AMME1705 · Introduction to Electromechanical Systems
Electronics Foundations: Voltage, Current, Resistance & Power
The opening week of AMME1705 Introduction to Electromechanical Systems at the University of Sydney defines the four quantities every later circuit rests on — voltage (V, volts), current (I, amperes), resistance (R, ohms) and power (P, watts) — and the laws that tie them together. The core is Ohm's law V = I·R, its power companions P = V·I = I²R = V²/R, the series and parallel combination rules, conductance G = 1/R, and the voltage divider. Master these and you can already analyse most of the DC circuits this unit's 33% final exam asks you to work by hand.
What this chapter covers
- 01Name and unit the four quantities: voltage (V), current (I), resistance (R), power (W) — and why voltage is always a difference and current always a flow
- 02Apply Ohm's law V = I·R and its rearrangements I = V/R and R = V/I with consistent SI units
- 03Use conductance G = 1/R (siemens) as the reciprocal view of resistance
- 04Compute electric power three equivalent ways: P = V·I = I²R = V²/R, and see why a resistor turns all of it into heat
- 05Collapse resistors in series (R_eq = R1 + R2 + ...) using Kirchhoff's voltage law
- 06Collapse resistors in parallel (1/R_eq = 1/R1 + 1/R2 + ..., two-resistor shortcut R1R2/(R1+R2)) using Kirchhoff's current law
- 07Design and read a voltage divider: V_out = V_in·R2/(R1+R2), and know when loading breaks it
- 08Hit the specified-precision numeric answers the exam demands, showing the relation and carrying units
Series network: equivalent resistance, current, divider output and power
- +1(a) Equivalent resistance. Resistors in series add: R_eq = R1 + R2 = 1 kΩ + 2 kΩ = 3 kΩ.
- +1(b) Supply current. Ohm's law on the whole loop: I = V / R_eq = 9 V / 3000 Ω = 0.003 A = 3.0 mA.
- +1(c) Output across R2. Voltage divider: V_out = V_in · R2/(R1+R2) = 9 × 2/3 = 6.0 V (check: I·R2 = 3 mA × 2 kΩ = 6.0 V; R1 drops the remaining 3.0 V).
- +1(d) Total power. P = V · I = 9 V × 0.003 A = 0.027 W = 27 mW (check: I²R_eq = 0.003² × 3000 = 27 mW).
Key terms
- Voltage (potential difference), V
- The energy per unit charge between two points, measured in volts (V) — the 'push' that drives current. It is always a difference between two nodes, quoted relative to a chosen 0 V reference (ground), never an absolute value at one point.
- Current, I
- The rate of flow of electric charge through an element, measured in amperes (A); 1 A = 1 coulomb per second. Current flows through a component, whereas voltage appears across it.
- Resistance and conductance
- Resistance R (ohms, Ω) is the opposition to current under an applied voltage. Conductance G = 1/R (siemens, S) is its reciprocal — how easily current flows. A 2 kΩ resistor has G = 1/2000 = 0.5 mS.
- Ohm's law
- For an ohmic resistor, V = I·R, with rearrangements I = V/R and R = V/I. It holds where R is constant (resistors) but not for non-ohmic parts such as diodes, whose drop is roughly fixed near 0.7 V.
- Electric power
- The rate of energy delivery or dissipation, P = V·I (watts, W); substituting Ohm's law gives P = I²R and P = V²/R. In a resistor all of it becomes heat, which is why resistors carry a power rating.
- Series vs parallel
- Series (one current path): the same current flows and voltages add, so R_eq = R1 + R2 + ... (Kirchhoff's voltage law). Parallel (shared two nodes): the same voltage appears and currents add, so 1/R_eq = 1/R1 + 1/R2 + ... (Kirchhoff's current law), giving an R_eq smaller than the smallest branch.
- Voltage divider
- Two series resistors across V_in with the output tapped at the middle node: V_out = V_in·R2/(R1+R2), where R2 is the resistor V_out is read across. Valid only while negligible current is drawn at the node; a load in parallel with R2 lowers V_out (loading error).
Electronics Foundations: Voltage, Current, Resistance & Power FAQ
Can AI help me with Ohm's law and series/parallel circuit problems in AMME1705?
Yes — used well, an AI tutor like Sia is a strong study aid for this unit. Paste a circuit or a worked step and Sia can explain, step by step, how to pick the right form of Ohm's law, combine resistors, or set up a voltage divider, and it can generate fresh practice questions with your own numbers so you build the method rather than memorise one answer. Treat it as a way to understand and check your reasoning: it will not sit your quizzes or exam for you and cannot promise a particular mark or grade, and you should always confirm any numeric result against the specified-precision the exam demands.
Is the AMME1705 final exam open or closed book?
The two official sources differ: the assessment table describes the paper-based, 2-hour final (worth 33%) as closed book, while the exam cover sheet describes a restricted-open book exam that permits a calculator and one A4 double-sided handwritten note sheet. Because the wording has differed, prepare a note sheet with the core Week-1 equations (Ohm's law, power, series/parallel, the divider) but confirm the exact open/closed-book status on your current Canvas assessment page before the exam.
What is the difference between resistors in series and in parallel?
Series means one current path: the same current flows through every resistor, the voltage drops add up to the supply, and the equivalent resistance is the sum (R_eq = R1 + R2 + ...), always larger. Parallel means the resistors share the same two nodes: they see the same voltage, their branch currents add, and the reciprocals add (1/R_eq = 1/R1 + 1/R2 + ...), so the equivalent resistance is smaller than the smallest branch. A quick check: two equal resistors give 2R in series and R/2 in parallel.
Studying with AI? Sia — free AI electrical engineering tutor works through AMME1705 step by step.
Exam move
Treat Week 1 as the toolkit you will reuse all semester, and drill one reflex for every DC question: simplify, then apply Ohm, then find power. First collapse the network — add resistances in series, add reciprocals (and invert once more) in parallel, and remember the parallel result must be smaller than the smallest branch. Then apply the right form of Ohm's law: know V, want I, use I = V/R. Finally reach for the power form that avoids extra work — I²R when the current is known, V²/R when the voltage drop is known. Rehearse the voltage divider until V_out = V_in·R2/(R1+R2) is automatic, always putting the resistor you read the output across on top. Build a one-page equation sheet (Ohm, power, series/parallel, divider, G = 1/R), practise to the exam's specified precision (for example 'to the nearest 1 mA'), and always show the relation and carry the units — the marker rewards the correct formula and unit as much as the final digit. The whole paper runs at about 1.5 minutes per mark, so a clean, formula-first method is what keeps you on time.
