G5C: Series and Parallel Circuits – Ham Radio General License Study Guide
G5C covers how passive components — resistors, capacitors, and inductors — combine when wired together in series and parallel configurations, and how transformers use mutual inductance to transform voltage, current, and impedance.
The exam draws from topics including how to calculate the total resistance of parallel resistors, how total capacitance changes when capacitors are added in series or parallel, how total inductance changes when inductors are added in series or parallel, how to calculate transformer output voltage from the turns ratio, why a step-up transformer's primary uses heavier wire, what happens when a signal is applied to the secondary winding of a step-down transformer, and how to calculate the turns ratio required for impedance matching.
Resistors in Series and Parallel
Resistors in series simply add: Rtotal = R1 + R2 + R3 + ...
Resistors in parallel use the reciprocal formula: 1/Rtotal = 1/R1 + 1/R2 + 1/R3 + ... The total resistance of parallel resistors is always less than the smallest individual resistor value.
10 Ω, 20 Ω, and 50 Ω in parallel:
1/Rt = 1/10 + 1/20 + 1/50 = 0.1 + 0.05 + 0.02 = 0.17
Rt = 1 ÷ 0.17 = ~5.9 Ω
100 Ω and 200 Ω in parallel:
Rt = (100 × 200) ÷ (100 + 200) = 20,000 ÷ 300 = ~67 Ω
Capacitors in Series and Parallel
Capacitors in parallel add directly: Ctotal = C1 + C2 + C3 + ... Adding capacitors in parallel increases total capacitance. To increase the capacitance of a circuit, add a capacitor in parallel.
Capacitors in series use the reciprocal formula: 1/Ctotal = 1/C1 + 1/C2 + ... The total capacitance of series capacitors is less than the smallest individual value.
Two 5.0 nF capacitors and one 750 pF (= 0.75 nF) in parallel:
Ctotal = 5.0 + 5.0 + 0.75 = 10.750 nF
Three 100 µF capacitors in series:
1/Ct = 1/100 + 1/100 + 1/100 = 3/100
Ct = 100 ÷ 3 = ~33.3 µF
20 µF and 50 µF in series:
Ct = (20 × 50) ÷ (20 + 50) = 1000 ÷ 70 = ~14.3 µF
Inductors in Series and Parallel
Inductors in series add directly: Ltotal = L1 + L2 + L3 + ... Adding inductors in series increases total inductance. To increase the inductance of a circuit, add an inductor in series.
Inductors in parallel use the reciprocal formula: 1/Ltotal = 1/L1 + 1/L2 + ... The total inductance of parallel inductors is less than the smallest individual value.
20 mH and 50 mH in series:
Ltotal = 20 + 50 = 70 mH
Three 10 mH inductors in parallel:
1/Lt = 1/10 + 1/10 + 1/10 = 3/10
Lt = 10 ÷ 3 = ~3.3 mH
| Component | Series combination | Parallel combination | To increase total value, add in... |
|---|---|---|---|
| Resistors | Add directly | Reciprocal formula | Series |
| Capacitors | Reciprocal formula | Add directly | Parallel |
| Inductors | Add directly | Reciprocal formula | Series |
Transformers
A transformer uses mutual inductance to transfer energy between two windings. When AC voltage is applied to the primary winding, its changing current creates a changing magnetic field in the core. This changing field induces a voltage in the secondary winding.
The relationship between primary and secondary voltage follows the turns ratio:
V2 / V1 = N2 / N1
A transformer with 500 turns on the primary and 1500 turns on the secondary (a 1:3 step-up ratio) produces an output voltage of 120 × (1500/500) = 120 × 3 = 360 volts.
In a voltage step-up transformer, the secondary voltage is higher than the primary voltage, but the secondary current is proportionally lower (power is conserved). The primary therefore carries more current than the secondary, which is why the primary winding uses larger-diameter wire — it must handle the higher current without excessive resistive losses.
If a signal is applied to the secondary winding of a 4:1 step-down transformer instead of the primary, the transformer now acts in reverse. The signal is stepped up by a factor of 4 — the output voltage (appearing at what was formerly the primary) is the input voltage multiplied by 4.
Transformers also transform impedance by the square of the turns ratio:
Z1 / Z2 = (N1 / N2)²
To match a 600-ohm antenna feed point impedance to a 50-ohm coaxial cable, the required turns ratio is √(600/50) = √12 ≈ 3.5 to 1.
G5C Practice Questions
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