RG's E-Learning Centre · Book a free demo class · WhatsApp +91-9379416374

EMF, internal resistance and terminal voltage: Class 12 practice

A cell of emf E and internal resistance r drives I = E/(R + r) through an external resistor R. Read the method, follow the three worked examples, then try the six questions.

What you should know first

  • Ohm's law

The idea behind emf, internal resistance and terminal voltage

A cell of emf E and internal resistance r drives I = E/(R + r) through an external resistor R. Terminal voltage V = E − Ir. Cells in series add emfs and internal resistances.

Step-by-step method

  1. Total resistance = R + r.
  2. I = E/(R + r).
  3. V = E − Ir = IR.

Worked example 1 (Basic)

A cell of emf 9 V and internal resistance 2 Ω drives current through 1 Ω. Find the current.

  1. I = 9/(1 + 2)
  2. = 3 A
  3. Answer: 3 A

Worked example 2 (Standard)

For the same cell (5 V, 1.5 Ω) and 12 Ω load, find the terminal voltage.

  1. I = 0.37 A
  2. V = E − Ir = 4.444 V
  3. Answer: 4.444 V

Worked example 3 (Challenge)

4 identical cells (6 V, 1 Ω each) are in series with 17 Ω. Find the current.

  1. Total emf 24 V, internal 4 Ω
  2. I = 24/(17 + 4) = 1.143 A
  3. Answer: 1.143 A

Common mistakes and misconceptions

  • Ignoring internal resistance
  • Using E for the terminal voltage

Practice questions

  1. (Basic) A cell of emf 6 V and internal resistance 1 Ω drives current through 3 Ω. Find the current.
  2. (Basic) A cell of emf 2 V and internal resistance 1.5 Ω drives current through 5 Ω. Find the current.
  3. (Standard) For the same cell (4 V, 1 Ω) and 18 Ω load, find the terminal voltage.
  4. (Standard) For the same cell (5 V, 1 Ω) and 11 Ω load, find the terminal voltage.
  5. (Challenge) 4 identical cells (10 V, 2 Ω each) are in series with 17 Ω. Find the current.
  6. (Challenge) 2 identical cells (3 V, 1 Ω each) are in series with 5 Ω. Find the current.

Answer key with full working

  • Q1: I = 6/(3 + 1) → = 1.5 A → Answer: 1.5 A
  • Q2: I = 2/(5 + 1.5) → = 0.308 A → Answer: 0.308 A
  • Q3: I = 0.211 A → V = E − Ir = 3.789 V → Answer: 3.789 V
  • Q4: I = 0.417 A → V = E − Ir = 4.583 V → Answer: 4.583 V
  • Q5: Total emf 40 V, internal 8 Ω → I = 40/(17 + 8) = 1.6 A → Answer: 1.6 A
  • Q6: Total emf 6 V, internal 2 Ω → I = 6/(5 + 2) = 0.857 A → Answer: 0.857 A

Syllabus note

This skill is listed under "Current Electricity" for Class 12 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 emf, internal resistance and terminal voltage in?

"Current Electricity", Class 12 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.

Book a free demo class