Your Progress — Chapter 4: Electricity
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MCQI. Multiple Choice Questions1 mark each
Q.1 Which of the following is correct?
✓ Answer: (B) Rate of change of charge is current.
Q.2 SI unit of resistance is
✓ Answer: (C) ohm (Ω)
Q.3 In a simple circuit, why does the bulb glow when you close the switch?
✓ Answer: (B) Closing the switch completes the circuit.
Q.4 Kilowatt hour is the unit of
✓ Answer: (C) electrical energy
FillII. Fill in the Blanks1 mark each
| # | Statement (Answer in bold) |
|---|---|
| 1 | When a circuit is open, current cannot pass through it. |
| 2 | The ratio of the potential difference to the current is known as resistance. |
| 3 | The wiring in a house consists of parallel circuits. |
| 4 | The power of an electric device is a product of voltage and current. |
| 5 | LED stands for Light Emitting Diode. |
T/FIII. True or False1 mark each
| # | Statement | Answer | Correction (if False) |
|---|---|---|---|
| 1 | Ohm’s law states the relationship between power and voltage. | False | Ohm's law states the relationship between current and voltage: V = IR. |
| 2 | MCB is used to protect house hold electrical appliances. | True | MCB protects domestic circuits by tripping when excess current flows. |
| 3 | The SI unit for electric current is the coulomb. | False | The SI unit for electric current is ampere (A). Coulomb is the SI unit of electric charge. |
| 4 | One unit of electrical energy consumed is equal to 1000 kilowatt hour. | False | One unit of electrical energy consumed is equal to 1 kilowatt hour (1 kWh). |
| 5 | The effective resistance of three resistors connected in series is lesser than the lowest of the individual resistances. | False | The effective resistance of resistors connected in series is greater than the highest of the individual resistances. |
MatchIV. Match the Following1 mark each
| Column A | Column B |
|---|---|
| electric current | ampere |
| potential difference | volt |
| specific resistance | ohm metre |
| electrical power | watt |
| electrical energy | joule |
A&RV. Assertion & Reasoning2 marks each
Q.1
Assertion: Electric appliances with a metallic body have three wire connections.
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✓ Answer
Identify the purpose of the third (earth) wire: provide low-resistance path to ground to protect users. Therefore Assertion true (safety earthing required), Reason false (not about reducing heating).
Q.2
Assertion: In a simple battery circuit the point of highest potential is the positive terminal of the battery.
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✓ Answer
Assertion: True - positive terminal is the point of highest potential in a simple battery circuit. Reason: False - conventional current flows away from the highest potential, not towards it. Therefore the correct relation is: Assertion true, Reason false.
Q.3
Assertion: LED bulbs are far better than incandescent bulbs.
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✓ Answer
Assertion: True - LEDs are superior in efficiency, lifetime and heat management. Reason: True - LEDs consume less electrical power for the same light output compared to incandescent bulbs. Since lower power consumption (and higher efficiency) is the direct cause of the advantages listed, the reason correctly explains the assertion. Answer choice: a).
ShortVI. Short Answer Questions2 marks each
Q.1
Define the unit of current.
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✓ Answer
Definition: 1 ampere = 1 coulomb of charge passing a point per 1 second.
Formula: I = Q/t. Thus 1 A = 1 C / 1 s. (Symbol for ampere: A.)
Formula: I = Q/t. Thus 1 A = 1 C / 1 s. (Symbol for ampere: A.)
Q.2
What happens to the resistance, as the conductor is made thicker?
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✓ Answer
Start from R = ρL/A. For fixed ρ and L, R proportional to 1/A. For a circular wire A = pir^2, so increasing thickness (radius) increases A and thus reduces R. Conclusion: Resistance decreases when the conductor is made thicker.
Q.3
Why is tungsten metal used in bulbs, but not in fuse wires?
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✓ Answer
Tungsten metal is primarily used for the filaments in electric bulbs because of its exceptionally high melting point (around 3422 °C) and its ability to withstand high temperatures without vaporizing quickly. When electric current passes through the tungsten filament, it heats up to a very high temperature, causing it to glow and emit light. In contrast, fuse wires are designed to protect electrical circuits from overcurrents. They are made of materials with low melting points, such as alloys of lead and tin. When an excessive current flows through the fuse wire, it heats up rapidly and melts, breaking the circuit and preventing damage to appliances or potential fire hazards. Therefore, tungsten's high melting point makes it suitable for bulbs, while a low melting point is essential for fuse wires.
Q.4
Name any two devices, which are working on the heating effect of the electric current.
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✓ Answer
Two common household devices that operate based on the heating effect of electric current are the electric iron and the electric toaster. In both appliances, a high-resistance wire, typically made of a nichrome alloy (an alloy of nickel and chromium), is used. When an electric current passes through this nichrome coil, its high resistance causes it to heat up considerably due to the dissipation of electrical energy as heat (Joule heating). This generated heat is then utilized for the intended purpose: to press clothes in an electric iron or to toast bread in a toaster.
ShortVI. Short Answer Questions2 marks each
Q.1
Define electric potential and potential difference.
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✓ Answer
Electric potential at a point is the work done per unit positive charge in bringing the charge from infinity to that point. Formula: V = W/q. Its SI unit is volt; 1 V = 1 J C^-1.
Potential difference between two points is the work done per unit positive charge in moving the charge from one point to another. Formula: V = W/q, measured in volt.
Potential difference between two points is the work done per unit positive charge in moving the charge from one point to another. Formula: V = W/q, measured in volt.
Q.2
What is the role of the earth wire in domestic circuits?
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✓ Answer
The earth wire plays a crucial safety role in domestic electrical circuits, particularly for appliances with metal casings. Its primary function is to provide a low-resistance path for current to flow to the ground in the event of a fault. If the live wire accidentally comes into contact with the metal body of an appliance, the casing becomes live, posing a severe risk of electric shock to anyone who touches it. The earth wire connects the metal casing to the earth. In such a fault scenario, a large current flows from the live wire, through the casing and the earth wire, to the ground. This high fault current quickly causes the fuse to blow or the circuit breaker (MCB) to trip, immediately disconnecting the power supply and isolating the appliance, thereby preventing electric shock and protecting the user.
Q.3
State Ohm’s law.
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✓ Answer
Ohm's law states that, at constant temperature, the current through a conductor is directly proportional to the potential difference across its ends. Thus V ∝ I, or V = IR, where R is the resistance of the conductor.
Q.4
Distinguish between the resistivity and conductivity of a conductor.
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✓ Answer
Resistivity (ρ): It is the resistance of a conductor of unit length and unit area of cross-section. Formula: ρ = RA/l. Unit: ohm metre (Ω m). It measures how strongly a material opposes current.
Conductivity (σ): It is the reciprocal of resistivity. Formula: σ = 1/ρ. Unit: siemens per metre (S m^-1). It measures how easily a material allows current to pass.
Conductivity (σ): It is the reciprocal of resistivity. Formula: σ = 1/ρ. Unit: siemens per metre (S m^-1). It measures how easily a material allows current to pass.
Q.5
What connection is used in domestic appliances and why?
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✓ Answer
Parallel: Voltage across each branch = V (supply voltage). Currents through branches may differ depending on resistance. Practical consequence: independent operation and equal supply voltage to all appliances. Unit of voltage = volt (V).
LongVIII. Long Answer Questions5 marks each
Q.1
With the help of a circuit diagram derive the formula for the resultant resistance of three resistances connected: a) in series and b) in parallel
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✓ Answer
Series derivation: I is common, V = V1+V2+V3. Using V = IR and Vi = I Ri: IRs = I R1 + I R2 + I R3 -> Rs = R1 + R2 + R3.
Parallel derivation: V is common, total current I = I1 + I2 + I3 = V/R1 + V/R2 + V/R3 = V(1/R1 + 1/R2 + 1/R3). But I = V/Rp, so 1/Rp = 1/R1 + 1/R2 + 1/R3.
Parallel derivation: V is common, total current I = I1 + I2 + I3 = V/R1 + V/R2 + V/R3 = V(1/R1 + 1/R2 + 1/R3). But I = V/Rp, so 1/Rp = 1/R1 + 1/R2 + 1/R3.
Q.2
R1, R2 and R3 are connected in series. Here current through them is same, but voltage is different
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✓ Answer
V = V1 + V2 + V3. Using V = IR and Vi = I Ri: IR = I R1 + I R2 + I R3 => IR = I(R1+R2+R3) => Rs = R1 + R2 + R3.
Q.4
When a number of resistors are connected in series, their effective resistance is equal to the sum of the individual resistances.
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✓ Answer
For series of n resistors: Vtotal = V1 + V2 + ... + Vn = I(R1 + R2 + ... + Rn). Thus Rs = R1 + R2 + ... + Rn. Special case: if R1 = R2 = ... = Rn = R, then Rs = nR.
Q.5
The effective resistance in a series combination is greater than the highest of the individual resistances.
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✓ Answer
Example: Two resistors R1 and R2 in series: Rs = R1 + R2. Since R2 > 0, Rs = R1 + R2 > R1; similarly Rs > R2. Generalizes to n resistors: Rs = sum of Ri > any individual Ri.
Q.2
R1, R2 and R3 are connected in parallel. Here current through them is different, but voltage is same.
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✓ Answer
I_total = I1 + I2 + I3
But I_total = V/RP and Ii = V/Ri
So V/RP = V(1/R1 + 1/R2 + 1/R3)
Divide both sides by V => 1/RP = 1/R1 + 1/R2 + 1/R3
But I_total = V/RP and Ii = V/Ri
So V/RP = V(1/R1 + 1/R2 + 1/R3)
Divide both sides by V => 1/RP = 1/R1 + 1/R2 + 1/R3
Q.4
When a number of resistors are connected in parallel, the reciprocal of the effective resistance is equal to the sum of the reciprocals of the individual resistances
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✓ Answer
General derivation: For n resistors in parallel, voltage V is same across each. Total current I = sum of Ii = V(sum of 1/Ri). Define RP by I = V/RP so V/RP = V(sum of 1/Ri) => 1/RP = sum of 1/Ri.
Identical resistors case: If R1 = R2 = ... = Rn = R, then 1/RP = n(1/R) => RP = R/n.
Identical resistors case: If R1 = R2 = ... = Rn = R, then 1/RP = n(1/R) => RP = R/n.
Q.5
The equivalent (or) effective resistance in a parallel combination is less than the lowest of the individual resistances.
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✓ Answer
Proof for n resistors: 1/RP = 1/R1 + 1/R2 + ... + 1/Rn.
Let Rmin = min(R1,...,Rn). Each term 1/Ri ≥ 1/Rmin, so sum > 1/Rmin when n>1. Hence 1/RP > 1/Rmin => RP < Rmin.
Example (two resistors): RP = (R1 R2)/(R1 + R2). Since denominator > each R, RP is less than the smaller of R1 and R2.
Let Rmin = min(R1,...,Rn). Each term 1/Ri ≥ 1/Rmin, so sum > 1/Rmin when n>1. Hence 1/RP > 1/Rmin => RP < Rmin.
Example (two resistors): RP = (R1 R2)/(R1 + R2). Since denominator > each R, RP is less than the smaller of R1 and R2.
Q.2
a) What is meant by electric current? b) Name and define its unit. c) Which instrument is used to measure the electric current? How should it be connected in a circuit?
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✓ Answer
a) If charge Q passes a cross-section in time t, current I = Q/t.
b) 1 A = 1 C/1 s.
c) Connect ammeter in series. Practical note: ammeter should have low internal resistance and proper polarity.
b) 1 A = 1 C/1 s.
c) Connect ammeter in series. Practical note: ammeter should have low internal resistance and proper polarity.
Q.3
a) State Joule’s law of heating. b) An alloy of nickel and chromium is used as the heating element. Why? c) How does a fuse wire protect electrical appliances?
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✓ Answer
a) Derivation/use: H proportional to I^2, H proportional to R, H proportional to t => H = k I^2 R t; in SI units k = 1 so H = I^2 R t (J).
b) Nichrome properties: high resistivity -> good heating for given length, high melting point -> doesn't melt during normal use, resists oxidation -> long life.
c) Fuse operation: Fuse in series experiences full circuit current. If current > rated value, heat produced (I^2Rt) melts the fuse wire, opening the circuit and preventing damage.
b) Nichrome properties: high resistivity -> good heating for given length, high melting point -> doesn't melt during normal use, resists oxidation -> long life.
c) Fuse operation: Fuse in series experiences full circuit current. If current > rated value, heat produced (I^2Rt) melts the fuse wire, opening the circuit and preventing damage.
Q.4
Explain about domestic electric circuits. (circuit diagram not required)
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✓ Answer
- Supply from the local distribution transformer/service line is brought to the house and first passes through the energy meter (records consumption).
- After the meter there is a main switch to disconnect the house supply when needed.
- From the main switch the supply goes to the distribution board (main box) which contains fuses or MCBs to protect circuits.
- Three conductors are used: Live (L) - carries the supply (colour: red or brown), Neutral (N) - return path (colour: black or blue), Earth (E) - safety/ground (colour: green or green-yellow).
- Branch circuits are protected separately: typically lighting and low-power outlets on 5 A circuits, high-power appliances (iron, heater, air‑conditioner, refrigerator) on 15 A or appropriate rated circuits.
- On overcurrent (fault or overload) the fuse blows or the MCB trips, disconnecting the faulty circuit and protecting wiring and appliances.
- After the meter there is a main switch to disconnect the house supply when needed.
- From the main switch the supply goes to the distribution board (main box) which contains fuses or MCBs to protect circuits.
- Three conductors are used: Live (L) - carries the supply (colour: red or brown), Neutral (N) - return path (colour: black or blue), Earth (E) - safety/ground (colour: green or green-yellow).
- Branch circuits are protected separately: typically lighting and low-power outlets on 5 A circuits, high-power appliances (iron, heater, air‑conditioner, refrigerator) on 15 A or appropriate rated circuits.
- On overcurrent (fault or overload) the fuse blows or the MCB trips, disconnecting the faulty circuit and protecting wiring and appliances.
Q.5
a) What are the advantages of LED TV over the normal TV? b. List the merits of LED bulb.
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✓ Answer
a) LED TV:
- Brighter and higher contrast images.
- Slim and lightweight design.
- Uses less electrical power -> energy saving.
- Longer operational life and better reliability.
b) LED bulb:
- No filament -> less heat loss; more efficient.
- Lower power consumption compared to incandescent and often CFLs.
- Long life and instant illumination.
- Environmentally safer (no mercury) and lower maintenance.
- Brighter and higher contrast images.
- Slim and lightweight design.
- Uses less electrical power -> energy saving.
- Longer operational life and better reliability.
b) LED bulb:
- No filament -> less heat loss; more efficient.
- Lower power consumption compared to incandescent and often CFLs.
- Long life and instant illumination.
- Environmentally safer (no mercury) and lower maintenance.
NumericalVII. Numerical Problems3 marks each
Q.1
An electric iron consumes energy at the rate of 420 W when heating is at the maximum rate and 180 W when heating is at the minimum rate. The applied voltage is 220 V. What is the current in each case?
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✓ Answer
Maximum: I = P/V = 420 W / 220 V = 1.909 A approx. 1.91 A.
Minimum: I = 180 W / 220 V = 0.818 A approx. 0.818 A.
Minimum: I = 180 W / 220 V = 0.818 A approx. 0.818 A.
Q.2
A 100 watt electric bulb is used for 5 hours daily and four 60 watt bulbs are used for 5 hours daily. Calculate the energy consumed (in kWh) in the month of January.
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✓ Answer
100 W bulb: 100 W x 5 h/day x 31 days = 15,500 Wh = 15.5 kWh.
Four 60 W bulbs: total power = 4 x 60 W = 240 W. Energy = 240 W x 5 h/day x 31 days = 37,200 Wh = 37.2 kWh.
Total energy = 15.5 kWh + 37.2 kWh = 52.7 kWh.
Four 60 W bulbs: total power = 4 x 60 W = 240 W. Energy = 240 W x 5 h/day x 31 days = 37,200 Wh = 37.2 kWh.
Total energy = 15.5 kWh + 37.2 kWh = 52.7 kWh.
Q.3
A torch bulb is rated at 3 V and 600 mA. Calculate it’s
▾
✓ Answer
Given V = 3.0 V, I = 600 mA = 0.600 A.
(a) Power: P = V x I = 3.0 x 0.600 = 1.8 W.
(b) Resistance: R = V / I = 3.0 / 0.600 = 5.0 Ω.
(c) Energy for 4 h: E = P x t = 1.8 W x 4 h = 7.2 Wh = 0.0072 kWh.
(a) Power: P = V x I = 3.0 x 0.600 = 1.8 W.
(b) Resistance: R = V / I = 3.0 / 0.600 = 5.0 Ω.
(c) Energy for 4 h: E = P x t = 1.8 W x 4 h = 7.2 Wh = 0.0072 kWh.
Q.4
A piece of wire having a resistance R is cut into five equal parts.
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✓ Answer
a) Let original resistance = R (Ω). Each piece length = L/5 => resistance of each piece = R' = R/5.
b) Five equal resistances R/5 in parallel:
1/Rp = 5 x 1/(R/5) = 5 x (5/R) = 25/R => Rp = R/25 (Ω).
c) Series resistance Rs = 5 x (R/5) = R (Ω).
Ratio Rs : Rp = R : (R/25) = 25 : 1.
b) Five equal resistances R/5 in parallel:
1/Rp = 5 x 1/(R/5) = 5 x (5/R) = 25/R => Rp = R/25 (Ω).
c) Series resistance Rs = 5 x (R/5) = R (Ω).
Ratio Rs : Rp = R : (R/25) = 25 : 1.
Q.1
Two resistors when connected in parallel give the resultant resistance of 2 ohm; but when connected in series the effective resistance becomes 9 ohm. Calculate the value of each resistance.
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✓ Answer
Given: Rp = 2 Ω, Rs = R1 + R2 = 9 Ω.
Parallel formula: Rp = (R1 R2)/(R1+R2) => R1 R2 = Rp (R1+R2) = 2 9 = 18 Ω^2.
Let R1 = x => R2 = 9 - x.
x(9 - x) = 18 => 9x - x^2 = 18 => x^2 - 9x + 18 = 0.
Solve quadratic: (x - 3)(x - 6) = 0 => x = 3 Ω or 6 Ω.
Hence the resistances are 3 Ω and 6 Ω (order interchangeable).
Parallel formula: Rp = (R1 R2)/(R1+R2) => R1 R2 = Rp (R1+R2) = 2 9 = 18 Ω^2.
Let R1 = x => R2 = 9 - x.
x(9 - x) = 18 => 9x - x^2 = 18 => x^2 - 9x + 18 = 0.
Solve quadratic: (x - 3)(x - 6) = 0 => x = 3 Ω or 6 Ω.
Hence the resistances are 3 Ω and 6 Ω (order interchangeable).
Q.2
How many electrons are passing per second in a circuit in which there is a current of 5 A?
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✓ Answer
Given I = 5 A = 5 C/s, take t = 1 s => total charge Q = I t = 5 C.
Charge of one electron e = 1.6 x 10^-19 C.
Number of electrons n = Q/e = 5 / (1.6 x 10^-19) = 3.125 x 10^19 electrons (per second).
Charge of one electron e = 1.6 x 10^-19 C.
Number of electrons n = Q/e = 5 / (1.6 x 10^-19) = 3.125 x 10^19 electrons (per second).
Q.3
A piece of wire of resistance 10 ohm is drawn out so that its length is increased to three times its original length. Calculate the new resistance.
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✓ Answer
Given initial R = 10 Ω. Let initial length = L and area = A.
R = ρL/A = 10 Ω.
New length = 3L. With constant volume, new area = A/3.
R' = ρ(3L)/(A/3) = 9 (ρL/A) = 9R = 9 x 10 Ω = 90 Ω.
R = ρL/A = 10 Ω.
New length = 3L. With constant volume, new area = A/3.
R' = ρ(3L)/(A/3) = 9 (ρL/A) = 9R = 9 x 10 Ω = 90 Ω.
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