5.3 Series/Parallel Networks & Supply Internal Resistance

5.3 Series/Parallel Networks & Supply Internal Resistance

Quick Answer: Series: R_T = ΣR, same I, voltages share. Parallel: 1/R_T = Σ(1/R), same V, currents share. Reduce combinations group by group. An unloaded divider gives V_out = V_in × R₂/(R₁+R₂). With internal resistance r, V_terminal = E − I r, so heavier loads pull the terminals down.

Topic 3.6 finishes with the networks you actually meet on aircraft wiring and bench power supplies: strings, branches, dividers, and real sources that are not ideal voltage fonts.

Series Resistance Networks

For n resistances in series:

R_T = R₁ + R₂ + … + R_n

Worked Series Example

R₁ = 10 Ω, R₂ = 22 Ω, R₃ = 8 Ω on E = 40 V (ideal).

1. R_T = 10 + 22 + 8 = 40 Ω.
2. I = 40/40 = 1 A.
3. Drops: 10 V, 22 V, 8 V (sum 40 V).

Parallel Resistance Networks

1/R_T = 1/R₁ + 1/R₂ + … + 1/R_n

For two resistors: R_T = (R₁ R₂)/(R₁ + R₂).

R_T is always less than the smallest branch resistance.

Worked Parallel Example

R₁ = 6 Ω, R₂ = 12 Ω, R₃ = 4 Ω across 24 V.

1. 1/R_T = 1/6 + 1/12 + 1/4 = 0.1667 + 0.0833 + 0.25 = 0.5 ⇒ R_T = 2 Ω.
2. I_T = 24/2 = 12 A.
3. Branch currents: 4 A, 2 A, 6 A (sum 12 A).

Combination (Series–Parallel) Reduction

Worked Combination Example

Circuit: R₁ = 5 Ω in series with a parallel pair R₂ = 20 Ω and R₃ = 30 Ω; supply E = 50 V ideal.

1. Parallel pair: R₂₃ = (20 × 30)/(20 + 30) = 600/50 = 12 Ω.
2. R_T = 5 + 12 = 17 Ω.
3. I_total = 50/17 ≈ 2.941 A.
4. Drop on R₁: V₁ = 2.941 × 5 ≈ 14.71 V.
5. Parallel voltage: V₂₃ = 50 − 14.71 ≈ 35.29 V (also 2.941 × 12).
6. I₂ = 35.29/20 ≈ 1.765 A, I₃ = 35.29/30 ≈ 1.176 A (sum ≈ 2.941 A).

Always reduce from the inside out: combine pure parallel or series islands, redraw, repeat.

Voltage Dividers

Two resistors in series across V_in form a divider. Taking the output across the bottom resistor R₂:

V_out = V_in × R₂ / (R₁ + R₂) (unloaded)

Worked Unloaded Divider

V_in = 28 V, R₁ = 3 kΩ, R₂ = 1 kΩ, output across R₂.

V_out = 28 × 1000/(3000+1000) = 28 × 0.25 = 7 V.

Loaded Divider Trap

If a load R_L = 1 kΩ connects across R₂, R₂ and R_L are in parallel first:

R₂∥L = (1×1)/(1+1) = 0.5 kΩ.

Then V_out = 28 × 0.5/(3 + 0.5) = 28 × 0.5/3.5 = 4 V.

Loading pulled the tap from 7 V down to 4 V. Exam stems that ignore the load after mentioning it are testing whether you rebuild the parallel equivalent.

Supply Internal Resistance and Terminal Voltage

A real DC source is modelled as an ideal EMF E in series with internal resistance r. External load R_L connects to the terminals.

I = E / (R_L + r)

V_terminal = I R_L = E − I r = E × R_L/(R_L + r)

Worked Internal-r Example 1

Battery E = 24 V, r = 0.5 Ω, load R_L = 7.5 Ω.

1. I = 24 / (7.5 + 0.5) = 24/8 = 3 A.
2. V_T = 3 × 7.5 = 22.5 V.
3. Internal drop: I r = 1.5 V; check 24 − 1.5 = 22.5 V.

Worked Internal-r Example 2 — Heavier Load

Same battery, now R_L = 1.5 Ω.

1. I = 24 / (1.5 + 0.5) = 24/2 = 12 A.
2. V_T = 12 × 1.5 = 18 V.
3. Internal drop 6 V.

Terminal voltage fell from 22.5 V to 18 V as load current rose. That is why a weak battery (higher effective r) may show near-rated open-circuit voltage on a high-impedance meter yet collapse when a starter or pump demands high current.

Worked Internal-r Example 3 — Finding r from Measurements

Open-circuit voltage measured E = 28.0 V. Under a 10 A load, terminals read 26.5 V.

I r = E − V_T = 1.5 V ⇒ r = 1.5/10 = 0.15 Ω.

This measurement pattern appears in Module 3 reasoning even when the algebra is short: open-circuit ≈ EMF; loaded sag reveals r.

Linking Networks to Internal r

Suppose E = 30 V, r = 1 Ω, and the external network is 4 Ω in series with a parallel pair of 12 Ω and 6 Ω.

1. Parallel: R_p = (12×6)/(12+6) = 4 Ω.
2. External R_L = 4 + 4 = 8 Ω.
3. I = 30/(8+1) = 3.333 A.
4. V_T = 30 − 3.333×1 ≈ 26.67 V (also 3.333 × 8).
5. Series 4 Ω drop ≈ 13.33 V; parallel voltage ≈ 13.33 V; currents 13.33/12 ≈ 1.11 A and 13.33/6 ≈ 2.22 A.

One problem used series/parallel reduction, Ohm’s law, KCL at the parallel node, and internal-resistance sag together—the full topic 3.6 skill set.

Exam Priority Checklist

1. Label series vs parallel vs mixed; redraw after each reduction.
2. For dividers, ask whether the output is loaded.
3. Never treat battery terminal voltage as equal to EMF under load unless r = 0 or I ≈ 0.
4. Sense-check: adding a parallel path always lowers R_T; adding series always raises R_T; loading a divider always lowers (or leaves equal) V_out compared with the unloaded tap.

Series/parallel arithmetic plus internal-r awareness turns Ohm and Kirchhoff from isolated laws into a complete DC-circuit toolkit for CAAS Module 3.