🔗 Series/Parallel + Voltage Divider

The previous chapter gave us one resistor. Real circuits rarely stop there. They put resistors in chains and branches, then ask two bookkeeping questions:

1. Where can charge go?
2. How much energy does it give up along each path?

Those questions lead to series, parallel and the voltage divider.

By the end, you can

1. Explain why series resistors share one current but divide voltage
2. Explain why parallel resistors share one voltage but divide current
3. Use the voltage-divider equation to choose a safe ADC input voltage
4. Recognize loading: when the thing you connect changes the divider you calculated

Series: one path, one current

Put resistors in series and there is only one path for charge. The same current must pass through each part, because there is nowhere else for it to go. The resistances add:

$$R_\text{total} = R_1 + R_2 + R_3 + \cdots$$

The voltage does not vanish all at once. It drops piece by piece. The larger resistor gets the larger share of the voltage because the same current through a larger resistance requires a larger drop.

That is the voltage divider hiding in plain sight.

Parallel: one voltage, many currents

Put resistors in parallel and both ends of each resistor touch the same two nodes. That means each branch sees the same voltage. The current splits according to branch resistance: low resistance gets more current, high resistance gets less.

For parallel resistors, conductance adds:

$$\frac{1}{R_\text{total}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \cdots$$

For two resistors:

$$R_\text{total} = \frac{R_1R_2}{R_1 + R_2}$$

The total is always less than the smallest branch, because you have added another way for current to flow.

The divider: a deliberate fraction

Two resistors in series create a chosen fraction of an input voltage:

$$V_\text{out} = V_\text{in}\frac{R_2}{R_1 + R_2}$$

Here $R_1$ is the upper resistor from input to output, and $R_2$ is the lower resistor from output to ground. The output is the fraction of the total resistance that sits below the tap.

This is everywhere:

• Battery monitoring: scale a 4.2 V LiPo down into a 3.3 V ADC range.
• Level shifting: reduce a 5 V signal for a 3.3 V input, when speed and loading allow it.
• Sensor front ends: turn resistance changes into voltage changes.
• Feedback networks: tell a regulator what output voltage it has made.

Drive it

Use the calculator to move between series, parallel and divider thinking. Do not only enter round numbers. Try ratios.

Try these moves:

1. Set $R_1 = R_2$. The divider output is half the input.
2. Make $R_2$ much larger than $R_1$. The output rises toward the input.
3. Make two equal resistors parallel. The total becomes half of one resistor.

?A divider has R1 = 10 kΩ from 5 V to Vout and R2 = 10 kΩ from Vout to ground. What is Vout before loading?

• A 2.5 V
• B 5 V
• C 0 V
• D 10 V

The divider is only correct if nobody steals current

A divider equation assumes the output node is observed gently. Connect a low-resistance load and the load sits in parallel with $R_2$. Now the lower leg is no longer just $R_2$, and the output droops.

That is called loading. It is not a corner case. A multimeter has input resistance. An ADC input has capacitance and sampling current. A sensor input has leakage. A long wire has capacitance. If the divider impedance is too high, the measurement circuit becomes part of the divider.

The fix is not always "use smaller resistors." Smaller resistors waste more current. The design trade is between measurement stiffness and power:

• Low divider resistance: less loading error, more wasted battery current.
• High divider resistance: less wasted current, more sensitivity to leakage, ADC sampling and noise.

?Why can a voltage divider give the wrong output after you connect it to another circuit?

• A The load can sit in parallel with the lower resistor and change the effective ratio
• B Ohm's law stops working when there are two resistors
• C Parallel resistance is always larger than either resistor
• D The output node stores voltage permanently

↘ The divider equation from one current

In an unloaded divider, the same current flows through both resistors:

$$I = \frac{V_\text{in}}{R_1 + R_2}$$

The output is the voltage across $R_2$:

$$V_\text{out} = IR_2$$

Substitute the first equation into the second:

$$V_\text{out} = V_\text{in}\frac{R_2}{R_1 + R_2}$$

No memorized triangle is needed. The formula is just "same current through a series chain" plus "voltage across the lower resistor."

Practice 1 core

A fully charged single-cell LiPo is 4.2 V. You want the ADC input below 3.3 V. Choose $R_1 = 10\ \text{k}\Omega$ and $R_2 = 27\ \text{k}\Omega$. What output voltage do you expect before loading?

↘ Show worked solution

$$V_\text{out} = 4.2\frac{27}{10 + 27} \approx 4.2 \times 0.73 \approx 3.07\ \text{V}$$

That is below 3.3 V, so the ratio is plausible. You still need to check ADC input requirements and leakage, especially if the divider values get much larger.

Practice 2 stretch

The divider above uses 10 kΩ and 27 kΩ. Estimate the divider current from a 4.2 V battery. Is it tiny enough to leave connected for a low-power device that sleeps most of the time?

↘ Show worked solution

The total resistance is 37 kΩ, so:

$$I = 4.2/37000 \approx 0.000114\ \text{A}$$

That is 114 microamps. For a low-power device trying to sleep at 10 microamps, the divider alone would dominate the sleep budget. Use larger resistors, switch the divider off between readings, or accept the battery-life cost deliberately.

Key takeaways

• ▸Series resistors share one current and divide voltage.
• ▸Parallel resistors share one voltage and divide current.
• ▸A voltage divider is a deliberate fraction: $V_\text{out}=V_\text{in}R_2/(R_1+R_2)$.
• ▸Loading changes the divider because the load becomes part of the resistor network.
• ▸Divider design is a trade between wasted current and measurement stiffness.