=\dfrac{3}{12}$ , so $R_{\text{T}}=4.0\,\Omega$ . The supply current is $I=\dfrac{V}{R_{\text{T}}}=\dfrac{18}{4.0}=4.5\,\text{A}$ . Hence $P=VI=18\times4.5=81\,\text{W}$ .

Tier 3 · Hard

1. A $3.0\,\Omega$ resistor is in series with a parallel pair of $6.0\,\Omega$ and $3.0\,\Omega$ resistors. The network is connected to a $20\,\text{V}$ supply. Calculate the energy transferred by the $6.0\,\Omega$ resistor in $90\,\text{s}$ .[5 marks]
Answer
- $9.6\times10^2\,\text{J}$
Method: The parallel resistance is $R_{\parallel}=\left(\dfrac{1}{6.0}+\dfrac{1}{3.0}\right)^{-1}=2.0\,\Omega$ . The total resistance is $3.0+2.0=5.0\,\Omega$ , so the supply current is $20/5.0=4.0\,\text{A}$ . The potential difference across the series $3.0\,\Omega$ resistor is $4.0\times3.0=12\,\text{V}$ , leaving $8.0\,\text{V}$ across the parallel pair. The current in the $6.0\,\Omega$ resistor is $8.0/6.0=1.33\,\text{A}$ . Therefore $E=IVt=1.33\times8.0\times90=9.6\times10^2\,\text{J}$ .

3.5.1.5 · Potential divider

For two series resistors, the output across $R_2$ is $V_{\text{out}}=V_{\text{in}}\dfrac{R_2}{R_1+R_2}$ when the output is unloaded.
A potentiometer (a divider with a movable contact) provides a continuously adjustable output potential difference from a fixed supply.
In a sensor divider, first identify whether the LDR or thermistor is the upper or lower resistor; this decides whether the output rises or falls with the stimulus.
An LDR's resistance falls as light intensity increases, and an NTC thermistor's resistance falls as temperature increases.
A common error is to calculate the potential difference across the wrong component or to ignore loading by a component connected across the output.

Tier 1 · Easy

1. Two equal resistors form an unloaded potential divider across a $12\,\text{V}$ supply. State the potential difference across either resistor.[1 mark]
Answer
- $6.0\,\text{V}$
Method: Equal series resistors share the supply potential difference equally, so $V=12/2=6.0\,\text{V}$ .

Tier 2 · Standard

1. An LDR is the lower component of a potential divider and a $3.0\,\text{k}\Omega$ fixed resistor is the upper component. The supply is $9.0\,\text{V}$ . Calculate the output across the LDR when its resistance is $9.0\,\text{k}\Omega$ in darkness and $1.0\,\text{k}\Omega$ in bright light.[3 marks]
Answer
- $6.8\,\text{V}$ in darkness
- $2.3\,\text{V}$ in bright light
Method: Use $V_{\text{out}}=9.0\dfrac{R_{\text{LDR}}}{3.0+R_{\text{LDR}}}$ with resistances in $\text{k}\Omega$ . In darkness, $V_{\text{out}}=9.0\times9.0/12.0=6.75\,\text{V}$ , which is $6.8\,\text{V}$ to two significant figures. In bright light, $V_{\text{out}}=9.0\times1.0/4.0=2.25\,\text{V}$ , which is $2.3\,\text{V}$ to two significant figures.

Tier 3 · Hard

1. An NTC thermistor is the lower component of an unloaded potential divider. The upper resistor is $4.70\,\text{k}\Omega$ and the supply is $12.0\,\text{V}$ . A controller switches when the output across the thermistor is $4.00\,\text{V}$ . Determine the thermistor resistance and divider current at switching. State how the output changes if the thermistor then becomes warmer.[5 marks]
Answer
- $2.35\,\text{k}\Omega$
- $1.70\,\text{mA}$
- The output decreases as the thermistor becomes warmer.
Method: Let the thermistor resistance be $R$ . Then $4.00=12.0\dfrac{R}{4700+R}$ . Hence $4(4700+R)=12R$ , so $R=2350\,\Omega=2.35\,\text{k}\Omega$ . The total resistance is $4700+2350=7050\,\Omega$ , giving $I=12.0/7050=1.70\times10^{-3}\,\text{A}=1.70\,\text{mA}$ . Warming an NTC thermistor lowers $R$ , so the fraction $R/(4700+R)$ and therefore the output decrease.

3.5.1.6 · Electromotive force and internal resistance

Electromotive force is energy supplied by a source per unit charge: $\varepsilon=\dfrac{E}{Q}$ .
For a source of internal resistance $r$ supplying an external resistance $R$ , $\varepsilon=I(R+r)$ and the terminal potential difference is $V=IR=\varepsilon-Ir$ .
The lost volts $Ir$ and internal power $I^2r$ increase when the current increases.
On a graph of terminal potential difference against current, the intercept is $\varepsilon$ and the gradient is $-r$ ; the minus sign is commonly omitted.
Emf is not a force and is not generally equal to terminal potential difference while the source supplies current.

Tier 1 · Easy

1. A cell of internal resistance $0.50\,\Omega$ supplies a $2.5\,\Omega$ resistor with a current of $2.0\,\text{A}$ . Calculate the emf of the cell.[2 marks]
Answer
- $6.0\,\text{V}$
Method: Use $\varepsilon=I(R+r)=2.0(2.5+0.50)=6.0\,\text{V}$ .

Tier 2 · Standard

1. A battery has emf $9.0\,\text{V}$ and internal resistance $0.80\,\Omega$ . It is connected to a $4.2\,\Omega$ load. Determine the current, terminal potential difference and lost volts.[3 marks]
Answer
- $1.8\,\text{A}$
- $7.6\,\text{V}$
- $1.4\,\text{V}$
Method: $I=\dfrac{\varepsilon}{R+r}=\dfrac{9.0}{4.2+0.80}=1.8\,\text{A}$ . The terminal potential difference is $V=IR=1.8\times4.2=7.56\,\text{V}\approx7.6\,\text{V}$ . The lost volts are $Ir=1.8\times0.80=1.44\,\text{V}\approx1.4\,\text{V}$ , and $7.56+1.44=9.0\,\text{V}$ checks the result.

Tier 3 · Hard

1. Measurements for a cell give terminal potential differences of $5.40\,\text{V}$ at $0.50\,\text{A}$ and $4.20\,\text{V}$ at $2.00\,\text{A}$ . Determine the emf and internal resistance. The cell is then connected to a $2.10\,\Omega$ resistor. Calculate the new current and terminal potential difference.[5 marks]
Answer
- $5.80\,\text{V}$
- $0.800\,\Omega$
- $2.00\,\text{A}$
- $4.20\,\text{V}$
Method: From $V=\varepsilon-Ir$ , the gradient of a $V$ against $I$ graph is $-r$ . Thus $-r=\dfrac{4.20-5.40}{2.00-0.50}=-0.800\,\Omega$ , so $r=0.800\,\Omega$ . Using the first reading, $\varepsilon=V+Ir=5.40+0.50\times0.800=5.80\,\text{V}$ . With $R=2.10\,\Omega$ , $I=\dfrac{5.80}{2.10+0.800}=2.00\,\text{A}$ . Hence $V=IR=2.00\times2.10=4.20\,\text{V}$ .