Class 12 Current Electricity Numericals: EMF, Internal Resistance and Power
A cell's EMF is not always the voltage across its external load. When a cell supplies current, some potential difference is associated with its internal resistance. Draw the external load and internal resistance in the same series loop before calculating.
Curriculum checked 2 October 2026 · original practice for selected skills.
PREDICT · CHANGE · EXPLAIN
Change the external load
A 6 V cell with internal resistance 1 Ω supplies a resistor R. The model is I=6/(R+1), with temperature and internal resistance held fixed.
The curve shows this simplified model over the labelled range. Read the exact point below; the sketch is not a measuring instrument.
Current: 1 A
Will increasing load resistance make the current rise or fall?
Now close the explanation and try the independent retest. Moving a slider alone does not check your understanding.
Before you start
You will need: Ohm's law, series resistance and electrical power.
Attempt each task before opening a hint. If you need help, reveal the first hint, then the setup, then compare your written steps with the solution. Finish with the retests without referring back.
Three ideas to keep in view
For a cell delivering current through load R with internal resistance r, I=E/(R+r). All resistance in the loop matters.
Terminal voltage while delivering current is E−Ir, equal to IR across the external load. The sign convention changes when considering a charging cell.
Power supplied by the ideal EMF is EI. It divides into I²R in the load and I²r internally; this is a useful consistency check.
Try the worked example first
P12-CUR-W1Cell and loadFocused practice
A 6 V cell with internal resistance 1 Ω supplies a 5 Ω resistor. Find current, terminal voltage and power delivered to the resistor.
Show your working
Need a starting hint?
Add internal and external resistance for the current.
Still stuck? Reveal the setup
I=6/(5+1).
Show worked solution+
1
Loop current
The total resistance is 6 Ω, giving I=1 A.
2
Load quantities
Terminal voltage is 1 × 5=5 V and load power is 1² × 5=5 W.
Answer: 1 A, 5 V, 5 W
Before moving on: can you explain why your method works, as well as give the answer?
Your turn: three different checks
Write a method as well as an answer. The tasks change the reasoning, not just the numbers.
P12-CUR-Q1Infer internal resistanceFocused practice
A cell of EMF 12 V has terminal voltage 10 V while delivering 2 A. Find its internal resistance.
Show your working
Need a starting hint?
Find the lost voltage E−V.
Still stuck? Reveal the setup
Ir=2 V.
Show worked solution+
1
Voltage drop
The internal drop is 12−10=2 V.
2
Divide by current
r=2/2 Ω.
Answer: 1 Ω
Before moving on: can you explain why your method works, as well as give the answer?
P12-CUR-Q2Power budgetFocused practice
A 9 V cell with r=1 Ω supplies R=2 Ω. Find the current and power dissipated internally.
Show your working
Need a starting hint?
Find total resistance first.
Still stuck? Reveal the setup
I=9/3; internal power is I²r.
Show worked solution+
1
Current
The current is 3 A.
2
Internal heat rate
3² × 1=9 W; load power is 18 W and source power is 27 W.
Answer: 3 A; 9 W internally
Before moving on: can you explain why your method works, as well as give the answer?
P12-CUR-Q3Wire geometryFocused practice
A uniform wire's length is doubled and its cross-sectional area halved, with resistivity unchanged. How does resistance change?
Show your working
Need a starting hint?
Use R=ρL/A.
Still stuck? Reveal the setup
The length contributes a factor 2 and reciprocal area another factor 2.
Show worked solution+
1
Apply both changes
Rnew=ρ(2L)/(A/2).
2
Compare
This simplifies to 4ρL/A.
Answer: Resistance becomes four times the original.
Before moving on: can you explain why your method works, as well as give the answer?
Catch a likely mistake
A tempting claim: “The resistor always receives the cell's full EMF.”
A delivering cell with non-zero internal resistance has an internal voltage drop.
Close the examples and try again
These changed questions test whether you can reconstruct the method. If you use a hint, record where you got stuck and retry later. Completing this small set does not establish full chapter mastery.
P12-CUR-R1Changed loadIndependent retest
A 9 V cell with r=1 Ω supplies R=8 Ω. Find I, terminal V and load power.
Show your working
Need a starting hint?
Use the whole loop to find I.
Still stuck? Reveal the setup
I=9/(8+1).
Show worked solution+
1
Current
I=1 A.
2
Load
V=8 V and P=8 W.
Answer: 1 A, 8 V, 8 W
Could you solve this without the earlier example? If not, revisit the first line where you got stuck.
P12-CUR-R2New voltage lossIndependent retest
An 8 V cell delivers 0.5 A at terminal voltage 7 V. Find r.
Show your working
Need a starting hint?
Subtract terminal voltage from EMF.
Still stuck? Reveal the setup
r=(8−7)/0.5.
Show worked solution+
1
Lost voltage
The internal drop is 1 V.
2
Resistance
1/0.5=2 Ω.
Answer: 2 Ω
Could you solve this without the earlier example? If not, revisit the first line where you got stuck.
Choose the next useful step
If a retest exposed the same error, rewrite the first incorrect step and explain its correction aloud. If both were independent, return to a mixed exercise or a missed paper question. A parent can ask what changed in the method rather than only asking for the answer.
Is this the same as the IGCSE electricity lesson?+
This set specifically adds cell EMF, internal resistance and terminal-voltage reasoning for Class 12. The IGCSE electricity lesson focuses on its own syllabus-labelled circuit quantities.
Can I print the questions and worked solutions separately?+
Yes. Use the two print buttons above. The question sheet leaves working space and hides solutions; the worked-solutions option includes the solution steps. Your browser can save either view as a PDF.
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