Two 1 kΩ resistors can make 2 kΩ or 500 Ω.
Connecting resistors in series lengthens the path, increasing total resistance and reducing source current. Connecting them in parallel adds a second path between the same two nodes, lowering equivalent resistance below the smallest branch.
Two components are in parallel only if both of their terminals connect to the exact same pair of nodes.
Three things to notice
- Parallel equivalent is always smaller. Adding any positive resistor in parallel offers another current path, so R_eq is strictly less than min(R1, R2).
- Open fault in series vs parallel. An open resistor in series stops all loop current (0 mA); an open branch in parallel reduces source current while the other branch keeps running.
- Never share one resistor across parallel LEDs carelessly. Slight Vf differences cause one LED to hog current; give each parallel LED branch its own resistor.
Detailed Worked Engineering Example & Step-by-Step Solution
Two 1 kΩ resistors divide 10 V into 5.00 V unloaded. Closing switch SW_L connects a 1 kΩ load in parallel with R2, collapsing the equivalent lower resistance to 500 Ω and pulling the output down to 3.33 V!
Vs = 10.0 V, R1 = 1000 Ω, R2 = 1000 Ω, Load R_L = 1000 Ω.
- 1. Unloaded (SW_L Open): V_out = Vs · R2 / (R1 + R2) = 10 × 1000 / 2000 = 5.00 V.
- 2. Loaded (SW_L Closed): R_eq = (R2 × R_L) / (R2 + R_L) = (1000 × 1000) / 2000 = 500 Ω.
- 3. New Output Voltage: V_out(loaded) = 10 × 500 / (1000 + 500) = 3.33 V (a 33.3% voltage droop!).
Conclusion: Rule of thumb: For < 10% droop, load resistance R_L must be at least 10× larger than R2.
Electronic Components Reference: How Each Part Works & Connects
An ideal DC source holds V(+) − V(−) = Vs regardless of load current. Real bench supplies have a current limit (CC mode) and small internal resistance.
How to connect: Connect nodeA (+) to your main supply rail (VCC) and nodeB (−) to the reference ground node (GND = 0 V). Never wire two different voltage sources directly in parallel across the same two nodes.
Worked Example: Example: A 9 V source connected across a 450 Ω load resistor drives I = 9 V / 450 Ω = 20 mA and delivers P = V · I = 9 V × 0.02 A = 180 mW.
Current through a resistor is strictly proportional to the voltage difference across its two terminals: I = (Va − Vb) / R. It is non-polarized.
How to connect: Place in series with loads (such as LEDs or Zener diodes) to limit current, or in pairs across a supply to form a voltage divider.
Worked Example: Example: To limit a 2.0 V LED to 10 mA from a 5 V supply: R = (5 − 2.0) / 0.01 A = 300 Ω (choose standard 330 Ω). Power: P = (0.01)² × 330 = 33 mW (safe for a 250 mW ¼ W resistor).
When wired as a rheostat (variable resistance between nodeA and nodeB), R_eff = R_total × (wiper / 100).
How to connect: Connect in series with a fixed minimum stopper resistor (e.g., 100 Ω) so turning the knob to 0% never shorts the supply.
Worked Example: Example: A 1 kΩ potentiometer set to α = 50% gives R_eff = 500 Ω. In series with a 220 Ω protection resistor, total resistance is 720 Ω.
An open switch drops the full available loop voltage across its open contacts while passing 0 mA. A closed switch passes loop current with negligible voltage drop (≈ 0 V).
How to connect: Place in series on the high-side (between VCC and load) to control power delivery to the circuit branch.
Worked Example: Example: In a 5 V LED circuit, opening SW1 makes I = 0 mA; measuring across SW1 with a voltmeter reads 5.00 V because the anode side sits at 5 V and the load side drops to 0 V.
Unlike a resistor, an LED has an exponential I–V curve. Once forward voltage reaches Vf, its dynamic resistance drops to a few ohms and it cannot limit its own current.
How to connect: Connect Anode (nodeA, triangle base) toward positive potential and Cathode (nodeB, bar) toward GND, ALWAYS in series with a current-limiting resistor.
Worked Example: Example: Red LED (Vf = 1.9 V) with R = 330 Ω on a 5.0 V supply: V_R = 5.0 − 1.9 = 3.1 V, so I = 3.1 V / 330 Ω = 9.39 mA (bright, safe operation below 20 mA max).
When V(Anode) − V(Cathode) ≥ 0.7 V, the depletion barrier collapses and current flows. When reverse-biased, only negligible leakage current flows.
How to connect: Place in series with the positive supply input for reverse-polarity protection, or across inductive loads (cathode to +V) as a flyback protection diode.
Worked Example: Example: A 5.0 V supply passes through a series 1N4007 protection diode (Vf = 0.7 V) to a 430 Ω resistor: downstream node voltage is 5.0 − 0.7 = 4.3 V, and current is 4.3 / 430 = 10.0 mA.
Common misconceptions & corrections
“Adding a resistor always reduces total circuit current.” — Adding a resistor in series reduces current, but adding a resistor in parallel increases total source current.
“Components drawn parallel on paper are electrically in parallel.” — Only node connectivity matters: both ends must tie to the same two electrical nodes.
Learning outcomes, prerequisites & transfer question
- Calculate equivalent resistance for series (R_eq = R1 + R2) and parallel (R_eq = 1 / (1/R1 + 1/R2)) networks.
- Apply KCL at nodes and KVL around loops to verify branch currents and node voltages.
- Isolate an open branch in a parallel network using total current and node voltage evidence.
Transfer check: Two 1 kΩ resistors are wired in parallel across 5.0 V, so expected source current is 10.0 mA. If you measure only 5.0 mA from the source while supply voltage stays 5.0 V, what fault occurred?
One of the two parallel 1 kΩ branches is open (disconnected jumper or lifted resistor leg), leaving only a single 1 kΩ branch drawing 5.0 mA.