Energy used / āĻŦā§āĻ¯ā§ąāĻšā§āϤ āĻļāĻā§āϤāĻŋ: E = 2 × 5 × 10 = 100 kWh
Cost / āĻā§°āĻ: Cost = 100 × 6 = âš600
Ans / āĻāϤā§āϤ⧰: Total cost = âš600 : āĻŽā§āĻ āĻŦāĻŋāĻĻā§āϝā§ā§ āĻŦāĻŋāϞ = âšā§Ŧā§Ļā§Ļ
Electricity (MCQs)
1. What is the SI unit of electric current ? āĻĒā§ā§°āĻļā§āύ: āĻŦā§āĻĻā§āϝā§āϤāĻŋāĻ āϧāĻžā§°āĻžā§° SI āĻāĻāĻ āĻāĻŋ ?
A. Volt | B. Coulomb | C. Ampere | D. Ohm
Ans / āĻāϤā§āϤ⧰: C
Explanation: SI unit of current is Ampere (A). āĻŦā§āϝāĻžāĻā§āϝāĻž: āϧāĻžā§°āĻžā§° SI āĻāĻāĻ āĻšā§āĻā§ āĻāĻŽā§āĻĒāĻŋāϝāĻŧāĻžā§°āĨ¤
2. The resistivity of a conductor depends on: āĻĒā§ā§°āĻļā§āύ: āĻĒā§°āĻŋāĻŦāĻžāĻšāĻā§° ā§°ā§āϧāĻā§āώāĻŽāϤāĻž āĻāĻŋāĻšā§° āĻāĻĒā§°āϤ āύāĻŋā§°ā§āĻā§° āĻā§°ā§ ?
A. Length | B. Mass | C. Temperature | D. Shape
Ans / āĻāϤā§āϤ⧰: C
Explanation: Resistivity depends on temperature. āĻŦā§āϝāĻžāĻā§āϝāĻž: ā§°ā§āϧāĻā§āώāĻŽāϤāĻž āϤāĻžāĻĒāĻŽāĻžāϤā§āϰāĻžā§° āĻāĻĒā§°āϤ āύāĻŋā§°ā§āĻā§° āĻā§°ā§āĨ¤
3. A current of 5 A flows for 2 minutes. Charge is: āĻĒā§ā§°āĻļā§āύ: ā§Ģ A āϧāĻžā§°āĻž ⧍ āĻŽāĻŋāύāĻŋāĻ āĻāϞāĻŋāϞ⧠āĻāϧāĻžāύ āĻāĻŋāĻŽāĻžāύ ?
A. 5 C | B. 12 C | C. 600 C | D. 300 C
Ans / āĻāϤā§āϤ⧰: C
Explanation: Q = I × t = 5 × 120 = 600 C āĻŦā§āϝāĻžāĻā§āϝāĻž: Q = I × t = ā§Ģ × ā§§ā§¨ā§Ļ = ā§Ŧā§Ļā§Ļ C
4. In a parallel circuit: āĻĒā§ā§°āĻļā§āύ: āϏāĻŽāĻžāύā§āϤ⧰āĻžāϞ āĻĒā§°āĻŋāĻĒāĻĨāϤ -
A. Voltage changes | B. Current changes | C. Power not constant | D. Voltage same
Ans / āĻāϤā§āϤ⧰: D
Explanation: Voltage remains same in parallel circuit. āĻŦā§āϝāĻžāĻā§āϝāĻž: āϏāĻŽāĻžāύā§āϤ⧰āĻžāϞāϤ āĻāϞā§āĻā§āĻ āĻāĻā§ āĻĨāĻžāĻā§āĨ¤
5. If wire length is doubled, resistance becomes: āĻĒā§ā§°āĻļā§āύ: āϤāĻžā§° āĻĻā§āĻāϞ āĻĻā§āĻŦāĻŋāĻā§āĻŖ āĻā§°āĻŋāϞ⧠⧰ā§āϧ āĻāĻŋāĻŽāĻžāύ āĻšāϝāĻŧ ?
A. R/2 | B. 2R | C. R/4 | D. 4R
Ans / āĻāϤā§āϤ⧰: D
Explanation: R ∝ L² → double length → 4R āĻŦā§āϝāĻžāĻā§āϝāĻž: āĻĻā§āĻāϞ āĻŦāĻĸāĻŧāĻžāϞ⧠⧰ā§āϧ ā§Ē āĻā§āĻŖ āĻšāϝāĻŧāĨ¤
6. Filament of electric bulb is made of: āĻĒā§ā§°āĻļā§āύ: āĻŦāĻžāϞā§āĻŦā§° āĻĢāĻŋāϞāĻžāĻŽā§āύā§āĻ āĻāĻŋāĻšā§ā§°ā§ āĻŦāύā§ā§ąāĻž āĻšāϝāĻŧ ?
A. Copper | B. Tungsten | C. Silver | D. Aluminium
Ans / āĻāϤā§āϤ⧰: B
Explanation: Tungsten has high melting point. āĻŦā§āϝāĻžāĻā§āϝāĻž: āĻāĻžāĻāώā§āĻā§āύ⧰ āĻāϞāύāĻžāĻāĻ āĻŦā§āĻāĻŋāĨ¤
7. Ohm’s Law is valid when: āĻĒā§ā§°āĻļā§āύ: Ohm ā§° āύāĻŋāϝāĻŧāĻŽ āĻā§āϤāĻŋāϝāĻŧāĻž āĻĒā§ā§°āϝā§āĻā§āϝ ?
A. Voltage constant | B. Current constant | C. Temperature constant | D. None
Ans / āĻāϤā§āϤ⧰: C
Explanation: Temperature must remain constant. āĻŦā§āϝāĻžāĻā§āϝāĻž: āϤāĻžāĻĒāĻŽāĻžāϤā§āϰāĻž āĻāĻā§ āĻĨāĻžāĻāĻŋāĻŦ āϞāĻžāĻā§āĨ¤
8. A 6Ω resistor with 12V battery gives current: āĻĒā§ā§°āĻļā§āύ: ā§ŦΩ ā§°ā§āϧ āĻ⧰⧠⧧⧍V āϤ āϧāĻžā§°āĻž āĻāĻŋāĻŽāĻžāύ ?
A. 1A | B. 2A | C. 0.2A | D. 1.5A
Ans / āĻāϤā§āϤ⧰: B
Explanation: I = V/R = 12/6 = 2A āĻŦā§āϝāĻžāĻā§āϝāĻž: I = ⧧⧍/ā§Ŧ = ⧍A
9. If voltage is doubled, power becomes: āĻĒā§ā§°āĻļā§āύ: āĻāϞā§āĻā§āĻ āĻĻā§āĻŦāĻŋāĻā§āĻŖ āĻā§°āĻŋāϞ⧠āĻļāĻā§āϤāĻŋ āĻāĻŋāĻŽāĻžāύ āĻšāϝāĻŧ ?
A. Half | B. Double | C. Same | D. Quadruple
Ans / āĻāϤā§āϤ⧰: D
Explanation: P ∝ V² → 2² = 4 times āĻŦā§āϝāĻžāĻā§āϝāĻž: V² āĻ āύā§āϏāĻžā§°ā§ ā§Ē āĻā§āĻŖ āĻšāϝāĻŧ
10. Best conductor of electricity is: āĻĒā§ā§°āĻļā§āύ: āĻāϤā§āϤāĻŽ āĻĒā§°āĻŋāĻŦāĻžāĻšāĻ āĻā§āύāĻā§ ?
A. Copper | B. Aluminium | C. Silver | D. Iron
Ans / āĻāϤā§āϤ⧰: C
Explanation: Silver has highest conductivity. āĻŦā§āϝāĻžāĻā§āϝāĻž: ā§°ā§āĻĒ (Silver) āϏ⧰ā§āĻŦāĻžāϧāĻŋāĻ āĻĒā§°āĻŋāĻŦāĻžāĻšā§āĨ¤
===========================================================================================
1. Ohm’s Law
A potential difference of 12 V produces a current of 2 A in a conductor. Find the resistance. āĻāĻāĻž āĻĒā§°āĻŋāĻŦāĻžāĻšāĻā§° āĻŽāĻžāĻā§ā§°ā§ ⧧⧍ V āĻŦāĻŋāĻā§ą āĻĒāĻžā§°ā§āĻĨāĻā§āϝ āĻĻāĻŋāϞ⧠⧍ A āĻŦāĻŋāĻĻā§āϝā§ā§ āϧāĻžā§°āĻž āĻĒā§ā§°āĻŦāĻžāĻšāĻŋāϤ āĻšāϝāĻŧāĨ¤ āĻĒā§ā§°āϤāĻŋā§°ā§āϧ āύāĻŋā§°ā§āĻŖāϝāĻŧ āĻā§°āĻāĨ¤
Given / āĻĻāĻŋāϝāĻŧāĻž āĻšā§āĻā§: V = 12 V, I = 2 A
Formula / āϏā§āϤā§ā§°: V = IR, R = V / I
Calculation / āĻāĻŖāύāĻž: R = 12 / 2 = 6 Ω
Ans / āĻāϤā§āϤ⧰: Resistance = 6 Ω, āĻĒā§ā§°āϤāĻŋā§°ā§āϧ = ā§Ŧ āĻāĻšā§āĻŽ
2. Current Calculation
A resistor of 10 Ω is connected to a 20 V battery. Find the current. ā§§ā§Ļ Ω āĻĒā§ā§°āϤāĻŋā§°ā§āϧ āĻĨāĻāĻž āĻāĻāĻž ā§°ā§āĻāĻŋāώā§āĻā§° ⧍ā§Ļ V āĻŦā§āϝāĻžāĻāĻžā§°ā§ā§° āϏā§āϤ⧠āϏāĻāϝā§āĻ āĻā§°āĻž āĻšā§āĻā§āĨ¤ āϧāĻžā§°āĻž āύāĻŋā§°ā§āĻŖāϝāĻŧ āĻā§°āĻāĨ¤
Given / āĻĻāĻŋāϝāĻŧāĻž āĻšā§āĻā§: V = 20 V, R = 10 Ω
Formula / āϏā§āϤā§ā§°: I = V / R
Calculation / āĻāĻŖāύāĻž: I = 20 / 10 = 2 A
Ans / āĻāϤā§āϤ⧰: Current = 2 A, āĻŦāĻŋāĻĻā§āϝā§ā§ āϧāĻžā§°āĻž = ⧍ āĻāĻŽā§āĻĒāĻŋāϝāĻŧāĻžā§°
3. Resistance in Series
Three resistors of 2 Ω, 3 Ω and 5 Ω are connected in series. Find the equivalent resistance.
⧍ Ω, ā§Š Ω āĻā§°ā§ ā§Ģ Ω āĻŽāĻžāύ⧰ āϤāĻŋāύāĻŋāĻāĻž ā§°ā§āĻāĻŋāώā§āĻā§° āĻļā§āĻāĻāϞ āϏāĻāϝā§āĻāϤ āĻāĻā§āĨ¤ āϏāĻŽāϤā§āϞā§āϝ āĻĒā§ā§°āϤāĻŋā§°ā§āϧ āύāĻŋā§°ā§āĻŖāϝāĻŧ āĻā§°āĻāĨ¤
Given / āĻĻāĻŋāϝāĻŧāĻž āĻšā§āĻā§: Râ = 2 Ω, Râ = 3 Ω, Râ = 5 Ω
Formula / āϏā§āϤā§ā§°: Râ = Râ + Râ + Râ
Calculation / āĻāĻŖāύāĻž: Râ = 2 + 3 + 5 = 10 Ω
Ans / āĻāϤā§āϤ⧰: Equivalent resistance = 10 Ω. āϏāĻŽāϤā§āϞā§āϝ āĻĒā§ā§°āϤāĻŋā§°ā§āϧ = ā§§ā§Ļ āĻāĻšā§āĻŽ
4. Electric Power
A bulb operates at 220 V and draws a current of 0.5 A. Find the electric power. āĻāĻāĻž āĻŦāĻžāϞā§āĻŦ ⧍⧍ā§Ļ V āĻŦāĻŋāĻā§ąāϤ āĻā§°ā§ ā§Ļ.ā§Ģ A āϧāĻžā§°āĻžāϤ āĻāĻžāĻŽ āĻā§°ā§āĨ¤ āĻŦāĻŋāĻĻā§āϝā§ā§ āĻā§āώāĻŽāϤāĻž āύāĻŋā§°ā§āĻŖāϝāĻŧ āĻā§°āĻāĨ¤
Given / āĻĻāĻŋāϝāĻŧāĻž āĻšā§āĻā§: V = 220 V, I = 0.5 A
Formula / āϏā§āϤā§ā§°: P = VI
Calculation / āĻāĻŖāύāĻž: P = 220 × 0.5 = 110 W
Ans / āĻāϤā§āϤ⧰: Electric power = 110 W : āĻŦāĻŋāĻĻā§āϝā§ā§ āĻā§āώāĻŽāϤāĻž = ā§§ā§§ā§Ļ ā§ąāĻžāĻ
Exam Tips (Very Important / āĻ āϤāĻŋ āĻā§ā§°ā§āϤā§āĻŦāĻĒā§ā§°ā§āĻŖ)
5. Electric Energy (Units Consumed)
A heater of 1000 W works for 2 hours. Find the energy consumed in kWh. ā§§ā§Ļā§Ļā§Ļ W āĻā§āώāĻŽāϤāĻžā§° āĻāĻāĻž āĻšāĻŋāĻāĻžā§° ⧍ āĻāĻŖā§āĻāĻž āĻāϞāĻŋāϞ⧠āĻāĻŋāĻŽāĻžāύ āĻŦāĻŋāĻĻā§āϝā§ā§ āĻļāĻā§āϤāĻŋ āĻā§°āĻ āĻšāϝāĻŧ (kWhāϤ) āύāĻŋā§°ā§āĻŖāϝāĻŧ āĻā§°āĻāĨ¤
Given / āĻĻāĻŋāϝāĻŧāĻž āĻšā§āĻā§: Power = 1000 W = 1 kW, Time = 2 h
Formula / āϏā§āϤā§ā§°: Energy = Power × Time
Calculation / āĻāĻŖāύāĻž: Energy = 1 × 2 = 2 kWh
Ans / āĻāϤā§āϤ⧰: Energy consumed = 2 units : āĻŦā§āĻ¯ā§ąāĻšā§āϤ āĻŦāĻŋāĻĻā§āϝā§ā§ āĻļāĻā§āϤāĻŋ = ⧍ āĻāĻāύāĻŋāĻ
6. Heating Effect of Electric Current
A current of 2 A flows through a resistor of 5 Ω for 10 s. Find the heat produced. ⧍ A āϧāĻžā§°āĻž ā§Ģ Ω āĻĒā§ā§°āϤāĻŋā§°ā§āϧ⧰ āĻŽāĻžāĻā§ā§°ā§ ā§§ā§Ļ āĻā§āĻā§āĻŖā§āĻĄ āĻĒā§ā§°āĻŦāĻžāĻšāĻŋāϤ āĻšāϝāĻŧāĨ¤ āĻā§āĻĒāύā§āύ āϤāĻžāĻĒ āύāĻŋā§°ā§āĻŖāϝāĻŧ āĻā§°āĻāĨ¤
Given / āĻĻāĻŋāϝāĻŧāĻž āĻšā§āĻā§: I = 2 A, R = 5 Ω, t = 10 s
Formula / āϏā§āϤā§ā§°: H = I²Rt
Calculation / āĻāĻŖāύāĻž:
H = (2)² × 5 × 10
H = 200 J
Ans / āĻāϤā§āϤ⧰: Heat produced = 200 J : āĻā§āĻĒāύā§āύ āϤāĻžāĻĒ = ⧍ā§Ļā§Ļ āĻā§āϞ
7. Electric Power (Using Current & Resistance)
A resistor of 8 Ω carries a current of 2 A. Find the power. ā§Ž Ω āĻĒā§ā§°āϤāĻŋā§°ā§āϧ⧰ āĻŽāĻžāĻā§ā§°ā§ ⧍ A āϧāĻžā§°āĻž āĻĒā§ā§°āĻŦāĻžāĻšāĻŋāϤ āĻšāϝāĻŧāĨ¤ āĻā§āώāĻŽāϤāĻž āύāĻŋā§°ā§āĻŖāϝāĻŧ āĻā§°āĻāĨ¤
Given / āĻĻāĻŋāϝāĻŧāĻž āĻšā§āĻā§: I = 2 A, R = 8 Ω
Formula / āϏā§āϤā§ā§°: P = I²R
Calculation / āĻāĻŖāύāĻž: P = (2)² × 8 = 32 W
Ans / āĻāϤā§āϤ⧰: Power = 32 W : āĻā§āώāĻŽāϤāĻž = ā§Šā§¨ ā§ąāĻžāĻ
8. Electric Power (Using Voltage & Resistance)
A bulb has resistance 100 Ω and works at 220 V. Find power. āĻāĻāĻž āĻŦāĻžāϞā§āĻŦā§° āĻĒā§ā§°āϤāĻŋā§°ā§āϧ ā§§ā§Ļā§Ļ Ω āĻā§°ā§ āĻŦāĻŋāĻā§ą ⧍⧍ā§Ļ VāĨ¤ āĻā§āώāĻŽāϤāĻž āύāĻŋā§°ā§āĻŖāϝāĻŧ āĻā§°āĻāĨ¤
Given / āĻĻāĻŋāϝāĻŧāĻž āĻšā§āĻā§: V = 220 V, R = 100 Ω
Formula / āϏā§āϤā§ā§°: P = V² / R
Calculation / āĻāĻŖāύāĻž:
P = (220)² / 100
P = 484 W
Ans / āĻāϤā§āϤ⧰: Power = 484 W : āĻā§āώāĻŽāϤāĻž = ā§Ēā§Žā§Ē ā§ąāĻžāĻ
10. Cost of Electricity (Electricity Bill)
A 2 kW heater runs for 5 hours daily for 10 days.Cost of electricity = âš6 per unit. Find total cost. ⧍ kW āĻšāĻŋāĻāĻžā§° āĻāĻāĻž āĻĻāĻŋāύ⧠ā§Ģ āĻāĻŖā§āĻāĻž, ā§§ā§Ļ āĻĻāĻŋāύ āĻāϞāĻŋāϞ⧠āĻŦāĻŋāĻĻā§āϝā§ā§ āĻŦāĻŋāϞ āĻāĻŋāĻŽāĻžāύ āĻš'āĻŦ ? (āĻĒā§ā§°āϤāĻŋ āĻāĻāύāĻŋāĻā§° āĻŽā§āϞā§āϝ = âš6)