Class 10 · Science · Chapter 11 · बिहार बोर्ड (BSEB)CBSE · NCERT 2026-27

Electricity

Electricity

How to use this page:
1. Read — Activities 11.1 to 11.6, diagram, worked example, board tip
2. Check — each lesson has its own questions; the number follows the lesson
3. Mastery ★ — all of that lesson correct. Redo the wrong ones
4. The signature picture is an Ohm graph: V on the y-axis, I on the x-axis, a straight line through the origin.

In NCERT this is chapter 11. The older Bihar book calls the same text chapter 12. Progress stays in this browser.

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  • 1 Charge and current — I = Q/t
  • 2 Potential difference — V = W/Q
  • 3 Ohm law — Activity 11.1
  • 4 What resistance depends on — Activities 11.2 and 11.3
  • 5 Resistors in series — Activities 11.4 and 11.5
  • 6 Resistors in parallel — Activity 11.6
  • 7 Heating effect and power — H = I²Rt
  • 8 The fuse and household safety
  • Chapter winner — every lesson at mastery ★

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1

Charge and current — I = Q/t

आवेश और धारा — I = Q/t · NCERT 11.1 · coulomb · ampere

New
Charge and currentCellCurrent flows only in a closed path
In series the same current passes through every resistor.
Only a closed path is a circuitNotes

When charge flows through a conductor there is an electric current in it. In a torch the cells make charge flow and the bulb glows. A switch makes a conducting link between the cell and the bulb. A continuous and closed path of current is an electric circuit. If the path breaks anywhere, or the switch opens, the current stops and the bulb goes out.

In a metal wire the electrons carry the charge. Current was first thought of as a flow of positive charge. So in a circuit the direction of current is taken opposite to the direction of the electrons — from the positive terminal of the cell towards the negative terminal.

The unit, and the electrons in one coulomb

If a net charge Q crosses a section in time t, the current is I = Q/t. The SI unit of charge is the coulomb (C). One electron carries 1.6 × 10⁻¹⁹ C, so 1 C contains nearly 6 × 10¹⁸ electrons.

One ampere is the current in which 1 coulomb of charge flows every second: 1 A = 1 C/s. Small currents are written in milliampere (1 mA = 10⁻³ A) or microampere (1 µA = 10⁻⁶ A). The ammeter is connected in series.

Worked exampleExample

Question: A current of 0.40 A flows in a wire for 5 minutes. Find the charge that flows.

Formula: Q = It

Substitution: t = 5 × 60 = 300 s. Q = 0.40 A × 300 s = 120 C.

10-second revision
  • I = Q/t and 1 A = 1 C/s
  • The direction of current is opposite to the electrons
  • 1 C has nearly 6 × 10¹⁸ electrons; the ammeter is in series
Board tip · BSEBBoard tip

For BSEB write the unit with the formula. “Ampere” alone, without 1 C/s, stays incomplete.

Board tip · CBSEBoard tip

CBSE asks for the number of electrons. 1 C divided by 1.6 × 10⁻¹⁹ C is nearly 6 × 10¹⁸.

Check your understandingall correct = mastery ★
1
The SI unit of electric current is —
Check
2
In a circuit the direction of electric current is taken opposite to the direction of the electrons.
Check
3
The number of electrons in one coulomb of charge is nearly ______.
Check
4
How is an ammeter connected in a circuit?
Check
5
How much charge flows if a current of 0.40 A flows for 5 minutes? Write the formula, the substitution and the unit.
Check2 marks
Next lesson →
2

Potential difference — V = W/Q

विभवांतर — V = W/Q · NCERT 11.2 · volt · voltmeter in parallel

New
Potential differenceV ∝ I
The straight line says: double the potential difference and the current doubles too.
Charge flows only when there is a difference of electric pressureNotes

Water does not flow by itself in a level tube. If one end is joined to a tank at a height, the pressure difference pushes the water. Electrons in a metal wire move only when there is a potential difference along the wire. A battery of one or more cells produces that difference. Chemical action inside a cell keeps the potential difference across the terminals even when no current is drawn. To keep the current going, the cell spends its chemical energy.

The potential difference between two points is the work done to move a unit charge from one point to the other. V = W/Q. The SI unit is the volt. One volt is the potential difference when 1 joule of work moves 1 coulomb: 1 V = 1 J/C. Potential difference is measured with a voltmeter, connected in parallel across the two points.

Worked exampleExample

Question: How much work is done in moving 2 C between two points at 15 V? How much energy does a 6 V battery give to each coulomb?

Formula: W = VQ

Substitution: W = 15 V × 2 C = 30 J.

For a 6 V battery, each 1 C gets W = 6 V × 1 C = 6 J.

10-second revision
  • V = W/Q and 1 V = 1 J/C
  • Voltmeter in parallel, ammeter in series
  • A 6 V battery gives 6 J to every coulomb
Board tip · BSEBBoard tip

A BSEB definition should contain both words — unit charge and work.

Board tip · CBSEBoard tip

CBSE asks the pair of meters. Ammeter in series, voltmeter in parallel — do not swap the pair.

Check your understandingall correct = mastery ★
1
1 volt is equal to —
Check
2
A voltmeter is connected in parallel across the points whose potential difference is to be measured.
Check
3
The formula for potential difference is V = ______.
Check
4
How much energy does a 6 V battery give to each coulomb of charge? Write the formula and the substitution.
Check2 marks
Next lesson →
3

Ohm law — Activity 11.1

ओम का नियम — क्रियाकलाप 11.1 · NCERT 11.4 · Activity 11.1 · V = IR

New
Activity 11.1 — nichrome wire, four cells, two metersActivity

Set up a nichrome wire XY of length about 0.5 m, with an ammeter, a voltmeter and four cells of 1.5 V each. Nichrome is an alloy of nickel, chromium, manganese and iron. The voltmeter is in parallel across the ends of XY. The ammeter measures the current in series.

With one cell, note the ammeter reading I and the voltmeter reading V. Repeat with two, three and four cells, and find V/I for each pair. The ratio comes out nearly the same each time. Plot V on the vertical axis and I on the horizontal axis. The graph is a straight line through the origin.

If the temperature stays the same, V is proportional to INotes

In 1827 Georg Simon Ohm found that the potential difference across a metal wire is proportional to the current through it, provided the temperature does not change. This is Ohm law. V ∝ I, or V = IR. R is the resistance of that wire.

Resistance is the opposition to the flow of charge. The unit is the ohm (Ω). 1 Ω = 1 V/1 A. Since I = V/R, doubling the resistance halves the current. A component that changes the current without changing the source voltage is a variable resistance. A rheostat is often used for this in a circuit.

Ohm graph — V vertical, I horizontalVIStraight line through the origin, slope = RIf the temperature stays the same, V/I stays constant. That is Ohm law.
The V–I graph of a nichrome wire. The slope is the resistance R = V/I. If the temperature changes, the same line does not hold.
Worked exampleExample

Question: The potential difference across a nichrome wire is 6.0 V and the current is 0.50 A. Find the resistance.

Formula: R = V/I

Substitution: R = 6.0 V / 0.50 A = 12 Ω.

Four cells of 1.5 V together give 6.0 V. If the temperature stays the same, V/I remains this 12 Ω.

10-second revision
  • Nichrome = Ni, Cr, Mn, Fe; wire about 0.5 m; four cells of 1.5 V
  • The V–I graph is a straight line through the origin; V vertical, I horizontal
  • V = IR, and 1 Ω = 1 V/1 A; the temperature must stay the same
Board tip · BSEBBoard tip

On a BSEB graph name the axes. Putting I on the vertical axis and V on the horizontal one loses marks.

Board tip · CBSEBoard tip

CBSE asks for the condition with the law. Without the sentence “temperature stays the same” the answer is incomplete.

Check your understandingall correct = mastery ★
1
What is the V–I graph of Ohm law like?
Check
2
Ohm law holds only when the temperature of the conductor stays the same.
Check
3
Nichrome is an alloy of nickel, chromium, manganese and ______.
Check
4
1 ohm is the resistance in which —
Check
5
A wire has 6.0 V across it and the current is 0.50 A. Find the resistance. Write the formula, the substitution and the unit.
Check2 marks
Next lesson →
4

What resistance depends on — Activities 11.2 and 11.3

प्रतिरोध किन बातों पर निर्भर है — क्रियाकलाप 11.2 और 11.3 · NCERT 11.5 · Activities 11.2, 11.3 · R = ρl/A

New
What resistance depends onCellCurrent flows only in a closed path
In series the same current passes through every resistor.
Activity 11.2 — different components, different currentsActivity

Take a nichrome wire, a torch bulb, a 10 W bulb, an ammeter of range 0–5 A, a plug key and connecting wires. Join four dry cells of 1.5 V each in series with the ammeter and leave a gap XY.

First connect the nichrome wire in the gap XY, plug the key and read the ammeter. Caution: take the plug key out after you have measured the current. Then read the current with the torch bulb, and after that with the 10 W bulb. The current is different for each component. A component of low resistance is a good conductor. A conductor with an appreciable resistance is called a resistor. An insulator of the same size offers a still higher resistance.

Activity 11.3 — double the length, half the currentActivity

Complete a circuit of a cell, an ammeter, a nichrome wire of length l marked (1), and a plug key. Note the current. Replace it by a nichrome wire of the same thickness but length 2l, marked (2). The ammeter reading falls to half.

A thicker nichrome wire of the same length l, marked (3), increases the current, because the area of cross-section is larger. A copper wire of the same length and the same area, marked (4), changes the current again. So the current depends on length, area and material.

R = ρl/A — resistivity is a property of the materialNotes

The resistance of a uniform metal wire is proportional to its length l and inversely proportional to its area of cross-section A. R = ρl/A. ρ (rho) is the resistivity. Its SI unit is Ω m. Metals and alloys have resistivity from about 10⁻⁸ Ω m to 10⁻⁶ Ω m. Insulators such as rubber and glass are of order 10¹² to 10¹⁷ Ω m. Both resistance and resistivity change with temperature.

The resistivity of an alloy is usually higher than that of its metals, and alloys do not oxidise readily at a high temperature. That is why the coils of irons and toasters are alloys. Tungsten is used for a bulb filament. Copper and aluminium are used for transmission lines. The table values need not be memorised. For a comparison, silver has the lowest resistivity, then copper. Iron has a lower resistivity than mercury, so iron is a better conductor than mercury.

Worked exampleExample

Question: A wire has resistance 8 Ω. A second wire of the same material has half the length and twice the area. Find the new resistance.

Formula: R = ρl/A. The material is the same, so ρ is the same.

Substitution: R2/R1 = (l2/l1) × (A1/A2) = (1/2) × (1/2) = 1/4.

R2 = 8 Ω × 1/4 = 2 Ω.

10-second revision
  • Double the length and the current is halved; a thicker wire gives more current
  • R = ρl/A, unit Ω m; take the key out after the reading
  • Alloys in heaters, tungsten in bulbs, copper and aluminium in lines
Board tip · BSEBBoard tip

For BSEB write the three factors — length, area, material. “A thick wire” alone is half an answer.

Board tip · CBSEBoard tip

CBSE identifies the better conductor by the lower resistivity. Iron is a better conductor than mercury because ρ is smaller.

Check your understandingall correct = mastery ★
1
A thick wire of the same material and the same length, compared with a thin wire, has —
Check
2
Doubling the length of the nichrome wire halves the ammeter reading.
Check
3
The SI unit of resistivity is ______.
Check
4
Why is the coil of an electric iron an alloy rather than a pure metal?
Check
5
A second wire of the material of an 8 Ω wire has half the length and twice the area. Find the new resistance.
Check3 marks
Next lesson →
5

Resistors in series — Activities 11.4 and 11.5

श्रेणीक्रम — क्रियाकलाप 11.4 और 11.5 · NCERT 11.6.1 · Activities 11.4, 11.5 · Rs = R1 + R2 + R3

New
Resistors in seriesCellCurrent flows only in a closed path
In series the same current passes through every resistor.
Activity 11.4 — move the ammeter, the current stays the sameActivity

Join three different resistors in series. Connect them to a battery, an ammeter and a plug key. For the trial you may use 1 Ω, 2 Ω, 3 Ω and a 6 V battery. Plug the key and read the ammeter.

Shift the ammeter to any place between the resistors. The reading stays the same. In a series combination the current is the same in every part. If one component breaks, the whole circuit opens and nothing works.

Activity 11.5 — the voltages addActivity

Connect a voltmeter across the ends X and Y of the same series combination. Plug the key, read the potential difference V, and compare it with the potential difference across the battery. Then place the voltmeter across the first resistor to read V1, across the second for V2 and across the third for V3.

You find V = V1 + V2 + V3. The current I is the same in each resistor, so V1 = IR1, V2 = IR2 and V3 = IR3. Adding gives IR = IR1 + IR2 + IR3, so Rs = R1 + R2 + R3. The equivalent resistance is greater than each single resistance.

Worked exampleExample

Question: 1 Ω, 2 Ω and 3 Ω are in series with a 6 V battery. Find the equivalent resistance, the current and the potential difference across each resistor.

Formula: Rs = R1 + R2 + R3, then I = V/Rs, and V1 = IR1.

Substitution: Rs = 1 Ω + 2 Ω + 3 Ω = 6 Ω. I = 6 V / 6 Ω = 1 A.

V1 = 1 A × 1 Ω = 1 V, V2 = 1 A × 2 Ω = 2 V, V3 = 1 A × 3 Ω = 3 V. The sum 1 + 2 + 3 = 6 V, equal to the battery.

10-second revision
  • In series the current is the same; moving the ammeter does not change the reading
  • V = V1 + V2 + V3 and Rs = R1 + R2 + R3
  • With 1 Ω, 2 Ω, 3 Ω and 6 V the current is 1 A; the voltages are 1 V, 2 V, 3 V
Board tip · BSEBBoard tip

In a BSEB numerical write Rs first, then I, then the separate voltages. A bare answer with no sum looks weak.

Board tip · CBSEBoard tip

CBSE asks why a bulb and a heater are not joined in series. They need different currents, and if one fails the other stops too.

Check your understandingall correct = mastery ★
1
When the ammeter is shifted to a point in between a series combination, the current —
Check
2
In series the total potential difference equals the sum of the potential differences across the separate resistors.
Check
3
The equivalent resistance in series is Rs = ______.
Check
4
The current through 1 Ω, 2 Ω and 3 Ω in series on 6 V is —
Check
5
In the circuit above, the potential difference across the 3 Ω resistor is —
Check
6
1 Ω, 2 Ω and 3 Ω are in series with a 6 V battery. Find the equivalent resistance, the current and the potential difference across 2 Ω.
Check3 marks
Next lesson →
6

Resistors in parallel — Activity 11.6

समांतर क्रम — क्रियाकलाप 11.6 · NCERT 11.6.2 · Activity 11.6 · 1/Rp = 1/R1 + 1/R2 + 1/R3

New
Resistors in parallelCellCurrent flows only in a closed path
In parallel the current splits. The voltage on every branch stays the same.
Activity 11.6 — three paths, one potential differenceActivity

Make a parallel combination XY of R1, R2 and R3. Connect it to a battery, a plug key and an ammeter. Connect a voltmeter in parallel with the combination. Plug the key and read the total current I and the potential difference V. Placing the voltmeter across each resistor still gives the same V. In parallel the potential difference across every resistor is the same.

Take the key out. Insert the ammeter in turn in the branches of R1, R2 and R3. The branch currents are I1, I2 and I3, and I = I1 + I2 + I3. By Ohm law I = V/Rp and I1 = V/R1. So 1/Rp = 1/R1 + 1/R2 + 1/R3. The equivalent resistance is smaller than the smallest resistance.

Why household appliances stay in parallelNotes

In series the current is the same everywhere. A bulb and a heater need different currents, so joining them in series is not practical. If one bulb fuses, the whole string goes out — finding the dead bulb in festival series lights takes a long time.

In a parallel connection each appliance gets the full potential difference of the source and each draws the current it needs. If one appliance is switched off, the others keep working. The total resistance falls, so the total current from the source can rise.

Worked exampleExample

Question: 2 Ω, 3 Ω and 6 Ω are in parallel with a 6 V battery. Find the equivalent resistance, the total current and the current in each branch.

Formula: 1/Rp = 1/R1 + 1/R2 + 1/R3, then I = V/Rp and I1 = V/R1.

Substitution: 1/Rp = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 1/1. So Rp = 1 Ω.

I = 6 V / 1 Ω = 6 A. I1 = 6/2 = 3 A, I2 = 6/3 = 2 A, I3 = 6/6 = 1 A. The sum 3 + 2 + 1 = 6 A.

10-second revision
  • In parallel V is the same and I = I1 + I2 + I3
  • 1/Rp = 1/R1 + 1/R2 + 1/R3; Rp is smaller than the smallest
  • 2 Ω, 3 Ω and 6 Ω in parallel give Rp = 1 Ω
Board tip · BSEBBoard tip

For BSEB show the sum for 1/Rp first, then invert it to Rp. A bare “1 ohm” without the fractions is incomplete.

Board tip · CBSEBoard tip

CBSE asks two advantages of parallel — full voltage to each appliance, and the others keep working if one is switched off.

Check your understandingall correct = mastery ★
1
In a parallel combination, what stays the same across every resistor?
Check
2
The equivalent resistance of a parallel combination is smaller than the smallest resistance.
Check
3
In parallel, 1/Rp = ______.
Check
4
The equivalent resistance of 2 Ω, 3 Ω and 6 Ω in parallel is —
Check
5
2 Ω, 3 Ω and 6 Ω are in parallel on 6 V. Find Rp and the total current. Write the formula and the substitution.
Check3 marks
Next lesson →
7

Heating effect and power — H = I²Rt

तापीय प्रभाव और शक्ति — H = I²Rt · NCERT 11.7–11.8 · joule · watt · kilowatt hour

New
Heating effect and powerCellCurrent flows only in a closed path
In series the same current passes through every resistor.
In a pure resistance all the energy becomes heatNotes

A cell spends energy to keep the current going. In a fan some of it becomes useful work and the rest becomes heat. In a circuit of resistors only, all the energy becomes heat. This is the heating effect of current. A heater, an iron, a toaster and a kettle work on it.

If a current I flows for time t through a resistance R at potential difference V, the energy supplied by the source is VIt. That energy is the heat. Putting V = IR from Ohm law gives H = I²Rt. This is Joule law of heating. The heat is proportional to the square of the current, to R when I is fixed, and to the time. The unit is the joule.

Power is the rate of energy. P = VI = I²R = V²/R. The unit is the watt. 1 W = 1 V × 1 A. The commercial unit is the kilowatt hour. 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J. We pay the board for energy, not for electrons. Electrons are not used up in the circuit.

Worked example — both joule and wattExample

Question: An iron of 44 Ω runs on 220 V for 10 s. Find the current, the power and the heat produced.

Formula: I = V/R, P = VI = I²R = V²/R, H = I²Rt.

Substitution: I = 220 V / 44 Ω = 5 A.

P = 220 V × 5 A = 1100 W. Check: I²R = 25 × 44 = 1100 W, and V²/R = 48400 / 44 = 1100 W.

H = I²Rt = (5 A)² × 44 Ω × 10 s = 11000 J. The same H = VIt = 220 × 5 × 10 = 11000 J.

10-second revision
  • H = I²Rt in joule; heat is proportional to I², R and t
  • P = VI = I²R = V²/R, unit watt
  • 1 kWh = 3.6 × 10⁶ J; electrons are not used up
Board tip · BSEBBoard tip

For BSEB write all three forms P = VI, I²R and V²/R together. One form loses marks.

Board tip · CBSEBoard tip

In a CBSE heat sum, dropping the square of I makes the number wrong. At 5 A write I² = 25 and then go on.

Check your understandingall correct = mastery ★
1
In Joule law the heat produced is proportional to —
Check
2
1 watt is the power of a device that draws 1 ampere at 1 volt.
Check
3
1 kilowatt hour is equal to ______ joule.
Check
4
What power does a 44 Ω iron take at 220 V?
Check
5
A 44 Ω iron runs at 220 V for 10 s. Find the power in watt and the heat in joule.
Check3 marks
Next lesson →
8

The fuse and household safety

फ्यूज और घरेलू सुरक्षा · NCERT 11.7.1 · fuse · 5 A · no wet hands

New
The fuse and household safetyCellCurrent flows only in a closed path
In series the same current passes through every resistor.
A fuse is a weak link on purposeNotes

The filament of a bulb is tungsten. Its melting point is 3380 °C, so it becomes hot enough to give light without melting. The filament is held on insulating supports. The bulb is filled with inactive nitrogen and argon so that the filament lasts longer. Most of the power of the filament becomes heat and a small part becomes light.

A fuse protects the circuit. It is placed in series with the appliance. The wire is a metal or alloy that melts at a suitable temperature, such as aluminium, copper, iron or lead. If the current rises above the rated value the wire melts and the circuit breaks. Household fuses are rated 1 A, 2 A, 3 A, 5 A and 10 A.

The cord of a heater does not glow while the heating element does, because the current is the same in both but the resistance of the element is much larger. In H = I²Rt a larger R gives more heat.

Do not put a thick wire in place of the fuseCaution

A thick copper wire used in place of a fuse does not melt in time. The extra current keeps heating the wires and can start a fire. Use a fuse of the rated value only.

Do not touch a switch with wet hands. In the activities, keep taking the plug key out after you measure the current. A 1 kW iron at 220 V draws about 4.55 A, so the fuse is 5 A, not 2 A or 3 A.

Worked exampleExample

Question: Which fuse will you choose for a 1000 W iron on 220 V?

Formula: I = P/V

Substitution: I = 1000 W / 220 V = 4.55 A.

4.55 A is a little under 5 A and more than 3 A. So among the rated fuses the one to use is 5 A. The fuse is in series with the iron.

10-second revision
  • A fuse is in series and melts on excess current to break the circuit
  • At 1000 W and 220 V the current is 4.55 A, so the fuse is 5 A
  • A thick wire in place of a fuse is dangerous; do not touch a switch with wet hands
Board tip · BSEBBoard tip

For BSEB write where the fuse sits — in series. If you write parallel, the answer becomes wrong.

Board tip · CBSEBoard tip

CBSE compares the cord and the heating element with I²R. The current is the same and the resistance is different.

Check your understandingall correct = mastery ★
1
How is a fuse connected in a circuit?
Check
2
It is safe to put a thick copper wire in place of a fuse.
Check
3
The suitable fuse for a 1000 W iron at 220 V is —
Check
4
Why does the cord of a heater not glow, while the heating element does?
Check2 marks
Question bank →

❓ Full question bank — with answers and explanations — 68 questions

No question is marked as a verified past paper. The BSEB set is a model for practice. CBSE items are CBSE-style, not a copy of any year’s paper.

Multiple choice

0/15
Pick one option. A wrong try brings a hint.
1
The correct relation for electric current is —
NCERT-style · practice1 mark
2
The direction of current in a circuit is taken —
Board-style (practice)1 mark
3
A voltmeter is connected —
Board-style (practice)1 mark
4
1 volt means —
NCERT-style · practice1 mark
5
In Ohm law, this stays constant if the temperature does not change —
Board-style (practice)1 mark
6
On the V–I graph the vertical axis carries —
Board-style (practice)1 mark
7
If the length of a wire is doubled and the area stays the same, the resistance —
Board-style (practice)1 mark
8
The unit of resistivity is —
Board-style (practice)1 mark
9
The equivalent of three resistors in series is —
NCERT-style · practice1 mark
10
In parallel, this stays the same —
Board-style (practice)1 mark
11
The equivalent of 2 Ω, 3 Ω and 6 Ω in parallel is —
Board-style (practice)1 mark
12
The correct form of Joule law is —
Board-style (practice)1 mark
13
1 kWh is equal to —
Board-style (practice)1 mark
14
The fuse for a 1000 W iron at 220 V should be —
Board-style (practice)1 mark
15
The cord of a heater does not glow because —
Board-style (practice)1 mark
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True or false

0/8
1
An ammeter is connected in parallel.
Board-style (practice)1 mark
2
1 Ω = 1 V/1 A.
Board-style (practice)1 mark
3
Ohm law stays exactly the same when the temperature changes.
Board-style (practice)1 mark
4
A thick wire of the same material and the same length has less resistance than a thin wire.
Board-style (practice)1 mark
5
In series the equivalent resistance is smaller than the largest resistance.
Board-style (practice)1 mark
6
In parallel the total current equals the sum of the branch currents.
Board-style (practice)1 mark
7
Electrons are used up in an electric circuit and the bill is for that consumption.
Board-style (practice)1 mark
8
A fuse is connected in series with the appliance.
Board-style (practice)1 mark
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Fill in the blanks

0/8
1
In I = Q/t, if t is in second and Q in coulomb, the unit of I is ______.
Board-style (practice)1 mark
2
In Ohm law, V = ______.
Board-style (practice)1 mark
3
The resistance formula is R = ρ ______.
Board-style (practice)1 mark
4
The equivalent of 1 Ω, 2 Ω and 3 Ω in series is ______ Ω.
Board-style (practice)1 mark
5
The equivalent of 2 Ω, 3 Ω and 6 Ω in parallel is ______ Ω.
Board-style (practice)1 mark
6
The Joule form of the heat is H = ______.
Board-style (practice)1 mark
7
1 watt = 1 volt × ______.
Board-style (practice)1 mark
8
Nichrome contains iron together with nickel, chromium and ______.
Board-style (practice)1 mark
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Match

0/2
1
Match the quantity with its unit.
Board-style (practice)2 marks
Column B: A. volt · B. ampere · C. watt · D. ohm
1. Current
2. Potential difference
3. Resistance
4. Power
2
Match the rule with its form.
NCERT-style · practice2 marks
Column B: A. 1/Rp = 1/R1 + 1/R2 + 1/R3 · B. V = IR · C. H = I²Rt · D. Rs = R1 + R2 + R3
1. Ohm
2. Series
3. Parallel
4. Joule
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Assertion–reason

0/5
Check both statements, then see whether the reason explains the assertion.
1

Assertion (A): The V–I graph of a metal wire is a straight line through the origin.

Reason (R): If the temperature stays the same, the potential difference is proportional to the current.

Board-style (practice)1 mark
2

Assertion (A): In series the current is the same in every resistor.

Reason (R): The series equivalent is smaller than the smallest resistance.

Board-style (practice)1 mark
3

Assertion (A): Household fans and bulbs are connected in parallel.

Reason (R): In parallel each appliance gets the full potential difference, and the others keep working if one is switched off.

NCERT-style · practice1 mark
4

Assertion (A): A thick wire is safer than a fuse.

Reason (R): A fuse melts above the rated current and breaks the circuit.

Board-style (practice)1 mark
5

Assertion (A): An ammeter is connected in series.

Reason (R): 1 kWh = 3.6 × 10⁶ J.

CBSE-style · competency-based (not a PYQ)1 mark
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Coefficient practice

0/4
These equations are not results of this chapter. They are only practice in filling coefficients.
1
This is coefficient practice, not a result of this chapter. Balance H2 + O2 → H2O.
Board-style (practice)1 mark
H2 + O2 → H2O
2
This is coefficient practice, not a reaction from the electricity chapter. Balance N2 + H2 → NH3.
Board-style (practice)1 mark
N2 + H2 → NH3
3
This is coefficient practice, not a result of this chapter. Balance CH4 + O2 → CO2 + H2O.
Board-style (practice)1 mark
CH4 + O2 → CO2 + H2O
4
This is coefficient practice, not a result of this chapter. Balance Fe + O2 → Fe2O3.
Board-style (practice)2 marks
Fe + O2 → Fe2O3
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Classify

0/2
1
Place each statement under series or parallel.
Board-style (practice)2 marks
Current the same in every resistor
Potential difference the same across every resistor
Rs = R1 + R2 + R3
1/Rp = 1/R1 + 1/R2 + 1/R3
2
Place each use in the right material class.
NCERT-style · practice2 marks
Bulb filament
Heating coil of an iron
Transmission line
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Very short answer

0/5
1
Define one ampere.
Board-style (practice)1 mark
2
Write Ohm law in one line.
NCERT-style · practice1 mark
3
What is resistivity? Write the unit.
Board-style (practice)2 marks
4
Define electric power and write one form.
Board-style (practice)2 marks
5
Write the job of a fuse wire in two lines.
Board-style (practice)2 marks
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Short answer

0/4
1
A current of 0.50 A flows for 10 minutes. Find the charge. Write the formula, the substitution and the unit.
Board-style (practice)3 marks
2
How does resistance depend on length and area? Give one observation from Activity 11.3.
NCERT-style · practice3 marks
3
Write two disadvantages of series, by comparing with parallel.
Board-style (practice)3 marks
4
A current of 2 A flows for 10 s in a 5 Ω resistor. Find the heat.
Board-style (practice)3 marks
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Long answer

0/3
1
Describe the circuit of Activity 11.1 and explain Ohm law with its graph.
Board-style (practice)5 marks
2
1 Ω, 2 Ω and 3 Ω are in series with 6 V. Find Rs, the current and the potential difference across each resistor. Show the formula and the unit at each step.
NCERT-style · practice5 marks
3
A 44 Ω iron runs at 220 V for 10 s. Find the current, the power in all three forms and the heat. Then state the energy in kWh for 2 hours.
Board-style (practice)5 marks
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BSEB model paper · practice

0/6

This model set is for practice. It is not an annual examination paper of any year.

1
Four cells of 1.5 V in series give a total potential difference of —
BSEB model · practice (not an annual paper)1 mark
2
The better conductor of iron and mercury is —
BSEB model · practice (not an annual paper)1 mark
3
The English name of a current controller is ______.
BSEB model · practice (not an annual paper)1 mark
4
How much energy does each coulomb receive from a 6 V battery?
BSEB model · practice (not an annual paper)2 marks
5
2 Ω, 3 Ω and 6 Ω are in parallel on 6 V. Find the current in each branch and the total current.
BSEB model · practice (not an annual paper)3 marks
6
Write Ohm law, the series formula and the parallel formula. Also write the meaning of each in one sentence.
BSEB model · practice (not an annual paper)5 marks
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CBSE-style questions

0/6

These are competency-based practice questions. They are not a past paper and not a copy of a CBSE paper.

1
A student put V on the horizontal axis and I on the vertical axis and still got a straight line through the origin. For the correct graph the student should —
CBSE-style · competency-based (not a PYQ)1 mark
2
Three resistors are 2 Ω, 3 Ω and 6 Ω. They are to be joined so that the equivalent is 1 Ω. The correct join is —
CBSE-style · competency-based (not a PYQ)1 mark
3

Assertion (A): The cord of a heater does not glow.

Reason (R): The current is the same in the cord and the element, but the cord has a very small resistance so the I²R heat is small.

CBSE-style · competency-based (not a PYQ)1 mark
4

Assertion (A): The equivalent resistance of a parallel combination becomes smaller.

Reason (R): In parallel the current in every branch is zero.

CBSE-style · competency-based (not a PYQ)1 mark
5
In a circuit the ammeter in series reads 0.50 A and the voltmeter in parallel reads 6.0 V. Find the resistance and say why the meters cannot be swapped.
CBSE-style · competency-based (not a PYQ)3 marks
6
A motor draws 5 A from a 220 V line. Find the power and the energy used in 2 hours, in kWh.
CBSE-style · competency-based (not a PYQ)3 marks
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🏛️ Board exam corner — Bihar Board (BSEB)— CBSE

Switch board with BSEB | CBSE above. The lessons follow the same NCERT chapter.

BSEB model · practiceBoard tip

This page has no verified annual-exam question, because no source page has been added. The model set below is practice in the board pattern.

🏛️ Model questions on one page →

The verified label will be used only when a source page for the question is available.

CBSE-style · competency-based

These are case and assertion-reason practice items. Do not treat them as past CBSE questions.

Open the CBSE-style questions →

🔁 Spaced review — today’s questions

Wrong questions return soon; correct ones return after a few days.

🧠 What you learned + equation sheet

What you learned

WhatKeep this
CurrentI = Q/t, मात्रक A
Potential differenceV = W/Q, मात्रक V
Ohm lawV = IR (ताप समान)
ResistivityR = ρl/A, मात्रक Ω m
Series and parallelRs = R1+R2+R3 ; 1/Rp = 1/R1+1/R2+1/R3
Heat and powerH = I²Rt ; P = VI = I²R = V²/R

The notes are original writing. The textbook was used only for activity order and numbers. “Verified” will be used only on a question that has a source page.