The name of 1 C/s.
The unit of current is the ampere. 1 A = 1 C/s.
Class 10 · Science · Chapter 11 · बिहार बोर्ड (BSEB)CBSE · NCERT 2026-27
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.
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आवेश और धारा — I = Q/t · NCERT 11.1 · coulomb · ampere
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.
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.
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.
For BSEB write the unit with the formula. “Ampere” alone, without 1 C/s, stays incomplete.
CBSE asks for the number of electrons. 1 C divided by 1.6 × 10⁻¹⁹ C is nearly 6 × 10¹⁸.
The name of 1 C/s.
The unit of current is the ampere. 1 A = 1 C/s.
True — current is taken in the direction of the flow of positive charge.
One electron carries 1.6 × 10⁻¹⁹ C.
Nearly 6 × 10¹⁸ electrons together make 1 C of charge.
It must measure the same current that passes through the device.
An ammeter is always connected in series. A voltmeter is connected in parallel.
Q = It. t = 300 s. Q = 0.40 A × 300 s = 120 C.
विभवांतर — V = W/Q · NCERT 11.2 · volt · voltmeter in parallel
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.
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.
A BSEB definition should contain both words — unit charge and work.
CBSE asks the pair of meters. Ammeter in series, voltmeter in parallel — do not swap the pair.
Work per charge.
1 V = 1 J/C. The 1 A = 1 C/s of current is a different statement.
True — the ammeter is in series and the voltmeter is in parallel.
Work divided by charge.
V = W/Q. The unit is the volt.
W = VQ = 6 V × 1 C = 6 J. So every coulomb receives 6 J of energy.
ओम का नियम — क्रियाकलाप 11.1 · NCERT 11.4 · Activity 11.1 · V = IR
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.
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.
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.
On a BSEB graph name the axes. Putting I on the vertical axis and V on the horizontal one loses marks.
CBSE asks for the condition with the law. Without the sentence “temperature stays the same” the answer is incomplete.
The graph of Activity 11.1.
The graph is a straight line through the origin. Potential difference V is on the vertical axis and current I on the horizontal axis.
True — if the temperature changes, R can change too.
The wire of Activity 11.1.
Nichrome contains nickel, chromium, manganese and iron.
R = V/I.
1 Ω = 1 V/1 A. The statement 1 A = 1 C/s defines the ampere, not the ohm.
R = V/I = 6.0 V / 0.50 A = 12 Ω. If the temperature stays the same, this ratio stays constant in Ohm law.
प्रतिरोध किन बातों पर निर्भर है — क्रियाकलाप 11.2 और 11.3 · NCERT 11.5 · Activities 11.2, 11.3 · R = ρl/A
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.
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.
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.
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.
For BSEB write the three factors — length, area, material. “A thick wire” alone is half an answer.
CBSE identifies the better conductor by the lower resistivity. Iron is a better conductor than mercury because ρ is smaller.
R is inverse to A.
A thick wire has a larger area, so its resistance is smaller. Resistivity is a property of the material and does not change with thickness.
True — in Activity 11.3 the resistance doubles and, on the same cell, the current falls to half.
From R = ρl/A, ρ = RA/l.
The unit of resistivity is ohm metre (Ω m).
It should not burn away at a high temperature.
An alloy has a higher resistivity and does not oxidise readily at a high temperature.
R = ρl/A. R2/R1 = (l2/l1) × (A1/A2) = (1/2) × (1/2) = 1/4. R2 = 8 Ω × 1/4 = 2 Ω.
श्रेणीक्रम — क्रियाकलाप 11.4 और 11.5 · NCERT 11.6.1 · Activities 11.4, 11.5 · Rs = R1 + R2 + R3
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.
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.
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.
In a BSEB numerical write Rs first, then I, then the separate voltages. A bare answer with no sum looks weak.
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.
10-second revision
Activity 11.4.
In series the current in every part is the same. Moving the ammeter does not change the reading.
True — V = V1 + V2 + V3. That is the result of Activity 11.5.
The three resistances add.
Rs = R1 + R2 + R3. It is greater than each single resistance.
First the sum is 6 Ω.
Rs = 6 Ω. I = 6 V / 6 Ω = 1 A.
The current is 1 A. V = IR.
V3 = 1 A × 3 Ω = 3 V. The three voltages 1 V, 2 V and 3 V add to 6 V.
Rs = 1 + 2 + 3 = 6 Ω. I = V/Rs = 6 V / 6 Ω = 1 A. Across 2 Ω, V2 = IR2 = 1 A × 2 Ω = 2 V.
समांतर क्रम — क्रियाकलाप 11.6 · NCERT 11.6.2 · Activity 11.6 · 1/Rp = 1/R1 + 1/R2 + 1/R3
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.
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.
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.
For BSEB show the sum for 1/Rp first, then invert it to Rp. A bare “1 ohm” without the fractions is incomplete.
CBSE asks two advantages of parallel — full voltage to each appliance, and the others keep working if one is switched off.
10-second revision
In Activity 11.6 the voltmeter gives the same reading on every branch.
In parallel the potential difference is the same. The currents add: I = I1 + I2 + I3.
True — 2 Ω, 3 Ω and 6 Ω in parallel are equivalent to 1 Ω, which is smaller than 2 Ω.
The reciprocals add, not the resistances themselves.
1/Rp = 1/R1 + 1/R2 + 1/R3.
1/2 + 1/3 + 1/6 = 1.
1/Rp = 1/2 + 1/3 + 1/6 = 1, so Rp = 1 Ω.
1/Rp = 1/2 + 1/3 + 1/6 = 1, so Rp = 1 Ω. I = V/Rp = 6 V / 1 Ω = 6 A.
तापीय प्रभाव और शक्ति — H = I²Rt · NCERT 11.7–11.8 · joule · watt · kilowatt hour
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.
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.
For BSEB write all three forms P = VI, I²R and V²/R together. One form loses marks.
In a CBSE heat sum, dropping the square of I makes the number wrong. At 5 A write I² = 25 and then go on.
10-second revision
H = I²Rt.
The heat is proportional to the square of the current, and equals I²Rt for a given time and resistance.
True — 1 W = 1 V × 1 A.
1000 W × 3600 s.
1 kWh = 3.6 × 10⁶ J.
First I = V/R = 5 A, then P = VI.
I = 220/44 = 5 A. P = 220 × 5 = 1100 W.
I = V/R = 220 V / 44 Ω = 5 A. P = VI = 220 × 5 = 1100 W. H = I²Rt = 25 × 44 × 10 = 11000 J.
फ्यूज और घरेलू सुरक्षा · NCERT 11.7.1 · fuse · 5 A · no wet hands
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.
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.
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.
For BSEB write where the fuse sits — in series. If you write parallel, the answer becomes wrong.
CBSE compares the cord and the heating element with I²R. The current is the same and the resistance is different.
10-second revision
If it melts, the whole current should stop.
A fuse is in series with the appliance, so that the circuit breaks when it melts.
False — a thick wire does not melt in time and the extra current can cause a fire.
I = P/V is about 4.55 A.
I = 1000/220 = 4.55 A. The next rated fuse above this is 5 A.
The current I is the same in both. The resistance of the heating element is much larger than that of the cord. In H = I²Rt a larger R gives more heat, so the element glows and the cord does not.
Pick a type. The 39 lesson checks are separate — each lesson has as many as its topic needs. All correct earns mastery ★.
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.
The rate of charge.
I = Q/t. 1 A = 1 C/s.
The old direction of positive charge.
The direction of current is taken opposite to the flow of electrons.
Across the points whose potential difference is needed.
A voltmeter is connected in parallel. An ammeter is connected in series.
Work divided by charge.
1 V = 1 J/C.
The slope of the graph.
V/I = R stays constant. The graph is a straight line through the origin.
Activity 11.1.
V is on the vertical axis and I is on the horizontal axis.
R = ρl/A.
R is proportional to length, so it doubles.
ρ = RA/l.
The SI unit of resistivity is Ω m.
Activity 11.5.
In series the resistances add: Rs = R1 + R2 + R3.
Activity 11.6.
In parallel the potential difference is the same and the currents add.
The sum of the reciprocals is 1.
1/Rp = 1/2 + 1/3 + 1/6 = 1, so Rp = 1 Ω.
The square of the current.
H = I²Rt. Power is P = VI = I²R = V²/R.
1000 × 3600.
1 kWh = 3.6 × 10⁶ J.
I = P/V is about 4.55 A.
The current is about 4.55 A, so a 5 A fuse is used.
H = I²Rt and I is the same.
The current is the same. The element has a larger resistance, so more heat is produced there.
False — an ammeter is connected in series. The voltmeter is connected in parallel.
True — this is the definition of the ohm.
False — the condition of the law is that the temperature stays the same.
True — R is inverse to the area.
False — series Rs is greater than each one. A smaller equivalent belongs to parallel.
True — I = I1 + I2 + I3.
False — electrons are not used up. The bill is for energy.
True — when it melts it breaks the path of the current.
The unit is the ampere.
1 C/s.
V = IR.
The temperature stays the same.
R = ρl/A.
Length on top, area below.
Rs = 6 Ω. On 6 V the current is 1 A.
A plain sum.
Rp = 1 Ω.
The sum of the reciprocals is 1.
H = I²Rt.
The square of the current.
1 W = 1 V × 1 A.
P = VI.
Nichrome is an alloy of nickel, chromium, manganese and iron.
The alloy of Activity 11.1.
Current is ampere, potential difference is volt, resistance is ohm, power is watt.
Ohm is V = IR, series is a sum, parallel is a sum of reciprocals, Joule is I²Rt.
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.
Both are true and R is the correct explanation of A.
Assertion (A): In series the current is the same in every resistor.
Reason (R): The series equivalent is smaller than the smallest resistance.
A is true. R is false — the series equivalent is greater than each one. The smaller equivalent belongs to parallel.
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.
Both are true and R is the correct reason.
Assertion (A): A thick wire is safer than a fuse.
Reason (R): A fuse melts above the rated current and breaks the circuit.
A is false. R is true — that is why a thin rated fuse is used.
Assertion (A): An ammeter is connected in series.
Reason (R): 1 kWh = 3.6 × 10⁶ J.
Both are true, but R does not explain A.
The same current and the sum of resistances belong to series. The same voltage and the sum of reciprocals belong to parallel.
The filament is tungsten, the heating coil is an alloy, and the transmission line is copper or aluminium.
One ampere is the current when one coulomb of charge flows in one second. 1 A = 1 C/s.
If the temperature stays the same, the potential difference across a conductor is proportional to the current through it, V = IR.
In R = ρl/A, ρ is the resistivity of the material. The unit is Ω m. It is a property of the material.
Power is the rate of using energy. P = VI. The unit is the watt and 1 W = 1 V × 1 A.
A fuse is connected in series with the appliance. If the current exceeds the rated value it melts and breaks the circuit.
Q = It. t = 10 × 60 = 600 s. Q = 0.50 A × 600 s = 300 C.
R = ρl/A. Resistance rises with length and falls as the area rises. In Activity 11.3, doubling the length halved the ammeter reading.
In series the appliances cannot have different currents, and if one fails all of them stop. In parallel each appliance gets the full potential difference, and the others keep working if one is switched off.
H = I²Rt = (2 A)² × 5 Ω × 10 s = 4 × 5 × 10 = 200 J.
The circuit has a nichrome wire XY about 0.5 m long, an ammeter, a voltmeter and four cells of 1.5 V. Nichrome is an alloy of nickel, chromium, manganese and iron. V and I are noted as cells are added one by one. V/I stays nearly constant. With 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 = IR. 1 Ω = 1 V/1 A.
Rs = R1 + R2 + R3 = 1 + 2 + 3 = 6 Ω. I = V/Rs = 6 V / 6 Ω = 1 A. V1 = IR1 = 1 V, V2 = 2 V, V3 = 3 V. The sum 6 V equals the battery.
I = V/R = 220/44 = 5 A. P = VI = 1100 W. I²R = 25 × 44 = 1100 W. V²/R = 48400/44 = 1100 W. H = I²Rt = 25 × 44 × 10 = 11000 J. In 2 hours the energy = 1.1 kW × 2 h = 2.2 kWh. 1 kWh = 3.6 × 10⁶ J.
This model set is for practice. It is not an annual examination paper of any year.
The battery of Activity 11.1.
4 × 1.5 V = 6 V.
A smaller ρ is a better conductor.
Iron has a lower resistivity than mercury, so iron is the better conductor.
It is called a rheostat.
It changes the resistance without changing the voltage source.
W = VQ = 6 V × 1 C = 6 J.
Each branch has 6 V. I1 = 6/2 = 3 A, I2 = 6/3 = 2 A, I3 = 6/6 = 1 A. Total I = 3 + 2 + 1 = 6 A. Rp = 1 Ω.
V = IR: if the temperature stays the same, potential difference is proportional to current. Rs = R1 + R2 + R3: in series the resistances add and the current stays the same. 1/Rp = 1/R1 + 1/R2 + 1/R3: in parallel the reciprocals add and the potential difference stays the same.
These are competency-based practice questions. They are not a past paper and not a copy of a CBSE paper.
The book graph keeps V on the vertical axis.
The Ohm graph keeps V on the vertical axis and I on the horizontal axis, and the line passes through the origin.
1/2 + 1/3 + 1/6 = 1.
If all three are in parallel, Rp = 1 Ω. In series the sum would be 11 Ω.
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.
Both are true and R is the correct reason.
Assertion (A): The equivalent resistance of a parallel combination becomes smaller.
Reason (R): In parallel the current in every branch is zero.
A is true. R is false — current flows in the branches and the total current is their sum.
R = V/I = 6.0 V / 0.50 A = 12 Ω. The ammeter must be in series so that it measures the whole current. The voltmeter must be in parallel so that it measures the difference between two points. If they are swapped, the ammeter will not carry the whole current and the voltmeter will break the circuit in the wrong way.
P = VI = 220 V × 5 A = 1100 W = 1.1 kW. Energy = 1.1 kW × 2 h = 2.2 kWh. In joule this is 2.2 × 3.6 × 10⁶ J, because 1 kWh = 3.6 × 10⁶ J.
Switch board with BSEB | CBSE above. The lessons follow the same NCERT chapter.
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.
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These are case and assertion-reason practice items. Do not treat them as past CBSE questions.
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What you learned
| What | Keep this |
|---|---|
| Current | I = Q/t, मात्रक A |
| Potential difference | V = W/Q, मात्रक V |
| Ohm law | V = IR (ताप समान) |
| Resistivity | R = ρl/A, मात्रक Ω m |
| Series and parallel | Rs = R1+R2+R3 ; 1/Rp = 1/R1+1/R2+1/R3 |
| Heat and power | H = 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.