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Samacheer Kalvi 10th Science Chapter 4 Long Answer Questions — Electricity with Answers

Long answer questions (5 marks) from Samacheer Kalvi Class 10 Science Chapter 4 Electricity with detailed answers covering Ohm's law, circuits, Joule's law, and domestic wiring.

Updated 25 May 20269 min read

Chapter 4 — Electricity: Long Answer Questions


1. State Ohm's Law. Describe an experiment to verify it. State its limitations.

Ohm's Law

Statement: The current (I) flowing through a metallic conductor is directly proportional to the potential difference (V) across its ends, provided the temperature and other physical conditions remain constant.

Mathematical form:

V ∝ I V = IR R = V/I

Where R = resistance (constant for a given conductor at constant temperature). Unit: ohm (Ω).


Experiment to Verify Ohm's Law

Apparatus: Battery, rheostat (variable resistor), ammeter, voltmeter, resistor (R), connecting wires, key.

Circuit: Battery and rheostat connected in series with the resistor. Ammeter connected in series (to measure current). Voltmeter connected in parallel across the resistor (to measure voltage).

Procedure: 1. Close the key; adjust rheostat to get a small current I₁ 2. Note the current reading (I₁) and voltage reading (V₁) 3. Increase the rheostat setting to get I₂; note V₂ 4. Repeat for several different settings 5. Calculate V/I for each reading

Observation:

V (volts)I (amperes)V/I (ohms)
1.00.52.0
2.01.02.0
3.01.52.0
4.02.02.0

V/I = constant = R for all readings.

A graph of V vs I is a straight line through the origin — confirming Ohm's Law.

Conclusion: V/I = R = constant, verifying Ohm's Law.


Limitations of Ohm's Law

1. Not applicable to semiconductors: Diodes and transistors have non-linear I–V characteristics 2. Not applicable at high temperatures: Resistance of metals increases with temperature; the I–V relationship becomes non-linear 3. Not applicable to electrolytes: I–V relationship is non-linear 4. Not applicable to gas discharge tubes: Current does not follow Ohm's Law in ionised gases


2. Derive expressions for equivalent resistance in series and parallel combinations. State their practical applications.

Series Combination

Setup: Resistors R₁, R₂, R₃ connected end-to-end in a single path.

Key property: Same current (I) flows through all resistors.

Derivation:

  • Voltage across R₁: V₁ = IR₁
  • Voltage across R₂: V₂ = IR₂
  • Voltage across R₃: V₃ = IR₃
  • Total voltage: V = V₁ + V₂ + V₃ = IR₁ + IR₂ + IR₃ = I(R₁ + R₂ + R₃)
  • But V = IR_equivalent
ThereforeR_s = R₁ + R₂ + R₃

The equivalent resistance in series is the sum of all resistances. It is always greater than any individual resistance.

Applications of series circuits:

  • Old decorative lights (if one bulb failed, all went out — a disadvantage)
  • Fuse wires (in series with the circuit)
  • Resistors for current limiting

Parallel Combination

Setup: All resistors connected between the same two points (multiple paths).

Key property: Same voltage (V) across all resistors.

Derivation:

  • Current through R₁: I₁ = V/R₁
  • Current through R₂: I₂ = V/R₂
  • Current through R₃: I₃ = V/R₃
  • Total current: I = I₁ + I₂ + I₃ = V/R₁ + V/R₂ + V/R₃ = V(1/R₁ + 1/R₂ + 1/R₃)
  • But I = V/R_equivalent
Therefore1/R_p = 1/R₁ + 1/R₂ + 1/R₃

The equivalent resistance in parallel is less than the smallest individual resistance.

Applications of parallel circuits:

  • Household wiring — every appliance gets the full supply voltage; failure of one does not affect others
  • Car electrical systems
  • When multiple paths for current are needed

3. State and explain Joule's Law of Heating. Derive the expression Q = I²Rt.

Joule's Law

Statement: The amount of heat produced in a conductor is: 1. Directly proportional to the square of the current (I²) 2. Directly proportional to the resistance (R) 3. Directly proportional to the time (t) for which current flows

Mathematical form:

Q = I²Rt

Where Q = heat produced (J), I = current (A), R = resistance (Ω), t = time (s)


Derivation

When a current I flows through a resistor R for time t:

The work done by the electric current (= electrical energy supplied) = charge × potential difference

  • Charge flowing in time t: q = It
  • Work done: W = qV = ItV

By Ohm's Law, V = IR:

  • W = It × IR = I²Rt

If all this electrical energy is converted to heat (in a purely resistive circuit):

Q = W = I²Rt

Alternative form (using V = IR):

Q = VIt = V²t/R

Applications of Heating Effect of Electric Current

DevicePrinciple
Electric ironNichrome wire gets hot due to I²Rt
Electric heater/room heaterNichrome coil radiates heat
Incandescent bulbTungsten filament heated to incandescence (over 2500°C)
Electric fuseThin wire melts when excessive current flows — protects circuit
Electric kettle / water heaterSubmerged heating element
ToasterNichrome coils

Why nichrome wire is used:

  • High resistivity → produces more heat for given current
  • High melting point → withstands high temperatures
  • Does not oxidise (corrode) at high temperatures

4. Describe the domestic electrical wiring system. Explain the role of fuse and earthing.

Domestic Electrical Wiring

In India, domestic supply is 220 V AC at 50 Hz (alternating current).

Three-wire system entering every house:

WireColourFunction
Live wire (L)Red or brownCarries current at 220 V
Neutral wire (N)Black or blueReturn path at 0 V (earthed at substation)
Earth wire (E)Green or yellow-greenSafety — connected to earth/ground

Distribution inside the house:

1. Main switch / meter — electricity meter measures consumption (in kWh/units) 2. Main fuse board / MCB panel — distributes to different circuits (lighting, fans, power points) 3. Separate circuits:

4. All circuits are connected in parallel so each gets 220 V and failure of one doesn't affect others

  • Lighting circuit: 5 A
  • Power circuit (appliances): 15 A

Role of Fuse

Fuse is a safety device made of a thin wire of low melting point alloy (tin-lead alloy) connected in series with the circuit.

Working
  • If current exceeds the safe limit (due to short circuit or overload), the fuse wire heats up rapidly (Joule's Law: Q = I²Rt)
  • The fuse wire melts (blows), breaking the circuit
  • This prevents damage to appliances and fire

Types:

  • Rewirable fuse: Old type; fuse wire replaced manually
  • Cartridge fuse: Enclosed in a ceramic body
  • MCB (Miniature Circuit Breaker): Modern equivalent; automatically trips and can be reset — no wire to replace

Fuse rating: Fuse must be rated slightly above the normal operating current of the circuit.


Role of Earthing

Earthing connects the metal body/casing of electrical appliances to the earth through the earth wire.

Purpose:

  • If the live wire accidentally touches the metal casing of an appliance (due to insulation failure), the metal body becomes live at 220 V
  • Without earthing, a person touching the appliance would get an electric shock (current flows through body to earth)
  • With earthing, the fault current flows harmlessly through the low-resistance earth wire to ground
  • The large fault current also blows the fuse, disconnecting the supply

Important: Earthing and fuse together provide two levels of protection — earthing provides a safe path; fuse disconnects the supply.


5. What is electric power? Derive its expressions and solve a problem on electricity billing.

Electric Power

Definition: Electric power is the rate at which electrical energy is consumed (or work is done) in an electric circuit.

Formula
P = W/t = VI = I²R = V²/R

Unit: Watt (W) = 1 J/s; larger unit: kilowatt (kW) = 1000 W

Derivation:

  • Work done in time t = charge × voltage = q × V
  • Power P = W/t = (qV)/t = (q/t) × V = IV
  • Substituting V = IR: P = I × IR = I²R
  • Substituting I = V/R: P = V × V/R = V²/R

Electricity Billing

Unit of electrical energy = 1 kWh (kilowatt-hour)

1 kWh is the energy consumed by a 1 kW appliance running for 1 hour.

1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J

Cost of electricity:

Number of units = Power (kW) × Time (hours) Cost = Number of units × Rate per unit

Worked Problem:

A household uses the following appliances daily:

  • 5 bulbs of 60 W each: 6 hours/day
  • 1 fan of 75 W: 10 hours/day
  • 1 TV of 120 W: 4 hours/day
  • 1 refrigerator of 150 W: 24 hours/day

Calculate the monthly electricity bill if the rate is ₹5 per unit.

Solution
AppliancePower (W)Hours/dayUnits/day (kWh)
5 bulbs5 × 60 = 300 W60.3 × 6 = 1.8
Fan75 W100.075 × 10 = 0.75
TV120 W40.12 × 4 = 0.48
Refrigerator150 W240.15 × 24 = 3.6

Total units per day = 1.8 + 0.75 + 0.48 + 3.6 = 6.63 kWh/day

Units per month (30 days) = 6.63 × 30 = 198.9 ≈ 199 units

Monthly bill = 199 × ₹5 = ₹995


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Frequently Asked Questions

Ohm's Law: The current flowing through a conductor is directly proportional to the potential difference across it, provided temperature and other physical conditions remain constant. V = IR, where V = voltage (V), I = current (A), R = resistance (Ω). Limitations: Ohm's Law does not apply to (1) semiconductors (diodes, transistors) — non-linear behaviour; (2) electrolytes; (3) filament bulbs at high temperatures (resistance changes with temperature); (4) gas discharge tubes.