Electricity – Numerical Questions


1. Ohm’s Law


A potential difference of 12 V produces a current of 2 A in a conductor. Find the resistance. āĻāϟāĻž āĻĒā§°āĻŋāĻŦāĻžāĻšāϕ⧰ āĻŽāĻžāĻœā§‡ā§°ā§‡ ⧧⧍ V āĻŦāĻŋāĻ­ā§ą āĻĒāĻžā§°ā§āĻĨāĻ•ā§āϝ āĻĻāĻŋāϞ⧇ ⧍ A āĻŦāĻŋāĻĻā§āĻ¯ā§ā§Ž āϧāĻžā§°āĻž āĻĒā§ā§°āĻŦāĻžāĻšāĻŋāϤ āĻšāϝāĻŧāĨ¤ āĻĒā§ā§°āϤāĻŋā§°ā§‹āϧ āύāĻŋā§°ā§āĻŖāϝāĻŧ āϕ⧰āĻ•āĨ¤


Given / āĻĻāĻŋāϝāĻŧāĻž āĻšā§ˆāϛ⧇: V = 12 V, I = 2 A


Formula / āϏ⧂āĻ¤ā§ā§°: V = IR, R = V / I


Calculation / āĻ—āĻŖāύāĻž: R = 12 / 2 = 6 Ω


Ans / āωāĻ¤ā§āϤ⧰: Resistance = 6 Ω, āĻĒā§ā§°āϤāĻŋā§°ā§‹āϧ = ā§Ŧ āĻ“āĻšā§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 āĻĨāĻ•āĻž āĻāϟāĻž ⧰⧇āϜāĻŋāĻˇā§āϟ⧰ ⧍ā§Ļ V āĻŦā§āϝāĻžāϟāĻžā§°ā§€ā§° āϏ⧈āϤ⧇ āϏāĻ‚āϝ⧋āĻ— āϕ⧰āĻž āĻšā§ˆāϛ⧇āĨ¤ āϧāĻžā§°āĻž āύāĻŋā§°ā§āĻŖāϝāĻŧ āϕ⧰āĻ•āĨ¤


Given / āĻĻāĻŋāϝāĻŧāĻž āĻšā§ˆāϛ⧇: V = 20 V, R = 10 Ω


Formula / āϏ⧂āĻ¤ā§ā§°: I = V / R


Calculation / āĻ—āĻŖāύāĻž: I = 20 / 10 = 2 A


Ans / āωāĻ¤ā§āϤ⧰: Current = 2 A, āĻŦāĻŋāĻĻā§āĻ¯ā§ā§Ž āϧāĻžā§°āĻž = ⧍ āĻāĻŽā§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 āφ⧰⧁ ā§Ģ Ω āĻŽāĻžāύ⧰ āϤāĻŋāύāĻŋāϟāĻž ⧰⧇āϜāĻŋāĻˇā§āϟ⧰ āĻļ⧃āĻ‚āĻ–āϞ āϏāĻ‚āϝ⧋āĻ—āϤ āφāϛ⧇āĨ¤ āϏāĻŽāϤ⧁āĻ˛ā§āϝ āĻĒā§ā§°āϤāĻŋā§°ā§‹āϧ āύāĻŋā§°ā§āĻŖāϝāĻŧ āϕ⧰āĻ•āĨ¤


Given / āĻĻāĻŋāϝāĻŧāĻž āĻšā§ˆāϛ⧇: R₁ = 2 Ω, R₂ = 3 Ω, R₃ = 5 Ω


Formula / āϏ⧂āĻ¤ā§ā§°: Rₛ = R₁ + R₂ + R₃


Calculation / āĻ—āĻŖāύāĻž: Rₛ = 2 + 3 + 5 = 10 Ω


Ans / āωāĻ¤ā§āϤ⧰: Equivalent resistance = 10 Ω. āϏāĻŽāϤ⧁āĻ˛ā§āϝ āĻĒā§ā§°āϤāĻŋā§°ā§‹āϧ = ā§§ā§Ļ āĻ“āĻšā§āĻŽ


4. Electric Power


A bulb operates at 220 V and draws a current of 0.5 A. Find the electric power. āĻāϟāĻž āĻŦāĻžāĻ˛ā§āĻŦ ⧍⧍ā§Ļ V āĻŦāĻŋāĻ­ā§ąāϤ āφ⧰⧁ ā§Ļ.ā§Ģ A āϧāĻžā§°āĻžāϤ āĻ•āĻžāĻŽ āϕ⧰⧇āĨ¤ āĻŦāĻŋāĻĻā§āĻ¯ā§ā§Ž āĻ•ā§āώāĻŽāϤāĻž āύāĻŋā§°ā§āĻŖāϝāĻŧ āϕ⧰āĻ•āĨ¤


Given / āĻĻāĻŋāϝāĻŧāĻž āĻšā§ˆāϛ⧇: V = 220 V, I = 0.5 A


Formula / āϏ⧂āĻ¤ā§ā§°: P = VI


Calculation / āĻ—āĻŖāύāĻž: P = 220 × 0.5 = 110 W


Ans / āωāĻ¤ā§āϤ⧰: Electric power = 110 W : āĻŦāĻŋāĻĻā§āĻ¯ā§ā§Ž āĻ•ā§āώāĻŽāϤāĻž = ā§§ā§§ā§Ļ ā§ąāĻžāϟ


Exam Tips (Very Important / āĻ…āϤāĻŋ āϗ⧁⧰⧁āĻ¤ā§āĻŦāĻĒā§‚ā§°ā§āĻŖ)



  • Always write the formula first

  • Power questions: Convert W to kW

  •  unit = 1 kWh (must remember)

  • Practice heating effect and electricity bill numericals daily


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 āĻ•ā§āώāĻŽāϤāĻžā§° āĻāϟāĻž āĻšāĻŋāϟāĻžā§° ⧍ āϘāĻŖā§āϟāĻž āϚāϞāĻŋāϞ⧇ āĻ•āĻŋāĻŽāĻžāύ āĻŦāĻŋāĻĻā§āĻ¯ā§ā§Ž āĻļāĻ•ā§āϤāĻŋ āĻ–ā§°āϚ āĻšāϝāĻŧ (kWhāϤ) āύāĻŋā§°ā§āĻŖāϝāĻŧ āϕ⧰āĻ•āĨ¤


Given / āĻĻāĻŋāϝāĻŧāĻž āĻšā§ˆāϛ⧇: Power = 1000 W = 1 kW, Time = 2 h


Formula / āϏ⧂āĻ¤ā§ā§°: Energy = Power × Time


Calculation / āĻ—āĻŖāύāĻž: Energy = 1 × 2 = 2 kWh


Ans / āωāĻ¤ā§āϤ⧰: Energy consumed = 2 units : āĻŦā§āĻ¯ā§ąāĻšā§ƒāϤ āĻŦāĻŋāĻĻā§āĻ¯ā§ā§Ž āĻļāĻ•ā§āϤāĻŋ = ⧍ āχāωāύāĻŋāϟ


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⧰ āĻŽāĻžāĻœā§‡ā§°ā§‡ ā§§ā§Ļ āϛ⧇āϕ⧇āĻŖā§āĻĄ āĻĒā§ā§°āĻŦāĻžāĻšāĻŋāϤ āĻšāϝāĻŧāĨ¤ āĻ‰ā§ŽāĻĒāĻ¨ā§āύ āϤāĻžāĻĒ āύāĻŋā§°ā§āĻŖāϝāĻŧ āϕ⧰āĻ•āĨ¤


Given / āĻĻāĻŋāϝāĻŧāĻž āĻšā§ˆāϛ⧇: I = 2 A, R = 5 Ω, t = 10 s


Formula / āϏ⧂āĻ¤ā§ā§°: H = I²Rt


Calculation / āĻ—āĻŖāύāĻž:
H = (2)² × 5 × 10
H = 200 J


Ans / āωāĻ¤ā§āϤ⧰: Heat produced = 200 J : āĻ‰ā§ŽāĻĒāĻ¨ā§āύ āϤāĻžāĻĒ = ⧍ā§Ļā§Ļ āϜ⧁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⧰ āĻŽāĻžāĻœā§‡ā§°ā§‡ ⧍ A āϧāĻžā§°āĻž āĻĒā§ā§°āĻŦāĻžāĻšāĻŋāϤ āĻšāϝāĻŧāĨ¤ āĻ•ā§āώāĻŽāϤāĻž āύāĻŋā§°ā§āĻŖāϝāĻŧ āϕ⧰āĻ•āĨ¤


Given / āĻĻāĻŋāϝāĻŧāĻž āĻšā§ˆāϛ⧇: I = 2 A, R = 8 Ω


Formula / āϏ⧂āĻ¤ā§ā§°: P = I²R


Calculation / āĻ—āĻŖāύāĻž: P = (2)² × 8 = 32 W


Ans / āωāĻ¤ā§āϤ⧰: Power = 32 W : āĻ•ā§āώāĻŽāϤāĻž = ā§Šā§¨ ā§ąāĻžāϟ


8. Electric Power (Using Voltage & Resistance)


A bulb has resistance 100 Ω and works at 220 V. Find power. āĻāϟāĻž āĻŦāĻžāĻ˛ā§āĻŦā§° āĻĒā§ā§°āϤāĻŋā§°ā§‹āϧ ā§§ā§Ļā§Ļ Ω āφ⧰⧁ āĻŦāĻŋāĻ­ā§ą ⧍⧍ā§Ļ VāĨ¤ āĻ•ā§āώāĻŽāϤāĻž āύāĻŋā§°ā§āĻŖāϝāĻŧ āϕ⧰āĻ•āĨ¤


Given / āĻĻāĻŋāϝāĻŧāĻž āĻšā§ˆāϛ⧇: V = 220 V, R = 100 Ω


Formula / āϏ⧂āĻ¤ā§ā§°: P = V² / R


Calculation / āĻ—āĻŖāύāĻž:
P = (220)² / 100
P = 484 W


Ans / āωāĻ¤ā§āϤ⧰: Power = 484 W : āĻ•ā§āώāĻŽāϤāĻž = ā§Ēā§Žā§Ē ā§ąāĻžāϟ


10. Cost of Electricity (Electricity Bill)


A 2 kW heater runs for 5 hours daily for 10 days.Cost of electricity = ₹6 per unit. Find total cost. ⧍ kW āĻšāĻŋāϟāĻžā§° āĻāϟāĻž āĻĻāĻŋāύ⧇ ā§Ģ āϘāĻŖā§āϟāĻž, ā§§ā§Ļ āĻĻāĻŋāύ āϚāϞāĻŋāϞ⧇ āĻŦāĻŋāĻĻā§āĻ¯ā§ā§Ž āĻŦāĻŋāϞ āĻ•āĻŋāĻŽāĻžāύ āĻš'āĻŦ ? (āĻĒā§ā§°āϤāĻŋ āχāωāύāĻŋāϟ⧰ āĻŽā§‚āĻ˛ā§āϝ = ₹6)










Energy used / āĻŦā§āĻ¯ā§ąāĻšā§ƒāϤ āĻļāĻ•ā§āϤāĻŋ: E = 2 × 5 × 10 = 100 kWh


Cost / āĻ–ā§°āϚ: Cost = 100 × 6 = ₹600


Ans / āωāĻ¤ā§āϤ⧰: Total cost = ₹600 : āĻŽā§āĻ  āĻŦāĻŋāĻĻā§āĻ¯ā§ā§Ž āĻŦāĻŋāϞ = ₹ā§Ŧā§Ļā§Ļ