Find voltage and impedance through a transformer, and total R, L and C in series and parallel.
Syllabus: Resistors, capacitors, and inductors in series and parallel; transformers
Compendium articles on the ideas in this lesson, beyond what the exam asks.
Practise the 14 pool questions for this group. Answers are shuffled, and every one comes with a one-line "why".
AC in the primary makes a changing magnetic field that cuts the secondary winding. That is mutual inductance. Voltage follows the turns, and it works from either winding.
Vout = Vin × (turns out ÷ turns in)RememberMore turns on the output side, more volts. Feed the secondary instead and the ratio reverses.
A transformer does not make power. Volts × amperes stays about the same on both sides, so the low-voltage side carries the high current. In a step-up transformer that is the primary, so it gets the thicker wire.
RememberHigher voltage means lower current. The high-current winding uses thicker wire.
A transformer also makes a load look different to the source, and the impedance ratio is the square of the turns ratio.
turns ratio = √(Z high ÷ Z low)600 Ω to 50 Ω: √(600 ÷ 50) = √12 = 3.46, about 3.5 to 1.
RememberImpedance ratio = turns ratio². Take the square root to find the turns.
In series the values simply add. In parallel there are more paths, so the total is smaller than the smallest part: add the reciprocals, then flip.
| Combination | Calculation | Total |
|---|---|---|
| 10 Ω ‖ 20 Ω ‖ 50 Ω | 1 ÷ (0.1 + 0.05 + 0.02) | 5.9 Ω |
| 100 Ω ‖ 200 Ω | 1 ÷ (0.01 + 0.005) | 67 Ω |
| 3 × 10 mH in parallel | 10 ÷ 3 | 3.3 mH |
| 20 mH + 50 mH in series | 20 + 50 | 70 mH |
RememberTo raise inductance, add an inductor in series. Equal parts in parallel: divide by how many.
Parallel plates add area, so capacitance adds. Series stacks the gaps, so the total is smaller than the smallest part. Convert to the same unit first.
| Combination | Calculation | Total |
|---|---|---|
| 5 nF + 5 nF + 750 pF, parallel | 5 + 5 + 0.75 | 10.75 nF |
| 3 × 100 µF in series | 100 ÷ 3 | 33.3 µF |
| 20 µF and 50 µF in series | 1 ÷ (0.05 + 0.02) | 14.3 µF |
RememberTo raise capacitance, add a capacitor in parallel.
Memory hookR and L: series adds. C is the odd one out: parallel adds.