Parallel Circuits
Imagine a motorway that splits into three lanes through a toll plaza, then merges back into one lane on the other side. Every lane starts at the same point and ends at the same point. A car entering any lane travels between exactly the same two places — but the lanes operate independently. A blockage in one lane does not stop traffic in the others. That is the essence of a parallel circuit.
In electronics, a parallel circuit is any arrangement where two or more components share the same two connection points. Each component forms its own independent branch between those points. Because every branch connects between identical nodes, every branch sees the same voltage. Current, however, splits — more flows through lower-resistance branches, less through higher-resistance branches. The total current drawn from the supply is the sum of all branch currents.
Parallel circuits are everywhere in electronics. Every electrical outlet in your home is wired in parallel. Every component in a transceiver is powered from parallel rails. Antenna matching networks use parallel elements to transform impedance. Understanding parallel circuits from the ground up is not optional — it is essential.
- What Makes a Circuit Parallel?
- Rule 1: Voltage Is the Same Across Every Branch
- Rule 2: Total Resistance Is Less Than Any Branch
- Rule 3: Total Current Is the Sum of Branch Currents
- Three Methods for Calculating Total Resistance
- Worked Example: Step by Step
- More Practice Examples
- Power in Parallel Circuits
- What Happens When Things Go Wrong
- Parallel Circuit Calculator
- Experiment: Parallel Resistors on a Breadboard
- Parallel Circuits in Ham Radio
A parallel circuit: all branches share the same two nodes, so all branches see the same voltage. Current splits between branches according to each branch's resistance.
View LargerWhat Makes a Circuit Parallel?
Before you can apply the parallel rules, you need to be able to identify a parallel connection when you see one — whether in a schematic, on a breadboard, or inside real equipment. The definition is precise: two components are in parallel if and only if both terminals of one component connect directly to both terminals of the other, with nothing else in between.
In a schematic, parallel branches appear as side-by-side paths between the same two wires or nodes. In real wiring, parallel components share the same pair of connection points with no other components between them and the nodes. The key test is this: can you trace a path from the positive terminal of the supply through each component and back to the negative terminal that does not pass through any other component? If yes for each branch independently, those branches are in parallel.
Where beginners often go wrong is confusing a parallel branch with a series connection. A series string has only one path — electrons must flow through every component in sequence. A parallel arrangement has multiple paths — electrons choose one path to follow. If removing one component would stop current in another, they are in series. If removing one component leaves the others still connected and operating, they are in parallel.
Another helpful check: in a parallel circuit, if you place a voltmeter across each branch in turn, every reading should be identical (equal to the supply voltage). If readings differ, something is not truly in parallel — there may be resistance in the wiring, a bad connection, or a component in series with one branch.
Rule 1: Voltage Is the Same Across Every Branch
This is the most fundamental rule of parallel circuits, and it follows directly from what a voltage actually is. Voltage is a difference in electrical potential between two points. In a parallel circuit, every branch connects between exactly the same two points — the same positive node and the same negative node. Since all branches connect between identical endpoints, they all see the same potential difference. There is no way for the voltage to be different across branches that share the same nodes.
This is completely different from a series circuit, where voltage divides across each component in proportion to its resistance. In a parallel circuit, every branch gets the full supply voltage regardless of its resistance. A 10 Ω branch and a 10 kΩ branch both connected in parallel across a 12 V supply will each have exactly 12 V across them — the 10 Ω branch will just carry far more current.
This rule has a very practical consequence: you can add or remove branches from a parallel circuit without changing the voltage seen by the remaining branches. Plug in a second lamp in your house — the first lamp's brightness does not change, because both lamps see the same line voltage. Switch off one branch of a transceiver — the other stages continue running at exactly the same supply voltage.
The rule also gives you a powerful diagnostic tool. If you measure different voltages across branches that should be in parallel, something is wrong. The most common cause is resistance in the wiring itself (especially in old or corroded connectors), which effectively puts a hidden series resistance in the line to one branch, creating a voltage divider. A poor solder joint, a bad switch contact, or a corroded connector can all cause this.
Rule 2: Total Resistance Is Less Than Any Branch
This rule surprises many beginners. Surely adding more resistors must increase resistance? In a series circuit, yes — adding resistors in series always increases total resistance. But in a parallel circuit, the opposite is true. Adding any additional branch always decreases total resistance.
Here is the physical reason. Resistance is the opposition to current flow. Adding a new branch provides an additional path for current. More paths means more current can flow for the same applied voltage. If more current flows for the same voltage, then by Ohm's Law (R = V / I), the effective resistance of the combination must be lower. You cannot add a path and make it harder for current to flow — you can only make it easier.
Think of it with conductance. Conductance is the reciprocal of resistance: G = 1 / R, measured in siemens (S). Conductance measures how easily current flows, rather than how much it is opposed. In a parallel circuit, conductances add directly — just like series resistances add directly. Every additional branch contributes its conductance to the total. The total conductance of a parallel circuit is:
where Gn = 1 / Rn
And since total resistance is the reciprocal of total conductance:
Notice that adding any term to the right side of that equation makes the sum larger, which makes 1/Rtotal larger, which makes Rtotal smaller. No matter how large the branch resistance is, adding it reduces total resistance — even if only by a tiny amount. Two 1 kΩ resistors in parallel give 500 Ω. Three give 333 Ω. Add a 1 MΩ resistor to a 10 Ω resistor in parallel and the result is still slightly below 10 Ω (about 9.9999 Ω) — the million-ohm branch barely contributes but it does contribute.
This is directly opposite to series circuits, where resistance always increases. Memorise which rule applies to which circuit type, and you will never confuse them:
| Property | Series Circuit | Parallel Circuit |
|---|---|---|
| Voltage | Divides — different across each component | Same across every branch |
| Current | Same through every component | Divides — different in each branch |
| Resistance | Adds directly — always increases | Reciprocal formula — always decreases |
| Adding a component | Increases total resistance | Decreases total resistance |
Rule 3: Total Current Is the Sum of Branch Currents
Current entering the junction at the top of a parallel circuit must equal current leaving the junction at the bottom. This is a statement of charge conservation — electrons do not accumulate at a junction or disappear from one. Every electron that arrives must leave. The total current from the supply splits between the available branches and all recombines at the far junction.
This is actually a statement of Kirchhoff's Current Law (KCL), which you will study in depth later in this module. For now, the key point is that each branch current is completely independent of the others. The current in the R1 branch does not depend on what R2 or R3 are doing — it depends only on the voltage across R1 (which is the supply voltage) and the value of R1 itself. Each branch obeys Ohm's Law independently:
This independence is a crucial advantage in practice. In a series circuit, adding a component changes the current through every other component. In a parallel circuit, adding a new branch does not change the current through any existing branch — it only adds to the total current from the supply. Each branch operates as if it were alone across the supply.
This is why your home's lighting and power sockets are wired in parallel, not in series. When you switch on an additional light, the other lights do not dim — each branch is independent. In a transceiver, switching on the transmitter section (which draws large current) does not starve the receiver's voltage regulator — both are parallel loads across the supply rail.
Three Methods for Calculating Total Resistance
There are three methods, each suited to different situations. Knowing all three saves time and reduces errors.
Method 1: The Reciprocal Formula (works for any number of resistors)
This is the general method. You calculate the reciprocal of each resistance, add them up, then take the reciprocal of the result. It is always correct but involves more steps.
Step 1: 1/Rtotal = 1/100 + 1/220 + 1/470
Step 2: 1/Rtotal = 0.01000 + 0.00455 + 0.00213 = 0.01668
Step 3: Rtotal = 1 / 0.01668 = 59.95 Ω
Check: 59.95 Ω is less than 100 Ω (the smallest branch). ✓
Method 2: Product-Over-Sum (two resistors only)
For exactly two resistors, there is a much faster shortcut that avoids the reciprocal steps entirely:
This formula is used so frequently — every time you combine two branches in a step-by-step reduction of a complex circuit — that you should memorise it. It only works for two resistors at a time, but you can apply it repeatedly to reduce a multi-branch parallel combination down to a single equivalent resistance.
Rtotal = (680 × 1200) / (680 + 1200)
Rtotal = 816,000 / 1880
Rtotal = 434 Ω
Check: 434 Ω is less than 680 Ω (the smallest branch). ✓
To reduce three resistors using this method: combine any two first using product-over-sum to get a single equivalent, then combine that equivalent with the third resistor using product-over-sum again.
Method 3: Equal Resistors (R divided by N)
If all resistors in the parallel combination are identical, the total resistance is simply one resistor's value divided by the number of resistors:
Rtotal = 1000 / 4 = 250 Ω
This is also useful for building specific resistance values. Want 50 Ω but only have 100 Ω resistors? Put two in parallel.
This method also explains why a dummy load (a precision 50 Ω resistive load used for testing transmitters) can be built from multiple higher-value resistors in parallel. Ten 500 Ω resistors in parallel give exactly 50 Ω. This allows you to spread the power dissipation across many small resistors instead of needing one large 50 Ω high-power component.
Worked Example: Step by Step
Let us work through a complete problem showing every step. A 9 V battery supplies three parallel branches: R1 = 330 Ω, R2 = 560 Ω, R3 = 1 kΩ (1000 Ω). Find the total resistance, each branch current, and the total current.
Step 1 — Find total resistance using the reciprocal formula:
1/Rtotal = 1/330 + 1/560 + 1/1000
1/Rtotal = 0.003030 + 0.001786 + 0.001000
1/Rtotal = 0.005816 S
Rtotal = 1 / 0.005816 = 171.9 Ω
Check: 171.9 Ω is less than 330 Ω. ✓
Step 2 — Find each branch current (all see 9 V):
IR1 = V / R1 = 9 / 330 = 0.02727 A = 27.3 mA
IR2 = V / R2 = 9 / 560 = 0.01607 A = 16.1 mA
IR3 = V / R3 = 9 / 1000 = 0.009000 A = 9.0 mA
Step 3 — Find total current by adding branch currents:
Itotal = 27.3 + 16.1 + 9.0 = 52.4 mA
Step 4 — Verify using Ohm's Law:
Itotal = V / Rtotal = 9 / 171.9 = 52.4 mA ✓
Observations:
R1 (lowest resistance) carries the most current: 27.3 mA out of 52.4 mA total = 52%.
R3 (highest resistance) carries the least: 9.0 mA = 17%.
The dominant branch (R1) determines most of the circuit's behaviour.
More Practice Examples
Two resistors are in parallel across a 12 V supply. R1 = 470 Ω. The total current drawn from the supply is 60 mA. What is R2?
Step 1: Find Rtotal from V and Itotal:
Rtotal = V / Itotal = 12 / 0.060 = 200 Ω
Step 2: Find IR1:
IR1 = 12 / 470 = 25.5 mA
Step 3: Find IR2:
IR2 = Itotal − IR1 = 60 − 25.5 = 34.5 mA
Step 4: Find R2:
R2 = V / IR2 = 12 / 0.0345 = 347.8 Ω ≈ 330 Ω (standard value)
A 50 Ω resistor is connected in parallel with a 47 kΩ resistor across a 5 V supply. What is the total resistance?
Using product-over-sum:
Rtotal = (50 × 47000) / (50 + 47000) = 2,350,000 / 47050 = 49.95 Ω
The 47 kΩ resistor barely affects the result. Its branch current is only 5 / 47000 = 0.106 mA, compared to 5 / 50 = 100 mA in the 50 Ω branch. In circuit analysis, when one branch resistance is more than 100 times smaller than another, you can often ignore the high-resistance branch entirely for a quick estimate.
This matters in practice: a multimeter's input resistance (typically 10 MΩ) connected across a 1 kΩ resistor barely changes the circuit. But a multimeter across a 5 MΩ resistor would halve the measured voltage — this is called meter loading and is why meter impedance matters.
You need a 33 Ω resistor but only have 68 Ω resistors in stock. How many in parallel give the closest value?
Two 68 Ω in parallel: 68 / 2 = 34 Ω (close, but 3% high)
Three 68 Ω in parallel: 68 / 3 = 22.7 Ω (too low)
Two in parallel gives 34 Ω — acceptable for most applications given resistor tolerances. Alternatively, try the product-over-sum approach with other available values: 47 Ω ∥ 100 Ω = (47 × 100)/(47 + 100) = 4700/147 = 31.97 Ω ≈ 32 Ω — even closer to 33 Ω.
This kind of improvisation is a real practical skill. You will encounter it when repairing old equipment where exact values are unavailable.
Power in Parallel Circuits
Each branch dissipates power independently. Since every branch sees the full supply voltage, you can calculate power in each branch using P = V² / R — the same voltage applies to every branch, making this formula particularly convenient.
The total power delivered by the supply equals the sum of all branch powers, which also equals Vsupply × Itotal. You can verify your calculations by summing branch powers and checking against the total.
PR1 = 9² / 330 = 81 / 330 = 245 mW
PR2 = 9² / 560 = 81 / 560 = 145 mW
PR3 = 9² / 1000 = 81 / 1000 = 81 mW
Total power = 245 + 145 + 81 = 471 mW
Verify: Ptotal = V × Itotal = 9 × 0.0524 = 472 mW ✓ (rounding difference)
Notice: R1 dissipates the most power even though each branch has the same voltage — it dissipates the most because it has the lowest resistance and carries the most current.
This has a practical consequence for component selection. In a parallel circuit, the lowest-resistance branch carries the most current and dissipates the most power. Always check the power rating of your lowest-resistance component. If you are paralleling power resistors to increase power handling, the parallel combination helps — each component now handles only a fraction of the total power.
For example, a dummy load for a 100 W transmitter built from ten 500 Ω resistors in parallel: each resistor dissipates 100 / 10 = 10 W. So you need ten 10 W resistors rather than one 100 W resistor. High-power resistors at 10 W are far more readily available and cheaper than 100 W units.
What Happens When Things Go Wrong
Understanding failure modes is just as important as understanding normal operation. Parallel circuits fail in two principal ways, and the outcomes are very different from each other — and very different from what happens in a series circuit.
Open-Circuit Failure (One Branch Breaks)
An open-circuit failure means one branch stops conducting — perhaps a resistor burns out, a wire breaks, or a switch opens. In a parallel circuit, the other branches are completely unaffected. They still connect between the same two nodes with the same supply voltage. They continue to operate exactly as before.
The only changes are: the total resistance increases (one fewer parallel path), and the total current drawn from the supply decreases (the broken branch no longer draws current). If you were to measure voltage across the open branch, you would read the full supply voltage — because the open branch still connects between the two supply nodes, just with no current flowing through it.
Before fault: 12 V supply, R1 = 100 Ω, R2 = 220 Ω, R3 = 470 Ω
Rtotal = 1 / (1/100 + 1/220 + 1/470) = 59.95 Ω
Itotal = 12 / 59.95 = 200 mA
After fault: R2 goes open-circuit
Rtotal = 1 / (1/100 + 1/470) = 82.5 Ω (only R1 and R3 remain)
Itotal = 12 / 82.5 = 145 mA
R1 still draws 120 mA; R3 still draws 25.5 mA. Neither is affected by R2's failure.
Voltage across the open R2: 12 V (full supply — the open gap is still between the two nodes).
Symptom: total current drops; supply voltage unchanged; other branches operate normally.
This is the key advantage of parallel wiring for loads that must continue operating even if one fails. House lighting is wired in parallel for exactly this reason — a burnt-out bulb does not take out the rest of the lights. Aircraft and spacecraft wire critical systems in parallel (often with redundancy) so that a single failure cannot disable a function.
Open-circuit fault in a parallel branch: R1 and R3 continue operating normally. Full supply voltage appears across the open branch, but no current flows through it.
View LargerShort-Circuit Failure (One Branch Collapses to Zero Resistance)
A short-circuit failure is the opposite — and far more dangerous. If one branch collapses to near-zero resistance (a failed capacitor, a shorted semiconductor, an insulation breakdown), it connects the two supply rails through essentially no resistance at all. This has catastrophic consequences.
By Ohm's Law, if resistance approaches zero and voltage is fixed, current approaches infinity. In practice, current will only be limited by the supply's internal resistance, the wiring resistance, and any protective device (fuse or circuit breaker). The supply voltage collapses because all of it appears across the near-zero resistance of the short, leaving nothing for the other branches.
Before fault: 12 V supply, R1 = 100 Ω, R2 = 220 Ω, R3 = 470 Ω (same circuit as before)
After fault: R2 goes short-circuit (0 Ω)
Rtotal = 1 / (1/100 + 1/0 + 1/470) — the 1/0 term is infinite, making Gtotal infinite
Rtotal = 0 Ω
Current from supply: limited only by supply internal resistance (could be hundreds of amperes)
Supply voltage: collapses toward 0 V (all voltage appears across short's near-zero resistance)
Voltage across R1 and R3: also collapses (they share the same nodes, which are now at nearly the same potential)
R1 and R3 stop operating.
Outcome: fuse blows, supply shuts down (if protected), or wiring overheats and potentially catches fire (if unprotected).
Short-circuit failures are the most dangerous electrical fault. Every power rail in electronics equipment must be protected by a fuse or circuit breaker rated appropriately. Fuses in parallel-fed circuits must be rated for the total expected current, not just the current of one branch. A single shorted branch can blow the supply fuse and disable the entire circuit — even though parallel circuits normally offer independent operation between branches.
In ham radio equipment, the most common short-circuit faults in parallel circuits are: failed electrolytic bypass capacitors that develop an internal short; shorted output transistors in audio amplifiers (which present a near-zero impedance load across the supply rail); and shorted rectifier diodes in power supplies. All of these will blow the main fuse if one is fitted.
Parallel Circuit Solver
Parallel Circuit Solver
Enter the supply voltage and up to four resistor values. Leave unused resistors blank. The calculator finds total resistance, total conductance, branch currents, total current, and power dissipated in each branch.
⚖ Experiment: Parallel Resistors on a Breadboard
This experiment directly confirms the three parallel rules: equal voltage across branches, additive currents, and total resistance less than the smallest branch. Predicting results before measuring is the key habit to build.
- 9 V battery with clip connector
- Three resistors: 470 Ω, 1 kΩ, 2.2 kΩ (±5% tolerance is fine)
- Breadboard and jumper wires
- Digital multimeter
- Predict first. Use the calculator above with V = 9 V, R1 = 470 Ω, R2 = 1000 Ω, R3 = 2200 Ω. Write down: predicted Rtotal, IR1, IR2, IR3, and Itotal. Keep these figures for comparison.
- Build the circuit. Connect one end of each resistor to the top power rail of the breadboard and the other end to the bottom rail. Connect the 9 V battery between the rails using the clip connector.
- Measure branch voltages. Set your multimeter to DC volts. Touch the probes across each resistor in turn. All three readings should be equal and approximately equal to the battery's terminal voltage (usually 8.5–9.1 V for a fresh 9 V battery). Record all three readings.
- Measure branch currents. To measure current in each branch, you must break that branch and insert the meter in series. Set meter to DC mA. Disconnect one end of R1 from the breadboard, connect one meter lead to the disconnected resistor leg and the other meter lead to the rail. Record IR1. Restore R1, then repeat for R2 and R3.
- Add the branch currents. Sum your three measured branch currents. This is your measured Itotal.
- Measure total current. Break the positive supply wire from the battery and insert the meter in series between the battery positive terminal and the top rail. Record total supply current. Compare to the sum you calculated in step 5.
- Calculate measured Rtotal. Divide your measured battery voltage by your measured total current: Rtotal = V / Itotal. Confirm this is less than 470 Ω.
- Simulate an open-circuit fault. Remove one resistor from the breadboard. Note that the other two branches continue to operate. Measure total current again — it should drop. Measure voltage across the empty branch — it reads the full supply voltage even though no resistor is there.
All three branch voltages are equal (within meter accuracy). The sum of branch currents equals total supply current. Total resistance (V/Itotal) is well below 470 Ω (the smallest branch). When you remove one resistor, the remaining branches are unaffected and full supply voltage appears across the open position. The 470 Ω branch carries approximately 4.7 times more current than the 2.2 kΩ branch — the ratio equals the inverse of the resistance ratio.
Parallel Circuits in Ham Radio
Power distribution rails
In a transceiver, every stage — receiver front end, IF amplifier, audio output, display, microprocessor — is connected in parallel across the DC supply rails (typically 13.8 V for a mobile rig). Adding or removing a stage does not change the supply voltage seen by the other stages, because all stages are independent branches across the same two rails. The power supply must deliver the sum of all branch currents. This is why transmit current draw is much higher than receive current draw — the power amplifier stage (a high-current parallel branch) is only active during transmission.
Bypass and decoupling capacitors
Bypass capacitors are connected in parallel with the DC supply at critical points on a circuit board. Their function is to provide a low-impedance AC path in parallel with the supply's wiring impedance. At RF or audio frequencies, the capacitor's reactance is very low — it behaves like a near-short circuit for AC. The parallel combination of the wiring inductance and the capacitor forms a low-impedance local reservoir of charge. When a transistor suddenly demands a burst of current during a signal peak, the capacitor supplies it instantly without the supply rail dropping. You will encounter decoupling capacitors (100 nF ceramic in parallel with 10 µF electrolytic) on virtually every amplifier stage in every piece of modern radio equipment.
Antenna matching networks
Shunt elements in L-match, pi-match, and T-match antenna tuner networks are components in parallel with the signal path. A shunt capacitor in an L-network, for example, presents a parallel impedance that transforms the load impedance seen at the input. The quarter-wave matching stub used on VHF/UHF antennas is a parallel resonant circuit. Analysing these networks requires confident handling of parallel impedances. You will revisit this topic extensively in the AC Circuit Theory module and the Filters and Impedance Matching module.
Dummy load construction
A high-power 50 Ω dummy load for a 100 W transmitter is typically built from multiple resistors in parallel. Ten 500 Ω, 10 W resistors in parallel give 50 Ω total resistance and handle 100 W total (10 W each). The parallel resistance formula Rtotal = R / N applies directly: 500 / 10 = 50 Ω. If you wanted a dummy load rated for 200 W with the same 500 Ω resistors, you would use 20 of them: 500 / 20 = 25 Ω — too low for a standard 50 Ω dummy load but useful for other impedances.
Speaker loads
Many ham transceivers provide a speaker output and a headphone socket wired in parallel. Two 8 Ω speakers in parallel present a 4 Ω load. If the audio amplifier's minimum load impedance is 8 Ω, connecting both simultaneously will push it below specification and may cause overheating or distortion. Knowing the parallel resistance formula lets you calculate the combined load immediately: (8 × 8) / (8 + 8) = 64 / 16 = 4 Ω — well within danger territory for a minimum-8 Ω amplifier.
Parallel capacitors in filter circuits
Capacitors wired in parallel have their values added directly (Ctotal = C1 + C2 + ... + Cn). This is the opposite of resistors — capacitor paralleling rules mirror series resistor rules, and series capacitor rules mirror parallel resistor rules. This sometimes causes confusion. The key is to think in terms of the fundamental quantities: capacitors in parallel present more plate area to charge, so total capacitance increases. Variable capacitors in tunable circuits are sometimes replaced with a fixed capacitor in parallel with a smaller trimmer — this allows fine adjustment of total capacitance within a small range around the fixed value.
Frequently Asked Questions
Why does parallel resistance always come out less than the smallest branch?
Every additional branch provides another path for current to flow. More paths means more total current flows for the same applied voltage. If more current flows for the same voltage, then R = V/I gives a lower resistance. You cannot add a path and make it harder for current to flow overall — you can only make it easier. Even a very high-resistance branch (say 1 MΩ added to a 10 Ω circuit) contributes a tiny additional current that reduces Rtotal marginally below 10 Ω. The total resistance can approach zero but can never reach zero with finite branch resistances, and it can never exceed the smallest branch.
Why does the formula for parallel capacitors look different from parallel resistors?
Capacitance and resistance behave inversely in many ways. For capacitors in parallel, Ctotal = C1 + C2 + ... — they add directly. For capacitors in series, you use the reciprocal formula that parallels the resistor parallel formula. This is because capacitance is proportional to plate area, and plates add when you connect capacitors in parallel. The pattern to remember: for resistors and capacitors, series and parallel follow opposite rules. Inductors behave identically to resistors in this respect.
What happens if one branch in a parallel circuit fails open?
Only that branch stops conducting. All remaining branches continue to operate normally — they still connect across the same supply voltage and draw the same current as before. Total resistance increases (one fewer path), total current falls, but each surviving branch is unaffected. The failed branch, though broken internally, still connects between the supply nodes, so you measure the full supply voltage across it. The open-circuit failure mode is the safe failure mode for parallel circuits. It is why residential wiring uses parallel branch circuits — one failed lamp does not cut power to the others.
What happens if one branch in a parallel circuit fails short?
A shorted branch connects the two supply rails through near-zero resistance. By Ohm's Law, this attempts to pull near-infinite current from the supply. In practice, current is limited only by the supply's internal resistance and wiring resistance — but can still reach hundreds of amps before a fuse opens. The supply voltage collapses to near zero because all the voltage appears across the shorted branch's tiny resistance, leaving none for the other branches. The surviving branches effectively lose their supply and stop working. A blown main fuse in a ham radio is often traced to a single shorted bypass capacitor or semiconductor creating a parallel short across the supply rail.
Can I measure the resistance of a parallel network without disconnecting it from the circuit?
Generally, no — not accurately. An ohmmeter works by injecting a small test current and measuring the resulting voltage. In a circuit with power applied, other current sources will interfere. Even with power removed, other parallel paths through the rest of the circuit will carry some of the ohmmeter's test current, giving a reading lower than the true value of the component you intend to measure. The reliable method is to disconnect at least one terminal of the component or network before measuring. If physical disconnection is impossible, measure voltage and current with the circuit operating and calculate R = V / I.
Is the product-over-sum formula just a shortcut, or is it more accurate than the reciprocal formula?
Both formulas are exactly equivalent mathematically — neither is more accurate. The product-over-sum formula (R1 × R2) / (R1 + R2) is derived algebraically from the reciprocal formula and gives identical results. The product-over-sum is simply faster when you have exactly two resistors, because it avoids the step of computing and summing reciprocals. For three or more resistors, the reciprocal formula is generally easier than repeatedly applying product-over-sum, though both approaches give the same answer.
Test Your Knowledge
Answer the questions below to check your understanding. Every answer can be found in the lesson above.