Series Resistor Calculator

Total series resistance, current and the voltage drop across each resistor, step by step

Find the current and voltage drops for 470, 1000 and 2200 ohm in series on 9 V

Size the series resistor for an LED with Vf = 2.1 V at 20 mA on a 12 V supply

Compare two 100 ohm resistors in series with the same two in parallel on 10 V

Find the voltage drop across a 220 ohm resistor carrying 18 mA

Adding Resistors in Series

Series components form one unbroken path, so the same current passes through every one of them and the resistances simply add:

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

Symbols and units:

$R_{\text{total}}$ — total series resistance, ohms (Ω)
$I$ — the single current shared by every element, amperes (A)
$V_n$ — the drop across element $n$ , volts (V)

The current follows from Ohm's law applied to the whole string, and each drop from Ohm's law applied to one element:

$I = \frac{V_{\text{supply}}}{R_{\text{total}}} \qquad V_n = I R_n$

Kirchhoff's voltage law guarantees the drops add back to the supply voltage, which is the check you should run every time.

The operating assumption: DC or steady-state resistive AC, ideal wires, and a source with no significant internal resistance. Reactive parts need impedance instead.

The mistake people make: adding a series total, then using the supply voltage again for a single resistor. Only $IR_n$ belongs to element $n$ .

Voltage Drop and the Divider Shortcut

You rarely need the current at all. Because the same $I$ flows everywhere, the supply voltage divides in direct proportion to resistance:

$V_n = V_{\text{supply}} \times \frac{R_n}{R_{\text{total}}}$

The largest resistor takes the largest share. Two equal resistors split the supply exactly in half.

This is also how a series resistor protects an LED. An LED is not ohmic: it holds a roughly fixed forward voltage $V_f$ and the resistor absorbs the rest, so the resistor is sized from the difference:

$R = \frac{V_{\text{supply}} - V_f}{I_f}$

with $I_f$ the wanted forward current in amperes.

Contrast this with parallel, where every branch sees the whole supply voltage and the currents add instead. Same two resistors, opposite behaviour: $2 \times 100\ \Omega$ gives $200\ \Omega$ in series but $50\ \Omega$ in parallel.

The mistake people make: sizing the LED resistor from the supply voltage rather than from $V_{\text{supply}} - V_f$ , which lets far too much current through.

Common Mistakes to Avoid

Adding series resistors as reciprocals — reciprocals belong to parallel. In series you just add.
Using the supply voltage in $V = IR$ for one element — use that element's own current and resistance.
Mixing prefixes — $2.2$ kΩ is $2200$ Ω; adding $470 + 1 + 2.2$ is meaningless.
Forgetting to check the sum — the individual drops must add back to the supply voltage. If they do not, the arithmetic is wrong.
Rounding the current too early — a current rounded to two figures propagates into every drop. Carry extra digits and round only the final answers.
Ignoring the power in each resistor — $P_n = I^2 R_n$ , and the biggest resistor in a series string dissipates the most.
Assuming an LED behaves like a resistor — its forward voltage is nearly constant, so it needs the difference formula, not a divider.

示例题目

Step 1:

R_{\text{total}} = 470\ \Omega + 1000\ \Omega + 2200\ \Omega = 3670\ \Omega

Step 2:

I = V/R_{\text{total}} = 9.0\ \text{V} \div 3670\ \Omega = 2.452 \times 10^{-3}\ \text{A} = 2.45\ \text{mA}

Step 3:

V_{470} = (2.452 \times 10^{-3}\ \text{A})(470\ \Omega) = 1.15\ \text{V}

Step 4:

V_{1000} = (2.452 \times 10^{-3}\ \text{A})(1000\ \Omega) = 2.45\ \text{V}

Step 5:

V_{2200} = (2.452 \times 10^{-3}\ \text{A})(2200\ \Omega) = 5.39\ \text{V}

Step 6: KVL check:

1.15 + 2.45 + 5.39 = 8.99 \approx 9.0\ \text{V}

Answer:

R_{\text{total}} = 3670

Ω,

I \approx 2.45

mA, drops

1.15

2.45

V and

5.39

Step 1: Voltage the resistor must absorb:

12\ \text{V} - 2.1\ \text{V} = 9.9\ \text{V}

Step 2:

R = 9.9\ \text{V} \div 0.020\ \text{A} = 495\ \Omega

Step 3: Nearest standard value above:

510\ \Omega

, chosen upward so the current cannot exceed

20

Step 4: Actual current:

I = 9.9\ \text{V} \div 510\ \Omega = 0.0194\ \text{A} = 19.4\ \text{mA}

Step 5:

P = I^2R = (0.0194\ \text{A})^2(510\ \Omega) = (3.764 \times 10^{-4}\ \text{A}^2)(510\ \Omega) = 0.192\ \text{W}

Step 6: A

0.25

W part is marginal at

77\%

of rating; a

0.5

W part runs comfortably

Answer:

R = 510

Ω standard value,

I \approx 19.4

mA,

P \approx 0.192

W — use a

0.5

W resistor

Step 1: Series:

R = 100\ \Omega + 100\ \Omega = 200\ \Omega

, so

I = 10\ \text{V} \div 200\ \Omega = 0.050\ \text{A}

Step 2: Each series resistor drops

V = (0.050\ \text{A})(100\ \Omega) = 5.0\ \text{V}

Step 3: Series total power:

P = VI = (10\ \text{V})(0.050\ \text{A}) = 0.50\ \text{W}

Step 4: Parallel:

R = (100)(100)/(100+100) = 50\ \Omega

, so each resistor sees the full

10

Step 5: Each parallel branch carries

10\ \text{V} \div 100\ \Omega = 0.10\ \text{A}

, total

0.20\ \text{A}

Step 6: Parallel total power:

P = (10\ \text{V})(0.20\ \text{A}) = 2.0\ \text{W}

, four times the series case

Answer: Series:

200

Ω,

0.050

5.0

V each,

0.50

W. Parallel:

50

Ω,

0.20

10

V each,

2.0

常见问题

Add the values directly: R_total = R1 + R2 + R3. Series resistance is always larger than the largest single resistor, which is the quick sanity check. Convert every value to ohms first so kilohms and ohms are not mixed.

Use V = IR with the current through that resistor, or take the shortcut V_n = V_supply x R_n / R_total. The drops across a series string always add back to the supply voltage.

Yes. A series path has nowhere for charge to go, so exactly the same current flows through every element. What differs is the voltage each one drops, in proportion to its resistance.

Subtract the LED forward voltage from the supply voltage, then divide by the wanted forward current in amperes: R = (V_supply - Vf)/If. Round up to the next standard value so the current stays at or below target, and check the resistor's power with P = I squared R.

Series Resistor Calculator

Total series resistance, current and the voltage drop across each resistor, step by step

Adding Resistors in Series

Voltage Drop and the Divider Shortcut

Common Mistakes to Avoid

示例题目

常见问题

How do I add resistors in series?

How do I calculate the voltage drop over a resistor?

Is the current the same through resistors in series?

How do I choose a series resistor for an LED?

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