Resistor
Current, p.d., power and resistance
Back to Higher hubCharge & current
Charge
Q = I t
charge = current × time
- Current is the rate of flow of charge — how many coulombs pass each second.
- Conventional current flows from + to − around the circuit.
- In a series loop the current is the same everywhere. At a junction the currents add (charge is conserved).
National 5 refresher — exam questions don't ask Q = I t on its own at Higher. It comes back in the Capacitors topic (charge stored for a constant charging current), so make sure it's automatic.
Circuit symbols used on this page (UK)
Variable resistor
Cell
Lamp
Switch
Voltmeter
Ammeter
LDR
Thermistor
Substitute & solve
A current of 0.25 A flows in a circuit for 2.0 minutes. Calculate the charge transferred in this time. 3 marks
Convert first
t = 2.0 × 60 = 120 s min → s
Q=I t
Q=0.25 × 120
Q=30 C
Rearrange — numbers in first
A charge of 72 C passes through the filament of a lamp in 1.0 minute. Calculate the current in the lamp. 3 marks
Convert first
t = 1.0 × 60 = 60 s min → s
Q=I t
72=I × 60numbers in first
I=72 ÷ 60now rearrange
I=1.2 A
Practice 1
A charge of 90 C passes a point in a circuit in 1.5 minutes. Calculate the current in the circuit. 3 marks
Answer
t = 1.5 × 60 = 90 s (minutes → seconds)
Q = It → 90 = I × 90 → I = 1.0 A
Potential difference, Ohm's law & resistance
Ohm's law
V = I R
p.d. = current × resistance
- Potential difference is the energy given to each coulomb of charge.
- Resistance opposes current. Rearrange Ohm's law as needed: R = V ÷ I.
- Ohmic conductor → V–I graph is a straight line through the origin (constant R).
- Non-ohmic (e.g. a lamp) → the line curves; resistance rises as it heats up.
Substitute & solve
The current in a 4.7 kΩ resistor is 2.0 mA. Calculate the potential difference across the resistor. 3 marks
Convert first
R = 4.7 × 1000 = 4700 Ω kΩ → Ω
I = 2.0 ÷ 1000 = 0.002 A mA → A
V=I R
V=0.002 × 4700
V=9.4 V
Rearrange — numbers in first
A 12 V supply of negligible internal resistance is connected across a resistor. The current in the resistor is 50 mA. Calculate the resistance of the resistor. 3 marks
Convert first
I = 50 ÷ 1000 = 0.050 A mA → A
V=I R
12=0.050 × Rnumbers in first
R=12 ÷ 0.050now rearrange
R=240 Ω
Practice 2
The current in a 1.0 kΩ resistor is 12 mA. Calculate the potential difference across the resistor. 3 marks
Answer
R = 1.0 × 1000 = 1000 Ω (kΩ → Ω)
I = 12 ÷ 1000 = 0.012 A (mA → A)
V = IR = 0.012 × 1000 = 12 V
Sig figs & notation: give your final answer with no more significant figures than the fewest in the data, and use scientific notation for very large or small values — marking instructions write answers like 1.2 × 10³ Ω.
"Show that" questions: start from the relationship, substitute the numbers, then state the target value with its unit — e.g. show that the current is 0.25 A: I = V ÷ R = 6.0 ÷ 24 = 0.25 A. Never start from the answer and work backwards — that scores zero.
Go further — uncertainty in R = V ÷ I
When two measured values are combined, the one with the largest percentage uncertainty is a good estimate of the percentage uncertainty in the result.
V = (10.0 ± 0.1) V → 0.1/10.0 = 1%
I = (0.50 ± 0.01) A → 0.01/0.50 = 2% (largest)
R = V ÷ I = 10.0 ÷ 0.50 = 20 Ω
uncertainty ≈ 2% of 20 = ± 0.4 Ω → R = (20.0 ± 0.4) Ω
Resistors in series & parallel
Series
RT = R1 + R2 + …
resistances add up
Parallel
1RT
=
1R1
+
1R2
+ …
total is less than the smallest branch
- Series: same current everywhere; the p.d.s add to the supply voltage.
- Parallel: same p.d. across each branch; the branch currents add.
- For parallel, remember to flip back at the end: you find 1/RT first, then invert.
Substitute & solve
A 2.0 kΩ resistor and a 3.0 kΩ resistor are connected in parallel. Calculate the total resistance of the combination. 3 marks
Convert first
R1, R2 = 2000 Ω, 3000 Ω kΩ → Ω
1/RT=1/2000 + 1/3000
1/RT=0.000833
RT=1 ÷ 0.000833flip back at the end
RT=1200 Ω (1.2 kΩ)
The total (1.2 kΩ) is less than the smallest branch (2.0 kΩ).
Rearrange — numbers in first
Two resistors connected in series have a total resistance of 10 kΩ. One of the resistors has a resistance of 6.8 kΩ. Determine the resistance of the other resistor. 3 marks
Convert first
RT = 10 × 1000 = 10000 Ω kΩ → Ω
RT=R1 + R2
10000=6800 + R2numbers in first
R2=10000 − 6800now rearrange
R2=3200 Ω (3.2 kΩ)
Mixed networks — reduce in stages
- Exam circuits usually mix series and parallel. Reduce the parallel part first, then add the series resistors — one stage at a time.
- Shortcut: n identical resistors in parallel give RT = R ÷ n — ten 220 Ω branches make 220 ÷ 10 = 22 Ω.
- To find the p.d. or power in one part of a circuit: reduce the network → find the total current → work back out to the part you want.
Substitute & solve
A 6.0 Ω resistor is connected in series with a parallel combination of a 12 Ω resistor and a 4.0 Ω resistor. Calculate the total resistance of the circuit. 5 marks
1/Rp=1/12 + 1/4.0parallel part first
1/Rp=0.333
Rp=1 ÷ 0.333 = 3.0 Ωflip back
RT=6.0 + 3.0now add the series R
RT=9.0 Ω
Several steps — work back out
A 20 Ω resistor is connected in series with a parallel combination of a 30 Ω resistor and a 60 Ω resistor, across a 12 V supply of negligible internal resistance. Determine the potential difference across the 60 Ω resistor. 5 marks
1/Rp=1/30 + 1/60 = 0.05parallel part first
Rp=20 Ωflip back
RT=20 + 20 = 40 Ω
I=V ÷ RT = 12 ÷ 40total current next
I=0.30 A
V=I Rp = 0.30 × 20work back out
V=6.0 Vsame across both branches
"Negligible internal resistance" in a question means: ignore the battery's own resistance. When it is not negligible, that's the next topic — Electrical sources & internal resistance. And a.c. questions reuse these exact circuit methods with rms values.
Practice 3
Two 4.0 kΩ resistors are connected in parallel. Calculate the total resistance of the combination. 3 marks
Answer
R₁ = R₂ = 4000 Ω (kΩ → Ω)
1/R_T = 1/4000 + 1/4000 = 2/4000 = 0.0005
R_T = 1 ÷ 0.0005 = 2000 Ω (2.0 kΩ)
Practice 4
A 1.0 kΩ resistor is connected in series with a parallel combination of a 3.0 kΩ resistor and a 6.0 kΩ resistor, across a 9.0 V supply of negligible internal resistance.
(a) Calculate the total resistance of the circuit. 5 marks
(b) Determine the current from the supply. 3 marks
Answer
(a) parallel part first: 1/R_p = 1/3000 + 1/6000 = 0.0005 (kΩ → Ω)
R_p = 1 ÷ 0.0005 = 2000 Ω
R_T = 1000 + 2000 = 3000 Ω (3.0 kΩ)
(b) I = V ÷ R_T = 9.0 ÷ 3000 = 0.003 A (3.0 mA)
Electrical power & energy
Power
P = I V = I2R =
V2R
pick the form from what you're given
Energy
E = P t
energy transferred = power × time
- Power is the energy transferred each second (watts = joules per second).
- All three power formulae give the same answer — choose the one matching your known values.
- For heat or energy over time, follow up with E = Pt.
Substitute & solve
The current in a 50 Ω heating element is 4.0 A. The heater operates for 2.0 minutes. Determine the energy transferred to the heater in this time. 5 marks
Convert first
t = 2.0 × 60 = 120 s min → s
P=I2 R
P=4.0² × 50square the current first
P=800 W
E=P t = 800 × 120
E=96000 J (96 kJ)
Rearrange — numbers in first
A 1.5 kW kettle is connected to the 230 V mains supply. Calculate the current in the kettle element. 3 marks
Convert first
P = 1.5 × 1000 = 1500 W kW → W
P=I V
1500=I × 230numbers in first
I=1500 ÷ 230now rearrange
I=6.5 A
Rated power — when the square root comes out
Substitute & solve
A heating element of resistance 23 Ω is connected to the 230 V mains supply. Calculate the power dissipated in the element. 3 marks
P=V2 ÷ Rgiven V and R — pick V²/R
P=230² ÷ 23square the voltage first
P=2300 W (2.3 kW)
Rearrange — numbers in first
A 2.2 kΩ resistor is rated at 0.25 W. Determine the maximum potential difference that can be applied across the resistor without exceeding its power rating. 3 marks
Convert first
R = 2.2 × 1000 = 2200 Ω kΩ → Ω
P=V2 ÷ R
0.25=V2 ÷ 2200numbers in first
V2=0.25 × 2200 = 550now rearrange
V=√550square root last
V=23 V
Practice 5
The current in a 2.0 kΩ resistor is 0.30 A. Calculate the power dissipated in the resistor. 3 marks
Answer
R = 2.0 × 1000 = 2000 Ω (kΩ → Ω)
P = I²R = 0.30² × 2000 = 0.09 × 2000 = 180 W
Practice 6
(a) A heater rated at 4.8 W has a resistance of 120 Ω. Calculate the current in the heater. 3 marks
(b) The current in a 2.2 MΩ resistor is 5.0 µA. Calculate the power dissipated in the resistor. 3 marks
Answer
(a) P = I²R → 4.8 = I² × 120 → I² = 0.04 → I = √0.04 = 0.20 A
(b) R = 2.2 × 10⁶ Ω (MΩ → Ω); I = 5.0 × 10⁻⁶ A (µA → A)
P = I²R = (5.0 × 10⁻⁶)² × 2.2 × 10⁶ = 5.5 × 10⁻⁵ W
Say it like the SQA: current in a component, p.d. (voltage) across it. Marking instructions give 0 marks for "voltage through" or "current across".
Justify it
1. A battery of negligible internal resistance is connected to a resistor in series with two identical resistors in parallel. A switch is connected in series with one of the parallel resistors. The switch is now closed. State whether the current from the battery increases, stays the same, or decreases. You must justify your answer. 2 marks
Answer
Increases. (1)
Closing the switch adds a parallel branch, so the total resistance of the
circuit decreases — I = V ÷ R_T, so the current increases. (1)
2. Two identical resistors connected in series with a battery are reconnected in parallel across the same battery. State whether the total power dissipated is greater than, equal to, or less than before. You must justify your answer. 2 marks
Answer
Greater than. (1)
Total resistance falls from 2R to R/2, and P = V² ÷ R_T,
so the power dissipated is greater (4 times as much). (1)
Potential dividers
Divider ratio
V1V2
=
R1R2
voltages share in the ratio of the resistances
Divider output
V2 =
R2R1 + R2
× Vs
p.d. across the bottom resistor (R2)
- R₁ is the top resistor, R₂ is the bottom resistor. The output V₂ is always the p.d. across the bottom resistor (R₂).
- Two resistors in series split the supply voltage in the ratio of their resistances.
- The bigger resistor takes the bigger share of the voltage.
- Sensor dividers: swap one resistor for an LDR (resistance falls as light rises) or a thermistor (resistance falls as temperature rises) so the output voltage changes with the surroundings.
- Recent Higher exam divider questions are numeric — sensor dividers are context you know from National 5, useful for understanding rather than a calculation you'll be set.
Relationships sheet: the SQA sheet writes the divider output across R1 — V1 = (R1 ÷ (R1 + R2)) × Vs — and V1/V2 = R1/R2. Same formula, different label: read the question carefully to see which resistor's p.d. is wanted.
Substitute & solve
A potential divider consists of R₁ = 3.0 kΩ (top) and R₂ = 7.0 kΩ (bottom) connected across a 12 V supply. Calculate the potential difference across R₂. 3 marks
V2=(R2 ÷ (R1 + R2)) × Vsboth in kΩ, so they cancel — no need to convert
V2=(7.0 ÷ 10.0) × 12
V2=8.4 V
Rearrange — numbers in first
Two resistors are connected in series across a 10 V supply. The potential difference across R₁ is 4.0 V and the resistance of R₂ is 3.0 kΩ. Determine the resistance of R₁. 4 marks
V2=10 − 4.0 = 6.0 Vfind the other p.d. first
V1 ÷ V2=R1 ÷ R2
4.0 ÷ 6.0=R1 ÷ 3.0numbers in first
R1=3.0 × (4.0 ÷ 6.0)now rearrange
R1=2.0 kΩ
Practice 7
A potential divider consists of a 2.2 kΩ resistor (top) and a 4.7 kΩ resistor (bottom) connected across a 9.0 V supply. Calculate the potential difference across the 4.7 kΩ resistor. 3 marks
Answer
Both in kΩ, so they cancel — no need to convert.
V₂ = (R₂ / (R₁ + R₂)) × V_s = (4.7 / (2.2 + 4.7)) × 9.0 = (4.7 / 6.9) × 9.0 = 6.1 V
Justify it
In a potential divider, the resistance of the bottom resistor R₂ is increased. State whether the output voltage V₂ increases, stays the same, or decreases. You must justify your answer. 2 marks
Answer
Increases. (1)
R₂ is now a bigger share of the total resistance,
so it takes a bigger share of the supply voltage. (1)
Two dividers across one supply
This circuit hasn't appeared directly in recent Higher papers — treat it as extension that sharpens your divider skills (and previews monitoring circuits).
Voltage "across the bridge"
V = VP − VQ
p.d. between the two midpoints
- Put two potential dividers across the same supply. Each midpoint voltage is just a divider output.
- The voltage "across the bridge" is the difference between the two midpoints: work out VP and VQ, then subtract.
- The reading is zero when the two dividers have the same ratio (R₁/R₂ = R₃/R₄).
Substitute & solve
Two potential dividers are connected across a 9.0 V supply: 2.0 kΩ over 4.0 kΩ on the left, and 6.0 kΩ over 3.0 kΩ on the right. Determine the potential difference between the midpoints P and Q. 5 marks
VP=(4.0 ÷ 6.0) × 9.0 = 6.0 Vleft divider first
VQ=(3.0 ÷ 9.0) × 9.0 = 3.0 Vthen the right divider
V=VP − VQ
V=6.0 − 3.0 = 3.0 V
Rearrange — numbers in first
Two potential dividers are connected across a 12 V supply. The left divider has two equal resistors. The voltmeter between the midpoints reads +1.5 V (P higher than Q). Determine VQ. 4 marks
VP=½ × 12 = 6.0 Vequal resistors → half the supply
V=VP − VQ
1.5=6.0 − VQnumbers in first
VQ=6.0 − 1.5now rearrange
VQ=4.5 V
Go further — try the simulator
V across the bridge = —
Practice 8
Two potential dividers are connected across a 10 V supply: 5.0 kΩ over 5.0 kΩ on the left, and 2.0 kΩ over 8.0 kΩ on the right. Determine the potential difference between the midpoints. 5 marks
Answer
V_P = (5/10)×10 = 5.0 V
V_Q = (8/10)×10 = 8.0 V
V = V_P − V_Q = 5.0 − 8.0 = −3.0 V (Q is 3.0 V higher than P)