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Mixed series–parallel circuits: Class 10 practice

Reduce a circuit step by step: combine parallel groups with 1/R = 1/R₁ + 1/R₂, then add series resistances. Read the method, follow the three worked examples, then try the six questions.

What you should know first

  • Ohm's law
  • Series and parallel rules

The idea behind mixed series–parallel circuits

Reduce a circuit step by step: combine parallel groups with 1/R = 1/R₁ + 1/R₂, then add series resistances. Current from the cell is V/R_total; it splits in parallel branches in inverse ratio of resistance.

Step-by-step method

  1. Combine the parallel part first.
  2. Add series resistors.
  3. I = V/R_total; split the current in parallel branches.

Worked example 1 (Basic)

2 Ω and 2 Ω are in parallel, and this is in series with 1 Ω. Find the total resistance.

  1. Parallel: 2×2/(2+2) = 1 Ω
  2. Total = 1 + 1 = 2 Ω
  3. Answer: 2 Ω

Worked example 2 (Standard)

The circuit (4 Ω ∥ 4 Ω, in series with 9 Ω) is connected to 11 V. Find the current from the cell.

  1. R = 11 Ω
  2. I = 11/11 = 1 A
  3. Answer: 1 A

Worked example 3 (Challenge)

In that circuit (24 V), what current flows through the 3 Ω resistor?

  1. I = 4 A
  2. Voltage across parallel part = 8 V
  3. I(3 Ω) = 8/3 = 2.67 A
  4. Answer: 2.67 A

Common mistakes and misconceptions

  • Adding parallel resistors directly
  • Assuming equal currents in unequal branches

Practice questions

  1. (Basic) 6 Ω and 12 Ω are in parallel, and this is in series with 5 Ω. Find the total resistance.
  2. (Basic) 3 Ω and 6 Ω are in parallel, and this is in series with 2 Ω. Find the total resistance.
  3. (Standard) The circuit (3 Ω ∥ 6 Ω, in series with 10 Ω) is connected to 48 V. Find the current from the cell.
  4. (Standard) The circuit (12 Ω ∥ 6 Ω, in series with 2 Ω) is connected to 6 V. Find the current from the cell.
  5. (Challenge) In that circuit (42 V), what current flows through the 20 Ω resistor?
  6. (Challenge) In that circuit (10 V), what current flows through the 4 Ω resistor?

Answer key with full working

  • Q1: Parallel: 6×12/(6+12) = 4 Ω → Total = 4 + 5 = 9 Ω → Answer: 9 Ω
  • Q2: Parallel: 3×6/(3+6) = 2 Ω → Total = 2 + 2 = 4 Ω → Answer: 4 Ω
  • Q3: R = 12 Ω → I = 48/12 = 4 A → Answer: 4 A
  • Q4: R = 6 Ω → I = 6/6 = 1 A → Answer: 1 A
  • Q5: I = 3 A → Voltage across parallel part = 12 V → I(20 Ω) = 12/20 = 0.6 A → Answer: 0.6 A
  • Q6: I = 1 A → Voltage across parallel part = 2 V → I(4 Ω) = 2/4 = 0.5 A → Answer: 0.5 A

Syllabus note

This skill is listed under "Electricity" for Class 10 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.

Related practice and notes

Frequently asked questions

Which chapter is mixed series–parallel circuits in?

"Electricity", Class 10 Physics (NCERT/CBSE). Other boards may place it differently.

How were the answers checked?

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.

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