Resistors in Series

What you’ll learn

Combine resistances along a single current path.

បូករេស៊ីស្តង់ក្នុងផ្លូវចរន្តតែមួយ។

Meaning and use

In a series path, the same steady current passes through each resistor. Voltage drops add to the applied total for the stated ideal circuit. Equivalent resistance is the sum of individual resistances. This rule does not apply if a junction creates parallel paths.

ក្នុងស៊េរី ចរន្តថេរដូចគ្នាឆ្លងកាត់រេស៊ីស្តង់នីមួយៗ។ ផលសងប៉ូតង់ស្យែលបូកគ្នា ហើយរេស៊ីស្តង់សមមូលជាផលបូក។

Formulas

Equivalent series resistance

\[R_s=R_1+R_2+\cdots\]

Current and voltage

\[I_1=I_2=I,\qquad V_{\mathrm{total}}=V_1+V_2+\cdots\]

Symbols and units

Symbol	Meaning	Unit or restriction
Rs, R₁, R₂	Equivalent and individual resistances	Ω
I	Steady current in the single path	A
Vtotal, V₁, V₂	Applied voltage and component drops	V

Conditions and limits

• Assume ideal connecting wires and account for all series resistances.
• Identify connection topology, not merely how components are drawn on a page.
• ប្រភពអគ្គិសនីក្នុងឧទាហរណ៍សន្មតថាឧត្តមគតិ។

Worked examples

Equivalent resistance and current

2 Ω and 4 Ω are in series across ideal 12 V.

1. Total resistance \(R_s=2+4=6\) Ω.
2. Current \(I=12/6=2\) A.

6 Ω; 2 A

រេស៊ីស្តង់បូកគ្នា រួចប្រើច្បាប់អូមលើសមមូល។

Voltage division

The same 2 A flows through 2 Ω and 4 Ω.

1. First drop \(V_1=2\times2=4\) V.
2. Second drop \(V_2=2\times4=8\) V.
3. Check \(4+8=12\) V.

4 V and 8 V

ចរន្តស្មើគ្នា តែតង់ស្យុងអាចខុសគ្នាតាម R។

Common mistakes

Instead of	Use	Why
Current divides at each resistor in a series path.	The same steady current passes through each.	There is only one path.
Every series resistor has equal voltage.	Voltage drops depend on resistance for a common current.	Equal voltage requires equal resistances in this model.

Practice

Solve these before revealing the worked solutions. Include units where relevant.

សាកល្បងដោះស្រាយមុនបើកមើលចម្លើយ។ សូមបង្ហាញជំហាន និងឯកតាដែលពាក់ព័ន្ធ។

1. Find equivalent series resistance of 3, 5, and 2 Ω.
2. A 10 Ω series combination is across 20 V. Find current.
3. At 2 A, find the drop across a 5 Ω resistor.

Answers and steps

Show solutions / បង្ហាញដំណោះស្រាយ

1. 10 Ω

   Add all three.

2. 2 A

   20/10 = 2.

3. 10 V

   V = IR = 2 × 5.

Quick recap

• Series means one current path.
• Add series resistances.
• Check voltage drops against the source total.