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Class 10 Physics: Electricity & Circuits

Electric current, potential difference, Ohm law (V = IR), resistance and resistivity factors, series and parallel equivalent resistance, Joule heating law, and electric power.

Class 10 Physics 5 Generated Practice Exercises NCERT Exercise Solutions

⚡ Key Formulas & Core Relations for Electricity & Circuits

V = I × R, R = ρ(L / A)
Series: R_s = R1 + R2 + R3
Parallel: 1/R_p = 1/R1 + 1/R2
Heat H = I²Rt = VIt = (V²/R)t
Power P = VI = I²R = V²/R (1 kWh = 3.6 × 10⁶ J)

NCERT Textbook Exercise Solutions

1 Solutions
NCERT Exercise - Q12

Textbook Exercise Problem

Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of 220 V line if the maximum allowable current is 5 A?

Step-by-Step Verified Solution:

  1. Step 1: Power of each bulb P = 10 W, Voltage V = 220 V.
  2. Step 2: Resistance of each bulb R = V² / P = (220)² / 10 = 48,400 / 10 = 4,840 Ω.
  3. Step 3: Maximum total current allowed I_max = 5 A at 220 V.
  4. Step 4: Minimum permissible equivalent resistance R_eq = V / I_max = 220 / 5 = 44 Ω.
  5. Step 5: For n identical bulbs in parallel: R_eq = R / n => 44 = 4840 / n.
  6. Step 6: n = 4840 / 44 = 110 bulbs.
Final Answer: 110 lamps can be connected in parallel safely
Key Concept / Exam Strategy: In parallel connection, total current is the sum of currents drawn by all branches.

5 AI-Generated Chapter Exercises (Foundation to HOTS)

5 Problems with Full Steps
Generated Exercise #1 Foundation
Question: An electric iron consumes energy at a rate of 840 W when heating is at the maximum rate and 360 W when the heating is at the minimum. The voltage is 220 V. Calculate the current and the resistance in each case.
Core Concept Tested: Power, voltage, current, and resistance Ohm relationships
Mentor Hint: I = P / V and R = V / I.

Step-by-Step Solution:

  1. Step 1: Maximum heating: I1 = 840 / 220 ≈ 3.82 A. R1 = 220 / 3.82 ≈ 57.60 Ω.
  2. Step 2: Minimum heating: I2 = 360 / 220 ≈ 1.64 A. R2 = 220 / 1.64 ≈ 134.15 Ω.
Verified Answer: Max: I = 3.82 A, R = 57.6 Ω; Min: I = 1.64 A, R = 134.15 Ω
Generated Exercise #2 Intermediate
Question: A wire of given material having length L and area of cross-section A has a resistance of 4 Ω. What would be the resistance of another wire of the same material having length L/2 and area of cross-section 2A?
Core Concept Tested: Proportionality of resistance to length and cross-sectional area
Mentor Hint: R = ρL/A. New wire has R' = ρ(L/2) / (2A) = (1/4)R.

Step-by-Step Solution:

  1. Step 1: Original wire: R = ρL/A = 4 Ω.
  2. Step 2: New wire length L' = L/2, area A' = 2A.
  3. Step 3: R' = ρ × (L/2) / (2A) = (1/4) × (ρL/A) = (1/4) × 4 = 1 Ω.
Verified Answer: 1 Ω
Generated Exercise #3 Intermediate
Question: Three resistors of 2 Ω, 3 Ω, and 6 Ω are given. How can they be connected to give a total resistance of: (a) 4 Ω, (b) 1 Ω?
Core Concept Tested: Hybrid series-parallel resistor configuration design
Mentor Hint: Combine 3 Ω and 6 Ω in parallel to get 2 Ω.

Step-by-Step Solution:

  1. Step 1: For (a) 4 Ω: Connect 3 Ω and 6 Ω in parallel: R_p = (3 × 6)/(3 + 6) = 18/9 = 2 Ω. Then connect this in series with the 2 Ω resistor: R_total = 2 + 2 = 4 Ω.
  2. Step 2: For (b) 1 Ω: Connect all three resistors in parallel: 1/R_p = 1/2 + 1/3 + 1/6 = (3 + 2 + 1)/6 = 6/6 = 1 => R_p = 1 Ω.
Verified Answer: (a) Connect 3 Ω and 6 Ω in parallel, then in series with 2 Ω. (b) Connect all three in parallel.
Generated Exercise #4 JEE-NEET Foundation
Question: An electric refrigerator rated 400 W operates 8 hours/day and a TV rated 100 W operates 6 hours/day. What is the cost of energy to operate them for 30 days at ₹ 6.00 per kWh?
Core Concept Tested: Commercial electricity bill billing units calculation in kWh
Mentor Hint: Total Energy (kWh) = (Watts × hours/day × 30) / 1000.

Step-by-Step Solution:

  1. Step 1: Refrigerator daily energy = 400 W × 8 h = 3200 Wh = 3.2 kWh.
  2. Step 2: TV daily energy = 100 W × 6 h = 600 Wh = 0.6 kWh.
  3. Step 3: Total daily energy = 3.2 + 0.6 = 3.8 kWh.
  4. Step 4: Total energy for 30 days = 3.8 × 30 = 114 kWh (Units).
  5. Step 5: Total cost = 114 × ₹ 6.00 = ₹ 684.00.
Verified Answer: ₹ 684.00
Generated Exercise #5 Advanced HOTS
Question: A cylindrical wire is stretched to double its original length without changing its volume or material. Calculate the percentage increase in its electrical resistance.
Core Concept Tested: Volume conservation and quadratic resistance scaling during wire stretching
Mentor Hint: Volume V = A × L is constant. When L -> 2L, A -> A/2.

Step-by-Step Solution:

  1. Step 1: Since volume is conserved: A1 × L1 = A2 × L2. Given L2 = 2L1 => A2 = A1 / 2.
  2. Step 2: Initial resistance R1 = ρL1 / A1.
  3. Step 3: New resistance R2 = ρ(2L1) / (A1/2) = 4 × (ρL1 / A1) = 4R1.
  4. Step 4: Percentage increase = [(R2 - R1) / R1] × 100% = [(4R1 - R1) / R1] × 100% = 300%.
Verified Answer: 300% increase (New resistance is 4 times original)