Lesson DC Circuits · Parallel
Parallel circuits
A parallel circuit has more than one path for current. Voltage is the same across each branch. Branch currents add to total current. Home receptacles and branch lighting are parallel so each load sees full supply voltage independently.
What we are trying to understand
If series is one hallway, parallel is a forked road from the same parking lot: same "pressure" (voltage) at the start of each road, but traffic (current) splits by how wide each road is (resistance).
Guiding question:
From the same two supply nodes, can current take more than one path?
If yes, those paths are in parallel between those nodes.
Why homes are wired in parallel
Imagine all lights in series: one burned lamp kills every room. Parallel wiring puts each lamp and receptacle across the same supply. Each gets (ideally) the same voltage; current depends on its own resistance.
If one branch opens, others can still operate—unlike series.
Three rules for parallel circuits
| Rule | In words | Formula hint |
|---|---|---|
| Voltage | Same across every branch | \(E_T = E_1 = E_2 = E_3\) |
| Current | Branch currents add | \(I_T = I_1 + I_2 + I_3\) |
| Resistance | Total is less than smallest branch | See next lesson for formulas |
These mirror series rules but swap roles: series shares I; parallel shares E.
Total resistance always drops
Each new parallel path gives current another way to flow. Total opposition decreases.
Analogy: opening another lane for water lowers overall resistance to flow—even if each lane has the same pipe size.
\(R_T\) is always less than the smallest individual branch resistance.
Solving a simple parallel circuit
Example: \(E_T = 120\,\mathrm{V}\), three branches \(R_1 = 20\,\Omega\), \(R_2 = 30\,\Omega\), \(R_3 = 60\,\Omega\).
First find \(R_T\) (methods in lesson 05). Suppose \(R_T = 10\,\Omega\). \[ I_T = 120/10 = 12\,\mathrm{A} \] \[ I_1 = 120/20 = 6\,\mathrm{A},\quad I_2 = 4\,\mathrm{A},\quad I_3 = 2\,\mathrm{A} \] Check: \(6 + 4 + 2 = 12\,\mathrm{A}\). Good.
Power in parallel
Total power is the sum of branch powers: \[ P_T = P_1 + P_2 + P_3 + \ldots \] Also \(P_T = E_T imes I_T\).
Each branch dissipates independently: \(P_n = E_T imes I_n = E_T^2 / R_n\).
Current divider concept preview
Every parallel network is a current divider. Fixed voltage across branches; current splits inversely with resistance—lower R draws more I.
Detailed formulas come in lesson 05; for now: fat branch (low R) carries more current.
Field tracing parallel paths
On a print:
- Find two common nodes (bus, neutral bar).
- List every device directly between them—that is one parallel group.
- Series elements inside a branch belong to that branch only.
A blown branch fuse kills that branch only; the bus voltage may still be fine—classic parallel symptom.
Field case
Situation. Three DC heaters on a 48 V bus: 10 Ω, 20 Ω, and 40 Ω. One heater fuse blows. The other two still run hot; the panel ammeter shows lower total current.
How to think with this lesson.
- Common 48 V across each heater path.
- Lost branch current = 48/10 = 4.8 A (the 10 Ω leg).
- Total current drops by that amount; bus V stays ~48 V.
- Replace fuse only after finding why the 10 Ω leg overcurrented.
Conclusion: parallel failures are local; series failures are global.
In the field
Symptom
One load dead; others normal; bus voltage OK
Where to look
Branch fuse, open conductor, bad contact on that leg only
Likely causes
- Open in one parallel branch
- not a main supply failure
What to measure
- V bus-to-bus
- I in each branch
- I total
What not to do
- Replace main fuse when one branch alone is open
Checklist
- I define parallel as multiple paths between same two nodes
- I state: same V, I adds, RT decreases
- I compute branch I from E/R
- I verify IT equals sum of branch currents
- I recognize home/receptacle wiring as parallel
- I trace branches from bus to bus on diagrams