The whole chapter in one place — read it, see it move, then test yourself. Clear notes, every formula you need, an interactive circuit you can build, and a quick quiz that tells you exactly what to revise.
Quick answerElectric current is the rate of flow of electric charge, I = Q/t, measured in ampere (A), where 1 A means 1 coulomb of charge passing a point each second. A current keeps flowing only in a closed circuit.
Electric current is simply the flow of electric charge. In a metal wire this charge is carried by tiny negative particles called electrons. When these charges move steadily through a conductor, we say an electric current is flowing.
Electric current is the rate of flow of electric charge. If a charge Q flows through any cross-section of a conductor in time t, the current I is given by I = Q / t. The SI unit of charge is the coulomb (C). One coulomb is the charge carried by about 6.25 x 1018 electrons.
The SI unit of current is the ampere (A). A current of one ampere flows when one coulomb of charge passes through a cross-section every second, so 1 A = 1 coulomb per second (1 C/s). Very small currents are written in smaller units:
1 milliampere (1 mA) = 10-3 A
1 microampere (1 µA) = 10-6 A
Current is measured with an ammeter, which is always connected in series in the circuit.
By convention, the direction of current is taken as the direction in which positive charge would flow. So conventional current flows from the positive (+) terminal to the negative (−) terminal of the cell through the outside circuit. Electrons actually move in the opposite direction, from − to +, but we still keep the old convention.
For a steady current to flow, the path must be complete. A closed circuit (cell, wires and device all joined with no gap) allows a continuous current. If a switch is opened or a wire breaks, the circuit becomes an open (broken) circuit and the current stops at once.
Worked example: A charge of 30 C flows through a bulb in 10 s. Find the current.
Given: Q = 30 C, t = 10 s.
Use I = Q / t = 30 C / 10 s.
I = 3 A. So a current of 3 ampere flows through the bulb.
Electric currentI = Q / tampere (A) · Q = charge in coulomb (C), t = time in second (s); current is the rate of flow of charge.
Definition of ampere1 A = 1 C / 1 sA · One ampere is one coulomb of charge flowing past a cross-section every second.
Remember
Electric current is the rate of flow of charge: I = Q/t. Its SI unit is the ampere (A).
1 A = 1 coulomb per second (1 C/s); 1 coulomb is the charge of about 6.25 x 10^18 electrons.
Conventional current flows from the + terminal to the - terminal, which is opposite to the actual direction of electron flow.
A steady current needs a closed circuit; an open or broken circuit stops the current immediately.
Current is measured by an ammeter connected in series; smaller units are mA (10^-3 A) and microampere (10^-6 A).
Potential Difference (Voltage)
Quick answerPotential difference between two points is the work done to move a unit positive charge from one point to the other. It is measured in volts (1 V = 1 J/C) and is what actually pushes current through a circuit.
Charges do not flow through a conductor on their own. They move only when there is a difference in electric potential between the two ends of the conductor. This is what a cell or battery provides.
The electric potential difference between two points is defined as the work done to move a unit charge from one point to the other. If work W is done to move a charge Q, then the potential difference is:
V = W / Q
Its SI unit is the volt (V), named after Alessandro Volta. From the definition, 1 volt = 1 joule per coulomb, i.e. 1 V = 1 J/C. So the potential difference between two points is 1 volt when 1 joule of work is done to move a charge of 1 coulomb between them.
A cell or battery does the work needed to maintain the potential difference across its terminals. It is this maintained potential difference that keeps driving the current around the circuit.
Potential difference is measured by an instrument called a voltmeter. Remember one key rule: a voltmeter is always connected in parallel, across the two points (or across the component) whose potential difference you want to measure.
Worked example: How much work is done in moving a charge of 2 C across two points having a potential difference of 12 V?
Given: Q = 2 C, V = 12 V.
Rearrange V = W/Q to get W = V x Q.
W = 12 V x 2 C = 24 J.
Potential differenceV = W / Qvolt (V) · W = work done (J), Q = charge moved (C)
Definition of the volt1 V = 1 J / 1 C1 volt = 1 joule per coulomb
Work done to move a chargeW = V x Qjoule (J) · rearranged from V = W/Q
Remember
Potential difference V = W/Q: it is the work done to move a unit charge between two points in a circuit.
SI unit is the volt (V); 1 V = 1 joule per coulomb (1 J/C).
A cell or battery maintains the potential difference that keeps the current flowing.
A voltmeter measures potential difference and is always connected in PARALLEL across the component.
Rearranged, work done to move charge is W = V x Q (answer comes out in joules).
Circuit Diagrams & Symbols
Quick answerIn a circuit diagram we draw every part with a fixed standard symbol. Remember: an ammeter goes in series (measures current, near-zero resistance) and a voltmeter goes in parallel (measures potential difference, very high resistance).
A circuit diagram is a simple drawing that shows how the parts of an electric circuit are joined. Instead of drawing a real cell or a real bulb, we use fixed standard symbols. These symbols are agreed upon everywhere, so any student, teacher or electrician reads the same diagram in the same way. They make a circuit quick to draw and easy to compare.
Here are the common symbols you must know for your board exam:
Electric cell — two parallel lines: the longer, thinner line is the positive (+) terminal and the shorter, thicker line is the negative (−) terminal.
Battery — a combination of two or more cells drawn in a row (several long–short line pairs joined together).
Plug key / switch (open) — a break shown open in the line; no current can flow.
Plug key / switch (closed) — the break is bridged; current can flow.
Wire and junction — a plain straight line is a wire; a solid dot where two wires meet shows a junction (joint). Wires that cross without a dot are not joined.
Fixed resistor — a plain rectangle, a resistor of resistance R.
Variable resistance / rheostat — a rectangle with an arrow across it (a sliding contact), used to change the resistance.
Electric bulb — a circle with a loop (the filament) inside it.
Ammeter — a circle with the letter A inside.
Voltmeter — a circle with the letter V inside.
Two meters need special care. An ammeter measures the current through a component, so it is always connected in series — the whole current must pass through it. A voltmeter measures the potential difference across a component, so it is always connected in parallel across that component.
Their resistances are chosen so that they barely disturb the circuit. An ideal ammeter has very low resistance (nearly zero), so, being in series, it hardly changes the current it is reading. An ideal voltmeter has very high resistance, so, being in parallel, it draws almost no current away from the component.
Remember
Standard symbols make circuit diagrams universal, so everyone reads them the same way.
Cell symbol: longer thin line = positive terminal; shorter thick line = negative terminal.
A dot at a crossing means the wires are joined; no dot means they only cross over.
Ammeter measures current and is connected in SERIES; voltmeter measures potential difference and is connected in PARALLEL.
Ideal ammeter has very low (near-zero) resistance; ideal voltmeter has very high resistance, so neither disturbs the circuit.
Ohm's Law & Resistance
Quick answerAt a constant temperature, the current through a conductor is directly proportional to the potential difference across it, so V = IR. Resistance R is the opposition to current, measured in ohms (Ω).
Ohm's law tells us how current and voltage are linked. It states that, at a constant temperature, the current I flowing through a conductor is directly proportional to the potential difference V across its ends. In symbols, V ∝ I. Removing the proportionality sign gives a constant, and that constant is the resistance R. So we get the well-known relation V = IR.
Resistance is the property of a conductor that opposes the flow of current through it. Rearranging Ohm's law, R = V / I. Its SI unit is the ohm, written with the Greek letter Ω (omega). By definition, 1 ohm = 1 volt / 1 ampere. In other words, a conductor has a resistance of 1 Ω if a potential difference of 1 V across it drives a current of 1 A.
If you plot a graph of V (on the y-axis) against I (on the x-axis) for such a conductor, you get a straight line passing through the origin. A conductor that obeys Ohm's law like this is called an ohmic conductor. The important point for exams: the slope of this V–I graph is equal to the resistance R. A steeper line means a larger resistance.
V ∝ I only when the temperature is kept constant.
The three useful forms are V = IR, I = V / R, and R = V / I.
A straight line through the origin on a V–I graph means the conductor is ohmic; its slope gives R.
Good conductors have low resistance; poor conductors and insulators have high resistance.
Worked example: A current of 0.3 A flows through a resistor of resistance 20 Ω. Find the potential difference across it. Using V = IR = 0.3 A × 20 Ω = 6 V. So a 6 V potential difference is needed to drive that current through the resistor.
Ohm's lawV = I x RV in volt (V), I in ampere (A), R in ohm (Omega) · Valid at constant temperature; V is directly proportional to I.
ResistanceR = V / IOmega (ohm) · R is the opposition to current; equals the slope of the V-I graph.
Definition of one ohm1 Omega = 1 V / 1 AOmega · A conductor has 1 ohm resistance if 1 V drives 1 A through it.
Current formI = V / RA (ampere) · Rearranged form of Ohm's law.
Remember
Ohm's law: at constant temperature, V is directly proportional to I, giving V = IR.
Resistance R = V / I is the opposition to current; SI unit is the ohm (Ω), where 1 Ω = 1 V/A.
For an ohmic conductor the V–I graph is a straight line through the origin, and its slope equals R.
Three exam-ready forms: V = IR, I = V/R, R = V/I.
Good conductors have low resistance; insulators have high resistance.
Worked check: I = 0.3 A, R = 20 Ω gives V = IR = 6 V.
What Resistance Depends On (Resistivity)
Quick answerThe resistance of a wire grows with its length and falls as it gets thicker, and it also depends on the material and the temperature. In short, R = rho x l / A, where rho (resistivity) is a fixed property of the material, measured in ohm metre.
The resistance of a wire is not the same for every wire. For a uniform conductor, its resistance R depends on four things:
Length (l):R is directly proportional to length. A longer wire has more resistance. Double the length, and the resistance doubles.
Area of cross-section (A):R is inversely proportional to area. A thicker wire has less resistance, because the charges have more room to flow.
Nature of the material: even for the same size, copper, aluminium, nichrome and others resist the current by different amounts.
Temperature: for most conductors, the resistance generally increases as the temperature rises.
The first three effects combine into one neat formula, R = ρl / A. Here ρ (rho) is the resistivity of the material. Resistivity is the resistance of a unit cube of the material (length 1 m, area 1 m2). Unlike resistance, it does not depend on the length or thickness of the wire — it is a property of the material itself. Its SI unit is the ohm metre (Ω m). Metals are good conductors with very low resistivity, of the order of 10-8 Ω m. Alloys have a somewhat higher resistivity than pure metals, but both are still far better conductors than insulators like rubber and glass, whose resistivity is extremely high. Resistivity also changes with temperature.
This is why different materials are chosen for different jobs:
Heating elements: alloys such as nichrome, manganin and constantan have higher resistivity than pure metals and do not oxidise (burn) easily even when red hot. So nichrome is used in electric heaters, irons and toasters.
Connecting wires and transmission: copper and aluminium have low resistivity, so they carry current with very little loss. That is why house wiring and long transmission lines use them.
Worked example: Find the resistance of a copper wire of length 2 m and area of cross-section 1 mm2 (= 1 × 10-6 m2), taking the resistivity of copper as 1.6 × 10-8 Ω m. Using R = ρl / A = (1.6 × 10-8 × 2) ÷ (1 × 10-6) = 3.2 × 10-2 Ω = 0.032 Ω. Such a tiny resistance is exactly why copper makes such good connecting wire.
Resistance of a uniform conductorR = rho x l / AOmega (ohm) · rho = resistivity, l = length, A = area of cross-section
Resistivity of the materialrho = R x A / lOmega m (ohm metre) · A property of the material; independent of the wire's length and thickness
Dependence on lengthR is directly proportional to lSame material and area: double the length, double the resistance
Dependence on area of cross-sectionR is inversely proportional to AA larger (thicker) cross-section gives a smaller resistance
Remember
Resistance of a uniform conductor: R = rho x l / A, where rho is resistivity.
R is directly proportional to length l and inversely proportional to area of cross-section A (thicker wire = less resistance).
Resistance also depends on the material and generally increases with temperature.
Resistivity rho is a property of the material (resistance of a unit cube); its SI unit is the ohm metre (Omega m).
Metals have very low resistivity (~10^-8 Omega m) and alloys a little higher; insulators have very high resistivity.
Alloys like nichrome/manganin/constantan (high rho, resist oxidation) are used in heating elements; copper and aluminium (low rho) are used for wires and transmission.
Resistors in Series and Parallel
Quick answerJoin resistors end to end (series) and the resistances add up, so R_s is bigger than the largest; connect them across the same two points (parallel) and the reciprocals add, so R_p is smaller than the smallest. Homes use parallel so every appliance gets the full voltage and its own switch.
Series combination. Here the resistors are joined end to end, so there is only one path for the current. Because of this, the same currentI flows through every resistor. The equivalent resistance is simply the sum, so it is larger than any single resistor: Rs = R1 + R2 + R3. Also, the potential differences across the resistors add up to the supply voltage: V = V1 + V2 + V3.
Worked example (series). Take R1 = 5 Ω, R2 = 10 Ω and R3 = 15 Ω joined in series across a 6 V battery. Then Rs = 5 + 10 + 15 = 30 Ω. The current is I = V / Rs = 6 / 30 = 0.2 A, and it is the same everywhere. The separate voltages are V1 = 0.2 × 5 = 1 V, V2 = 0.2 × 10 = 2 V and V3 = 0.2 × 15 = 3 V, which add up to 6 V — exactly the battery voltage.
Parallel combination. Here the resistors are connected across the same two points, so the same potential differenceV acts across each one. The current divides into branches, and the branch currents add up to the total current: I = I1 + I2 + I3. The equivalent resistance is found from the sum of the reciprocals: 1/Rp = 1/R1 + 1/R2 + 1/R3. Remember: Rp is always less than the smallest individual resistance.
Worked example (parallel). Take R1 = 2 Ω, R2 = 3 Ω and R3 = 6 Ω in parallel. Then 1/Rp = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1, so Rp = 1 Ω. Notice this is smaller than 2 Ω, the smallest resistor in the group. Try changing these values in the interactive circuit simulator to watch how Rs and Rp respond — it makes the rules click.
Why our homes use parallel wiring. Domestic circuits connect appliances in parallel, not in series, for three main reasons:
Each appliance gets the full supply voltage, so it runs at its rated value.
Appliances work independently — you can switch off the fan without turning off the lights.
Each appliance can have its own switch and can draw the current it needs; a heater and a bulb require very different currents. In a series circuit the same current would flow through everything, and a single break would stop it all.
Equivalent resistance in seriesR_s = R1 + R2 + R3Ω · Same current through each resistor; R_s is more than any single resistor.
Voltage in a series circuitV = V1 + V2 + V3V · Potential differences across the resistors add up to the source voltage.
Equivalent resistance in parallel1/R_p = 1/R1 + 1/R2 + 1/R3Ω · R_p is always less than the smallest individual resistance.
Current in a parallel circuitI = I1 + I2 + I3A · Same voltage across each branch; the branch currents add up to the total current.
Remember
Series: the same current flows through every resistor, and R_s = R1 + R2 + R3 (the equivalent resistance is greater than the largest resistor).
Series: the potential differences across the resistors add up to the source voltage (V = V1 + V2 + V3).
Parallel: the same voltage acts across each branch, and the branch currents add up to the total current (I = I1 + I2 + I3).
Parallel: 1/R_p = 1/R1 + 1/R2 + 1/R3, and R_p is always less than the smallest individual resistance.
Homes use parallel wiring so each appliance gets the full supply voltage, works independently, and has its own switch and suitable current.
Quick answerWhen current flows through a resistor, electrical energy turns into heat. The heat produced is given by Joule's law: H = I²Rt.
When an electric current flows through a resistor, the electrical energy supplied by the source is converted into heat. This is called the heating effect of electric current. The heat produced is given by Joule's law of heating, H = I2Rt. According to this law, the heat produced in a resistor is:
directly proportional to the square of the current (I2) for a given resistance;
directly proportional to the resistanceR for a given current;
directly proportional to the timet for which the current flows.
Here H is in joules (J), I in amperes (A), R in ohms (Ω) and t in seconds (s). This heat is really the work done by the current, H = VIt, which becomes I2Rt on using V = IR. Quick numerical: if a current of 5 A flows through a heater of resistance 20 Ω for 30 s, then H = I2Rt = (5)2 × 20 × 30 = 15000 J (15 kJ).
The heating effect is used in many everyday appliances. The electric iron, room heater, toaster and geyser use a heating element made of an alloy called nichrome. Nichrome is chosen because it has a high resistivity (so even a short element gives out a lot of heat), a high melting point (so the element does not melt even when it becomes red-hot) and it does not oxidise (burn) easily at high temperature.
In an electric bulb, the filament is made of tungsten because it has a very high melting point (about 3380 °C). This lets it be heated to a very high temperature so that it glows brightly without melting. The bulb is usually filled with chemically inactive gases such as nitrogen and argon to make the filament last longer.
An electric fuse is an important safety device. It is a short piece of wire made of a metal or alloy of low melting point, connected in series with the live wire. Every fuse is rated for a certain safe current. If the current in the circuit rises above this safe value (due to overloading or a short circuit), the fuse wire heats up, melts and breaks the circuit. This cuts off the current and protects the appliances and wiring from damage.
Joule's law of heatingH = I² x R x tjoule (J) · Heat produced when current I flows through resistance R for time t; I in A, R in ohm, t in s.
Heat as work done by the currentH = V x I x t = I² x R x tjoule (J) · Using V = I x R; all the electrical energy consumed in the resistor appears as heat.
Remember
Heating effect: when current passes through a resistor, electrical energy is converted into heat.
Joule's law of heating: H = I²Rt — heat is proportional to I², to R, and to time t; H is in joules.
Heating appliances (iron, heater, toaster, geyser) use a nichrome element: high resistivity, high melting point, and it does not oxidise at high temperature.
An electric bulb filament is made of tungsten because of its very high melting point, so it can glow without melting.
A fuse is a safety wire of low melting point placed in series with the live wire; it melts and breaks the circuit when the current exceeds the safe value.
Electric Power & Your Electricity Bill
Quick answerElectric power P = VI (also I²R or V²/R), measured in watts. Energy = power × time, and the electricity board bills you in kilowatt-hours (units), where 1 unit = 1 kWh = 3.6×10⁶ J and cost = units × rate per unit.
Electric power is the rate at which electrical energy is used up, or the rate at which electrical work is done. If a current I flows through a device across a potential difference V, then the power is P = VI.
Using Ohm's law (V = I R), we can write power in two more handy forms: P = I2R and P = V2 / R. Pick whichever one fits the values you are given.
The SI unit of power is the watt (W). One watt is the power used when a current of 1 A flows across a potential difference of 1 V, so 1 W = 1 joule/second. Larger powers are measured in kilowatt (kW), where 1 kW = 1000 W.
Every appliance carries a power rating, for example a bulb marked "60 W, 220 V". This tells you the bulb uses 60 J of energy every second when run at 220 V. Quick check: current drawn = P/V = 60/220 ≈ 0.27 A.
Electrical energy = power × time (E = P × t). The electricity board does not charge you in joules — it uses the kilowatt-hour (kWh), also called 1 unit. One unit is the energy used by a 1 kW appliance running for 1 hour, and 1 kWh = 3.6×106 J.
To estimate your bill:
Find the energy in units: units (kWh) = power (kW) × time (hours).
Multiply the units by the rate per unit in ₹: cost = units × rate.
Worked example. Suppose a 100 W (= 0.1 kW) bulb runs for 5 hours a day for 30 days. Energy = 0.1 kW × 5 h × 30 = 15 kWh = 15 units. Taking an example rate of ₹8 per unit (always use the actual rate printed on your own bill), cost = 15 × 8 = ₹120.
Electric powerP = V x Iwatt (W) · V = potential difference, I = current
Power from Ohm's lawP = I^2 x R = V^2 / RW · obtained by substituting V = IR
Watt (definition)1 W = 1 J/swatt · 1 kW = 1000 W
Electrical energyE = P x tjoule (J) or kWh
Commercial unit of energy1 kWh = 3.6 x 10^6 Jkilowatt-hour (kWh) · 1 kWh = 1 unit
Units consumedunits = power(kW) x time(hours)kWh
Cost of electricitycost = units x rate per unitRupees · rate per unit is taken from your electricity bill
Remember
Electric power P = VI; using Ohm's law it is also P = I²R = V²/R.
SI unit of power is the watt (W): 1 W = 1 J/s, and 1 kW = 1000 W.
A rating like "60 W, 220 V" means the appliance uses 60 J every second when run at 220 V.
Electrical energy = power × time; the commercial unit is the kilowatt-hour (kWh), also called 1 unit.
1 kWh = 3.6×10⁶ J.
Bill: units = power(kW) × time(hours); cost = units × rate per unit in ₹.
The formula sheet
Every formula in this chapter, in one place — screenshot it before your exam.
I = Q / t
Electric currentampere (A)
1 A = 1 C / 1 s
Definition of ampereA
V = W / Q
Potential differencevolt (V)
1 V = 1 J / 1 C
Definition of the volt
W = V x Q
Work done to move a chargejoule (J)
V = I x R
Ohm's lawV in volt (V), I in ampere (A), R in ohm (Omega)
R = V / I
ResistanceOmega (ohm)
1 Omega = 1 V / 1 A
Definition of one ohmOmega
I = V / R
Current formA (ampere)
R = rho x l / A
Resistance of a uniform conductorOmega (ohm)
rho = R x A / l
Resistivity of the materialOmega m (ohm metre)
R is directly proportional to l
Dependence on length
R is inversely proportional to A
Dependence on area of cross-section
R_s = R1 + R2 + R3
Equivalent resistance in seriesΩ
V = V1 + V2 + V3
Voltage in a series circuitV
1/R_p = 1/R1 + 1/R2 + 1/R3
Equivalent resistance in parallelΩ
I = I1 + I2 + I3
Current in a parallel circuitA
H = I² x R x t
Joule's law of heatingjoule (J)
H = V x I x t = I² x R x t
Heat as work done by the currentjoule (J)
P = V x I
Electric powerwatt (W)
P = I^2 x R = V^2 / R
Power from Ohm's lawW
1 W = 1 J/s
Watt (definition)watt
E = P x t
Electrical energyjoule (J) or kWh
1 kWh = 3.6 x 10^6 J
Commercial unit of energykilowatt-hour (kWh)
units = power(kW) x time(hours)
Units consumedkWh
cost = units x rate per unit
Cost of electricityRupees
Test yourself
Tap an answer to check it instantly — you'll see why it's right, and what to revise if it isn't.
0 correct · 0/12 answered
Q1Ohm's laweasy
A resistor of 20 Ω carries a current of 0.5 A. What is the potential difference across it?
By Ohm's law, V = I × R = 0.5 A × 20 Ω = 10 V.
Q2Series resistanceeasy
Three resistors of 2 Ω, 3 Ω and 5 Ω are joined in series. What is their equivalent resistance?
In series, equivalent resistance is the sum: R = 2 Ω + 3 Ω + 5 Ω = 10 Ω.
Q3Definitions & unitseasy
A potential difference of 1 volt between two points in a circuit means that:
Potential difference V = W/Q. So 1 volt = 1 joule/coulomb, meaning 1 J of work is done to move 1 C of charge between the two points. Option 0 states exactly this. Options 1 and 2 describe current/power (charge or energy per second), not potential difference. Option 3 describes Ohm's law (V=IR gives 1 V, but it's a specific condition, not the definition of the volt).
Q4Conventional currenteasy
In an electric circuit, the direction of conventional current is taken to be:
Conventional current is defined as the direction of flow of positive charge, which is opposite to the actual motion of electrons (which are negative). In the external circuit conventional current flows from the positive terminal to the negative terminal, while electrons drift from negative to positive. Option 0 states the same direction as electrons (wrong). Option 2 describes electron flow direction (negative to positive in external wires), not conventional current. Option 3 is wrong because a steady cell gives steady DC. Hence option 1 is correct.
Q5Fuse & safetyeasy
Which of these correctly describes an electric fuse used in a household circuit?
Per NCERT Class 10 (Electricity), a fuse is a safety device made from a thin wire of an alloy with a LOW melting point (e.g. tin-lead). When current exceeds the rated value, Joule heating (H = I^2Rt) melts the fuse wire and breaks the circuit. It must be connected in the LIVE (phase) wire so that, when it blows, it disconnects the live supply from the appliance, making it safe. Option 2 states low melting point + live wire. Options 0 and 3 (high melting point / thick copper, earth wire) are wrong; option 1 (neutral wire) is wrong because a fuse in the neutral would still leave the appliance connected to the live wire even after blowing.
Q6Parallel resistancemedium
A 6 Ω resistor and a 3 Ω resistor are connected in parallel. What is the equivalent resistance?
For resistors in parallel: 1/R = 1/6 + 1/3 = 1/6 + 2/6 = 3/6 = 1/2, so R = 2 Ω. Parallel equivalent resistance is always less than the smallest resistor (3 Ω), confirming 2 Ω.
Q7Energy & costmedium
An electric heater of 1000 W is used for 2 hours every day. At ₹5 per unit (kWh), what is the electricity cost for 30 days?
Power = 1000 W = 1 kW. Energy per day = 1 kW × 2 h = 2 kWh (2 units). Over 30 days = 2 × 30 = 60 units. Cost = 60 units × ₹5/unit = ₹300.
Q8Circuit instrumentsmedium
To correctly measure the current through a resistor and the potential difference across it, you should:
An ammeter has very low (ideally zero) resistance and must carry the same current as the resistor, so it goes in series. A voltmeter has very high (ideally infinite) resistance and must sample the potential difference across the resistor without drawing appreciable current, so it goes in parallel. Thus: ammeter in series, voltmeter in parallel — option 0. (Options with the roles swapped would short-circuit via the ammeter and block current via the series voltmeter.)
Q9Factors affecting resistancemedium
A uniform metal wire has resistance R. It is replaced by another wire of the same material, same thickness and same temperature, but twice the length. The resistance of the new wire is:
Resistance R = ρL/A. Same material (ρ) and same thickness (same cross-sectional area A), so R ∝ L. Doubling the length L → 2L gives new resistance R' = ρ(2L)/A = 2·(ρL/A) = 2R.
Q10Heating element materialmedium
Nichrome is used for the heating element of appliances like electric irons and toasters because it:
Heating elements need a material that dissipates heat (H = I^2 R t), so high resistivity is essential. Nichrome (~100 x 10^-8 ohm m, far higher than copper's ~1.6 x 10^-8 ohm m) has high resistivity, a high melting point (~1400 C), and does not oxidise/burn readily even when red-hot. This lets it get hot and glow without melting or corroding. Option 1 states high resistivity + non-oxidising, which is correct. Option 0 (low resistivity) is wrong (opposite of what's needed), option 2 (low melting point) is wrong (it has a high melting point), and option 3 (perfect insulator) is wrong (it conducts current).
Q11Heating effect (H=I²Rt)medium
The heating element of an electric iron has a resistance of 40 Ω and draws a current of 5 A. How much heat is produced in 30 s?
Heat produced H = I²Rt = (5 A)² × 40 Ω × 30 s = 25 × 40 × 30 = 30000 J.
Q12Power & resistancehard
An electric bulb is rated 100 W, 250 V. What is the resistance of its filament at this rating?
Resistance R = V²/P = (250 V)² / (100 W) = 62500 / 100 = 625 Ω. This uses P = V²/R rearranged. Option 0 (625 Ω) is correct.
NCERT solutions & previous-year questions
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NCERT questions 6
1What is meant by saying that the potential difference between two points is 1 V?Electric Potential and Potential Difference
The potential difference (V) between two points is defined as the work done (W) in moving a unit charge (Q) from one point to the other.
V = W / Q
So, the potential difference between two points is said to be 1 volt when 1 joule of work is done to move a charge of 1 coulomb from one point to the other.
That is, 1 V = 1 J / 1 C = 1 J C-1.
2How much current will an electric bulb draw from a 220 V source, if the resistance of the bulb filament is 1200 Ω? If a heater of resistance 100 Ω is connected to the same source, how much current will it draw?Ohm's Law
Given: Source voltage V = 220 V.
Formula (Ohm's law): I = V / R
For the bulb (R = 1200 Ω):
I = 220 / 1200 = 0.18 A (approximately)
For the heater (R = 100 Ω):
I = 220 / 100 = 2.2 A
Thus the heater, having lower resistance, draws a much larger current than the bulb from the same source.
3A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10⁻⁸ Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?Resistivity and Factors Affecting Resistance
Given: diameter d = 0.5 mm, so radius r = 0.25 mm = 0.25 × 10-3 m; resistivity ρ = 1.6 × 10-8 Ω m; resistance R = 10 Ω.
Area of cross-section: A = πr2 = 3.14 × (0.25 × 10-3)2 = 1.96 × 10-7 m2
If the diameter is doubled: the area of cross-section becomes 4 times (since A ∝ d2). As R ∝ 1/A, the new resistance becomes one-fourth of the original.
New resistance = 10 / 4 = 2.5 Ω.
4An electric lamp of resistance 100 Ω, a toaster of resistance 50 Ω and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?Parallel Combination of Resistors
Given: R1 = 100 Ω, R2 = 50 Ω, R3 = 500 Ω in parallel; V = 220 V.
The electric iron takes the same current (7.04 A) from the same 220 V source. Therefore, its resistance must equal the equivalent resistance:
Riron = V / I = 220 / 7.04 = 31.25 Ω, and current through it = 7.04 A.
5Why is a parallel arrangement (rather than a series arrangement) used in domestic electric circuits?Series and Parallel Circuits
A parallel arrangement is used in domestic circuits because of the following advantages:
Each appliance gets the full supply voltage (220 V) required for its proper working.
Each appliance can be operated independently with its own switch, without affecting the others.
If one appliance is switched off or gets damaged, the others continue to work, because the circuit is not broken for them.
The total (equivalent) resistance decreases in a parallel connection, so a larger current can be drawn as required.
Different appliances that need different currents can be run together from the same source.
In a series circuit, breaking of the circuit at any point stops all appliances, and each would receive only a share of the total voltage, so series is not suitable for domestic wiring.
6An electric heater of resistance 8 Ω draws 15 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater, and the total heat produced.Heating Effect of Current (Joule's Law)
Given: R = 8 Ω, I = 15 A, t = 2 hours = 2 × 3600 = 7200 s.
Rate at which heat is developed = power:
P = I2R = (15)2 × 8 = 225 × 8 = 1800 W (i.e. 1800 J per second)
Q1Why are the coils of electric toasters and electric irons made of an alloy (such as nichrome) rather than a pure metal? CBSE 20201 mark
Coils of electric toasters and irons are made of an alloy such as nichrome because:
An alloy has higher resistivity than pure metals, so it produces more heat for the same current.
It has a high melting point and does not melt at high temperatures.
It does not oxidise (burn) readily even when red hot.
Q2Two resistors of resistance 10 Ω and 15 Ω are connected in series to a 6 V battery of negligible internal resistance. Calculate the equivalent resistance of the circuit and the current flowing through it. CBSE 20192 marks
The same current of 0.24 A flows through both resistors, as they are in series.
Q3An electric bulb is rated 220 V and 100 W. When it is operated on a 110 V supply, what will be the power consumed by the bulb? CBSE 20233 marks
Given: Rated voltage V = 220 V, rated power P = 100 W, operating voltage V' = 110 V.
Step 1 — Find the resistance of the filament (resistance stays constant):
P = V2 / R, so R = V2 / P = (220)2 / 100 = 48400 / 100 = 484 Ω
Step 2 — Find the power at 110 V:
P' = V'2 / R = (110)2 / 484 = 12100 / 484 = 25 W
Thus the bulb consumes only 25 W at 110 V, which is one-fourth of its rated power (because power varies as the square of the voltage when resistance is constant).
Q4Three resistors of 5 Ω, 10 Ω and 30 Ω are connected in parallel to a battery of 12 V having negligible internal resistance. Calculate (a) the total (equivalent) resistance of the circuit, (b) the current through each resistor, and (c) the total current drawn from the battery. CBSE 20235 marks
Given: R1 = 5 Ω, R2 = 10 Ω, R3 = 30 Ω in parallel; V = 12 V. In a parallel circuit, the potential difference across each resistor is the same and equals 12 V.
(b) Current through each resistor (using I = V / R):
I1 = 12 / 5 = 2.4 A
I2 = 12 / 10 = 1.2 A
I3 = 12 / 30 = 0.4 A
(c) Total current from the battery:
I = I1 + I2 + I3 = 2.4 + 1.2 + 0.4 = 4 A
Check: I = V / R = 12 / 3 = 4 A, which matches the sum of the branch currents. Note that the equivalent resistance (3 Ω) is smaller than the smallest individual resistance (5 Ω), as expected for a parallel combination.
