Electric Circuit Fundamentals
This topic covers the fundamental principles of electric circuits, including components, key quantities like current and voltage, and the relationships that govern them. Mastering these concepts is crucial for analysing circuit behaviour and calculating energy and power, which are frequently tested.
Part of the ESAT Physics syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.
Key points
- In series circuits, the current is the same through all components, and the total voltage is the sum of the voltages across each component.Vtotal = V1 + V2 + ⋯
- In parallel circuits, the voltage across each branch is the same, and the total current is the sum of the currents in each branch.Itotal = I1 + I2 + ⋯
- The total resistance of resistors in parallel is always less than the resistance of the smallest individual resistor in the combination.
- A filament lamp's V-I graph is a curve bending towards the voltage axis, indicating its resistance increases as it gets hotter (see the diagram below).
- Special components have variable resistance: an NTC thermistor's resistance drops as temperature rises, while an LDR's resistance drops as light intensity increases.
- Ideal diodes have infinite resistance in one direction (reverse bias) and zero resistance in the other (forward bias), acting as one-way gates for current.
Diagram
› Why does this happen?
At low voltages
The filament is relatively cool, so its resistance is low. The curve is steep here - current rises almost in proportion to voltage, behaving nearly like an ohmic conductor.
At high voltages
The filament becomes very hot and glows. The extra thermal energy makes the metal's positive ions vibrate more vigorously, so the free electrons collide with them far more often. These collisions impede the flow of charge, so the resistance rises.
Working out the resistance
To find the resistance at any point on the graph, just read off the current (I) and the voltage (V) there and use R = V / I. Try a point low down on the curve and another higher up: the higher point gives a bigger R - that is the resistance increasing as the filament heats up.
Formulae
I = Q / t To find the current (I) when a certain amount of charge (Q) flows in a specific time (t).
V = I × R Ohm's Law, relating voltage (V), current (I), and resistance (R) for an ohmic component.
P = I × V To calculate the electrical power (P) dissipated by a component. Can be combined with Ohm's law to get P = I2 × R or P = V2 / R.
E = P × t = V × I × t To calculate the total energy (E) transferred by a component over a period of time (t).
Rtotal = R1 + R2 + ⋯ To find the total resistance of multiple resistors connected in series.
Definitions
- Electric Current (I)
- The rate of flow of electric charge. Measured in Amperes (A), where 1 A = 1 Coulomb per second.
- Voltage (V)
- Also known as potential difference. The work done or energy transferred per unit of charge. Measured in Volts (V), where 1 V = 1 Joule per Coulomb.
- Resistance (R)
- A measure of a component's opposition to the flow of electric current. Measured in Ohms (Ω).
- Direct Current (dc)
- Current that flows in only one direction. Produced by cells and batteries.
- Alternating Current (ac)
- Current that periodically reverses its direction. The mains electricity supply is ac.
Worked example
A circuit contains a 12 V cell. It is connected to a 4 kΩ resistor in series with a parallel arrangement of a 6 kΩ resistor and a 3 kΩ resistor. What is the potential difference across the 4 kΩ resistor?
- 1
First, calculate the equivalent resistance of the parallel section (Rp).
Use the formula 1/Rp = 1/R1 + 1/R2.
- 2 1/Rp = 1/(6 kΩ) + 1/(3 kΩ)
Find a common denominator:
1/Rp = 1/6 + 2/6 = 3/6 = 1/2Therefore, Rp = 2 kΩ.
- 3
Next, find the total resistance of the entire circuit (Rtotal) by adding the series resistor to the parallel equivalent.
- 4 Rtotal = Rseries + Rp = 4 kΩ + 2 kΩ = 6 kΩ
- 5
Now, calculate the total current flowing from the cell using Ohm's Law (I = V / Rtotal).
- 6 I = 12 V / 6 kΩ = 12 V / 6000 Ω = 2 × 10-3 A = 2 mA
- 7
This total current flows through the 4 kΩ series resistor.
Use Ohm's Law again to find the voltage across it (V = I × R).
- 8 V4kΩ = (2 × 10-3 A) × (4 × 103 Ω) = 8 V
Answer: 8 V
Common mistakes
- ×Forgetting to handle prefixes like 'milli' (mA), 'micro' (μA), and 'kilo' (kΩ) correctly during calculations. Always convert to base units (A, V, Ω) before applying formulas to avoid errors by factors of 1000.
- ×When calculating parallel resistance with the 1/R formula, a common mistake is to forget the final step of inverting the result to find R, not 1/R.
- ×Misinterpreting V-I graphs. Remember that the gradient of a V-I graph is resistance (R = V/I), whereas the gradient of an I-V graph is the reciprocal of resistance (1/R).
- ×Confusing series and parallel rules. Trace the current path: if it must go through all components sequentially, it's series. If the path splits and rejoins, it's parallel.
No-calculator tips
- ✓For two resistors in parallel, the product-over-sum rule `Rp = (R1 × R2) / (R1 + R2)` is often quicker than the reciprocal method.
- ✓When a question involves ratios (e.g., finding the voltage across one resistor in a series pair), use the potential divider principle: V1 = Vtotal × (R1 / Rtotal). This avoids calculating the current first.
- ✓If you have to divide awkward numbers, like 12 / 5, multiply the top and bottom by 2 to get 24 / 10, which is easily calculated as 2.4.
- ✓Before calculating, estimate your answer. For example, the total resistance of a parallel combination must be smaller than the smallest resistor in the branch. This is a quick sanity check for your final value.