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Series and Parallel Resistors, Dividers and the Loading Effect

7 min read · bitGateLab Team

Many circuits can be simplified before you solve them: several resistors are replaced by a single equivalent one, and the problem becomes a single-mesh circuit. That same idea gives rise to two of the most widely used circuits in electronics, the voltage divider and the current divider. Here you'll learn to reduce series and parallel combinations, to design dividers and to recognize the loading effect, the reason a correctly calculated divider sometimes delivers a different voltage from the one you expected.

Resistors in series

Two or more resistors are in series when the same current flows through all of them, with no node in between to divert part of it. The equivalent resistance is the sum:

Req = R1 + R2 + … + Rn

The series equivalent is always larger than the largest of the resistors. For example, 1 kΩ + 2.2 kΩ + 4.7 kΩ = 7.9 kΩ.

Resistors in parallel

Resistors are in parallel when they are all connected between the same pair of nodes and therefore have the same voltage. Here it is the conductances (G = 1/R, in siemens) that add:

1/Req = 1/R1 + 1/R2 + … + 1/Rn

For just two resistors there is a shortcut: Req = (R1 · R2) / (R1 + R2), "product over sum". The parallel equivalent is always smaller than the smallest of the resistors. Two equal resistors in parallel give half the value: 1 kΩ ∥ 1 kΩ = 500 Ω.

Example 1. Calculate 1 kΩ ∥ 2.2 kΩ ∥ 4.7 kΩ.

  1. Conductances in mS: 1/1 = 1, 1/2.2 = 0.4545, 1/4.7 = 0.2128.
  2. Sum: 1 + 0.4545 + 0.2128 = 1.6673 mS.
  3. Req = 1 / 1.6673 mS = 0.5998 kΩ ≈ 600 Ω, less than 1 kΩ, as it should be.

Mixed circuits are reduced piece by piece, from the inside out: first the groups that are clearly in series or in parallel, then the result with whatever comes next, until you are left with a single resistance.

Voltage divider

Two resistors in series connected to a source split the voltage in proportion to their values. The voltage across R2 (the lower one, connected to ground) is:

Vout = Vin · R2 / (R1 + R2)

It is used to obtain a reference voltage, to scale a large signal down to the input of a converter, or to read a resistive sensor such as a photoresistor or a thermistor.

+ − Vin R1 R2 RL Vout 0 V
Voltage divider R1–R2. If you connect a load RL to the output, it ends up in parallel with R2.

Example 2. With Vin = 12 V, R1 = 10 kΩ and R2 = 4.7 kΩ, and no load, what is Vout?

  1. Vout = 12 × 4.7 / (10 + 4.7) = 12 × 4.7 / 14.7.
  2. Vout = 56.4 / 14.7 = 3.84 V.
  3. Divider current: 12 V / 14.7 kΩ = 0.816 mA. Check: 0.816 mA × 4.7 kΩ = 3.84 V.

To design a divider, choose R2 and solve for the other resistor: R1 = R2 · (Vin/Vout − 1).

Current divider

When a current reaches two resistors in parallel, it splits, and most of it goes through the one with the lower resistance. For two branches:

I1 = Itotal · R2 / (R1 + R2)        I2 = Itotal · R1 / (R1 + R2)

Notice that the numerator holds the resistance of the other branch. With more than two branches it is easier to use conductances: Ik = Itotal · Gk / Gtotal.

Example 3. A current of 10 mA enters the parallel combination of R1 = 1 kΩ and R2 = 4 kΩ. How much flows through each one?

  1. I1 = 10 × 4 / (1 + 4) = 8 mA.
  2. I2 = 10 × 1 / 5 = 2 mA.
  3. Check: both have the same voltage, 8 mA × 1 kΩ = 8 V and 2 mA × 4 kΩ = 8 V. And 8 + 2 = 10 mA.

Loading effect

The voltage divider formula assumes that no current is drawn from the output. As soon as you connect something (another stage, a sensor, a meter), that load RL sits in parallel with R2, the effective lower resistance drops, and so does the output voltage. This is called the loading effect.

Example 4. A 10 V divider with R1 = R2 = 10 kΩ delivers 5 V with no load. What happens if you connect RL = 10 kΩ? And RL = 1 MΩ?

  1. With RL = 10 kΩ: R2 ∥ RL = 5 kΩ. Vout = 10 × 5 / (10 + 5) = 3.33 V. A third of the voltage is gone!
  2. With RL = 1 MΩ: R2 ∥ RL = 9.901 kΩ. Vout = 10 × 9.901 / 19.901 = 4.975 V, only 0.5 % low.

Key point: the divider behaves like a source with an internal resistance equal to R1 ∥ R2 (5 kΩ in the example). As a rule of thumb, the load should be at least 10 times larger than that resistance to keep the error below about 10 %; for errors under 1 %, about 100 times larger. The multimeter is a load too: if you measure a divider made of 1 MΩ and 1 MΩ with a 10 MΩ meter, you will read 4.76 V instead of 5 V.

Common mistakes

Try it in the simulator

The loading effect is much easier to understand when you watch it change in real time. Build the divider from Example 4 and try different loads.

  1. Place a 10 V DC source and two 10 kΩ resistors in series across its terminals.
  2. Connect the multimeter between the midpoint and ground and observe the 5 V.
  3. Connect a 10 kΩ load resistor in parallel with the lower one and watch the voltage drop to 3.33 V.
  4. Change the load to 100 kΩ and then to 1 MΩ; note down the voltage in each case.
  5. Place two resistors in parallel and measure the current in each branch to verify the current divider.

Exercises

  1. A 100 Ω resistor is in series with the parallel combination of 200 Ω and 300 Ω. Calculate the equivalent resistance.

    Show answer

    200 ∥ 300 = 60 000 / 500 = 120 Ω. Req = 100 + 120 = 220 Ω.

  2. You need 3.3 V from 9 V using a divider. If R2 = 3.3 kΩ, what must R1 be?

    Show answer

    R1 = 3.3 × (9/3.3 − 1) = 3.3 × 1.7273 = 5.7 kΩ. Check: 9 × 3.3 / (5.7 + 3.3) = 9 × 3.3 / 9 = 3.3 V.

  3. A current of 30 mA enters three resistors in parallel of 2 kΩ, 3 kΩ and 6 kΩ. How much flows through each one?

    Show answer

    Conductances: 0.5 + 0.3333 + 0.1667 = 1 mS, so Req = 1 kΩ and the common voltage is 30 mA × 1 kΩ = 30 V. Currents: 30/2 = 15 mA, 30/3 = 10 mA and 30/6 = 5 mA. Sum: 30 mA.

  4. A 12 V divider with R1 = R2 = 2 kΩ feeds a 2 kΩ load. What is the output voltage with and without the load?

    Show answer

    Without the load: 6 V. With the load: R2 ∥ RL = 1 kΩ, Vout = 12 × 1 / (2 + 1) = 4 V.

If the circuit cannot be reduced with series and parallel combinations, move on to nodal and mesh analysis.