Class 10 Science · Original EZT practice
Electricity (Class 10 Chapter 11): summary and practice
Current, voltage and resistance lock together through Ohm's law, and this chapter turns that into circuit skills: series and parallel combinations, the heating effect and electric power. All questions are original EZT tutor practice following the chapter's own section order.
8 original worked examples, 8 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 20 October 2026:
1. Current, potential difference and circuits
Current is charge per unit time: I = Q/t. Potential difference is work done per unit charge: V = W/Q. A circuit diagram uses standard symbols for the cell, battery, resistor, ammeter, voltmeter and switch.
2. Ohm's law and what resistance depends on
Ohm's law says V = IR at constant temperature. Resistance grows with length, shrinks with area and depends on the material: R = rho l/A.
3. Resistors in series and parallel
In series the resistances add: Rs = R1 + R2 + R3, and the same current flows through each. In parallel the reciprocals add: 1/Rp = 1/R1 + 1/R2 + 1/R3, and each branch gets the full voltage.
4. Heating effect and electric power
A current heats its conductor: H = I^2 R t = VIt. Power is P = VI = I^2 R. The commercial unit of energy is the kilowatt hour: 1 kWh = 3.6 x 10^6 J.
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Find resistance from V and I
Choose the method: R = V/I, straight from Ohm's law.
EZT practice question
A resistor has 12 V across it and carries 3 A. Find its resistance.
Show the worked solution
- R = V/I.
- R = 12/3.
Answer: 4 ohm.
Check: A 4 ohm resistor at 12 V carries 12/4 = 3 A - consistent.
Watch for this: Do not compute I x V; that is power, not resistance.
Now try it yourself
A resistor carries 2 A at 6 V. Find its resistance.
Check your answer
3 ohm.
R = 6/2 = 3 ohm.
Combine resistors in series
Choose the method: Add the resistances; the same current flows through each.
EZT practice question
A 4 ohm and a 2 ohm resistor are in series across a 6 V battery. Find the current and the voltage drop across each.
Show the worked solution
- Rs = 4 + 2 = 6 ohm.
- I = V/Rs = 6/6 = 1 A.
- Drops: 1 x 4 = 4 V and 1 x 2 = 2 V.
Answer: Current 1 A; drops of 4 V and 2 V, adding to the 6 V supply.
Check: The drops must add to the battery voltage: 4 + 2 = 6 V.
Watch for this: Do not divide the battery voltage by one resistor alone; use the series total.
Now try it yourself
Three 2 ohm resistors in series across 12 V - current?
Check your answer
2 A.
Rs = 6 ohm, so I = 12/6 = 2 A.
Combine two resistors in parallel
Choose the method: Add the reciprocals, then invert.
EZT practice question
A 12 ohm and a 4 ohm resistor are in parallel across 4.5 V. Find the equivalent resistance and the total current.
Show the worked solution
- 1/Rp = 1/12 + 1/4 = 1/12 + 3/12 = 4/12 = 1/3.
- Rp = 3 ohm.
- I = 4.5/3 = 1.5 A.
Answer: Rp = 3 ohm and total current 1.5 A.
Check: Parallel equivalents are smaller than the smallest branch: 3 ohm is under 4 ohm.
Watch for this: Do not average the resistances; add the reciprocals instead.
Now try it yourself
Find the parallel equivalent of 6 ohm and 3 ohm.
Check your answer
2 ohm.
1/Rp = 1/6 + 1/3 = 1/2, so Rp = 2 ohm.
Combine three resistors in parallel
Choose the method: Same reciprocal rule, three terms.
EZT practice question
Resistors of 2 ohm, 3 ohm and 6 ohm are in parallel. Find the equivalent resistance.
Show the worked solution
- 1/Rp = 1/2 + 1/3 + 1/6.
- That is 3/6 + 2/6 + 1/6 = 6/6 = 1.
Answer: Rp = 1 ohm.
Check: The result is smaller than the smallest branch, 2 ohm - correct.
Watch for this: Do not stop at 1/Rp = 1; remember to invert.
Now try it yourself
Two 4 ohm resistors in parallel - equivalent?
Check your answer
2 ohm.
1/Rp = 1/4 + 1/4 = 1/2, so Rp = 2 ohm.
Compute the heat produced
Choose the method: H = I^2 R t with t in seconds.
EZT practice question
A current of 2 A flows through a 10 ohm resistor for 60 s. Find the heat produced.
Show the worked solution
- H = I^2 R t.
- H = 2^2 x 10 x 60 = 4 x 10 x 60.
Answer: 2400 J.
Check: H = VIt gives the same: V = IR = 20 V, so 20 x 2 x 60 = 2400 J.
Watch for this: Do not use minutes directly; convert time to seconds first.
Now try it yourself
How much heat does 1.5 A produce in 8 ohm over 10 s?
Check your answer
180 J.
H = 1.5^2 x 8 x 10 = 2.25 x 80 = 180 J.
Estimate an electricity bill
Choose the method: Energy in kWh is power in kW times hours; cost needs a tariff.
EZT practice question
A 1.5 kW geyser runs 2 h a day for 30 days. Find the energy used and the cost at an illustrative tariff of Rs 6 per kWh.
Show the worked solution
- Energy = 1.5 kW x 2 h x 30 days = 90 kWh.
- Cost = 90 x 6 = Rs 540.
Answer: 90 kWh, costing Rs 540 at the illustrative Rs 6 per kWh tariff.
Check: 90 kWh is 90 x 3.6 x 10^6 = 3.24 x 10^8 J.
Watch for this: The tariff here is assumed for practice; real bills use your actual rate, and power must be in kW, not W, for kWh.
Now try it yourself
A 100 W bulb runs 10 h a day for 30 days - how many kWh?
Check your answer
30 kWh.
0.1 kW x 10 h x 30 days = 30 kWh.
Use R = rho l/A
Choose the method: Longer means more resistance; thicker means less.
EZT practice question
A wire's length is doubled and its area kept the same. What happens to its resistance?
Show the worked solution
- R = rho l/A, so R is proportional to l.
- Doubling l doubles R.
Answer: The resistance doubles.
Check: If the area had doubled instead, the resistance would halve.
Watch for this: Do not say resistance is unchanged because the material is the same; geometry matters too.
Now try it yourself
A wire is replaced by one of the same material, same length, but twice the area - new resistance?
Check your answer
Half the original.
R is inversely proportional to A.
Work with electric power
Choose the method: P = VI = I^2 R = V^2/R.
EZT practice question
A device draws 2 A from the 220 V mains. Find its power.
Show the worked solution
- P = VI.
- P = 220 x 2.
Answer: 440 W.
Check: In one hour it uses 0.44 kW x 1 h = 0.44 kWh.
Watch for this: Do not mix up W and kW when moving to energy in kWh.
Now try it yourself
Express 1 kWh in joules.
Check your answer
3.6 x 10^6 J.
1 kWh = 1000 W x 3600 s = 3.6 x 10^6 J.
Try a mixed set
Choose a method or explanation before opening each answer.
1. State Ohm's law.
Show answer and reason
V = IR at constant temperature.
It ties voltage, current and resistance into one relation.
2. Series combination formula?
Show answer and reason
Rs = R1 + R2 + R3.
The same current flows through each, so the drops add.
3. Parallel combination formula?
Show answer and reason
1/Rp = 1/R1 + 1/R2 + 1/R3.
Each branch takes the full voltage, so the currents add.
4. 1 kWh in joules?
Show answer and reason
3.6 x 10^6 J.
It is 1000 W running for 3600 s.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- A resistor carries 5 A at 20 V. Find its resistance.
- Find the parallel equivalent of 6 ohm and 3 ohm.
- A current of 1.5 A flows through 8 ohm for 10 s. Find the heat produced.
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Questions about this practice page
- What is potential difference?
- The work done per unit charge to move charge between two points: V = W/Q, measured in volts.
- Why are home appliances connected in parallel?
- In parallel each appliance gets the full 220 V and works independently. The chapter notes that in a series circuit the voltage divides and if one device fails, the rest stop working too.
- How do I read a circuit diagram?
- Each component has a standard symbol - cell, battery, resistor, ammeter in series, voltmeter in parallel, switch. Trace the loop from the battery's positive terminal around to the negative.
- What is the heating effect of current?
- A current dissipates energy as heat in a resistor: H = I^2 R t. It runs heaters, irons and the electric fuse.
- Are these the NCERT exercise questions for Electricity (Class 10)?
- No. EZT pages are original practice following the chapter's section order - use them to prepare for the textbook exercises, not as a substitute for your book.