Home · Schooling · CBSE · Class 10 · Science · Electricity · In-text questions (page 188)

NCERT Class 10 Science Electricity: In-text questions (page 188) Solutions

Step-by-step working for every question of In-text questions (page 188) in NCERT's Class 10 Science book, chapter “Electricity”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

1. In-text (page 188) Q1

   Judge the equivalent resistance in parallel of (a) 1 Ω and 10⁶ Ω (b) 1 Ω, 10³ Ω and 10⁶ Ω.

   1. (a) The formula for equivalent resistance in parallel is 1/R = 1/R₁ + 1/R₂.
      why?

      In parallel every resistor has the same potential difference, so the currents add, giving 1/R = 1/R₁ + 1/R₂ + …; the total is less than the smallest resistance.

   2. (a) Substituting the values, 1/R = 1/1 + 1/10⁶.
   3. (a) This gives 1/R = 1 + 0.000001, which is approximately 1.
   4. (a) Therefore, R ≈ 1 Ω.
   5. (b) The formula for three resistors in parallel is 1/R = 1/R₁ + 1/R₂ + 1/R₃.
      why?

      In parallel every resistor has the same potential difference, so the currents add, giving 1/R = 1/R₁ + 1/R₂ + …; the total is less than the smallest resistance.

   6. (b) Substituting the values, 1/R = 1/1 + 1/10³ + 1/10⁶.
   7. (b) This gives 1/R = 1 + 0.001 + 0.000001, which is approximately 1.
   8. (b) Therefore, R ≈ 1 Ω.

   Answer: (a) about 1 Ω; (b) about 1 Ω

   Report an error in this solution

2. In-text (page 188) Q2

   A 100 Ω lamp, a 50 Ω toaster and a 500 Ω water filter are in parallel on 220 V. What is the resistance of an iron on the same source that takes as much current as all three, and what current does it take?

   1. The total current drawn by the three appliances in parallel is the sum of the currents through each.
   2. Current through the 100 Ω lamp is I₁ = V/R₁ = 220/100 = 2.2 A.
   3. Current through the 50 Ω toaster is I₂ = V/R₂ = 220/50 = 4.4 A.
   4. Current through the 500 Ω water filter is I₃ = V/R₃ = 220/500 = 0.44 A.
   5. Total current I = I₁ + I₂ + I₃ = 2.2 + 4.4 + 0.44 = 7.04 A.
   6. The iron takes the same current, so I_iron = 7.04 A.
   7. Resistance of the iron is R = V/I = 220/7.04 = 31.25 Ω.
      why?

      Ohm's law: at a steady temperature the current through a conductor is proportional to the potential difference across it, so V = IR (R = V/I, I = V/R).

   Answer: Resistance of the iron is 31.25 Ω and the current it takes is 7.04 A.

   Report an error in this solution

3. In-text (page 188) Q3

   What are the advantages of connecting electrical devices in parallel with the battery instead of in series?

   1. In a parallel circuit, each device is connected directly across the battery terminals, so every device receives the full voltage of the battery.
   2. In a series circuit, the battery voltage is shared among all devices, so each device gets only a fraction of the total voltage.
   3. Because each device in parallel gets the full voltage, it can operate at its rated power and brightness, unlike in series where it may be dim or weak.
   4. If one device in a parallel circuit fails or is removed, the other devices continue to work because they have separate paths to the battery.
   5. In a series circuit, if one device fails, the circuit breaks and all other devices stop working.
   6. Parallel connections allow independent control of each device using individual switches, which is convenient in household wiring.

   Answer: Devices in parallel get full battery voltage, work independently, and one failing does not affect the others.

   Report an error in this solution

4. In-text (page 188) Q4

   How can resistors of 2 Ω, 3 Ω and 6 Ω be connected to give (a) 4 Ω (b) 1 Ω?

   1. (a) Connect the 3 Ω and 6 Ω resistors in parallel.
      why?

      In parallel every resistor has the same potential difference, so the currents add, giving 1/R = 1/R₁ + 1/R₂ + …; the total is less than the smallest resistance.

   2. The equivalent resistance of this parallel combination is (3 × 6) ÷ (3 + 6) = 18 ÷ 9 = 2 Ω.
      why?

      In parallel every resistor has the same potential difference, so the currents add, giving 1/R = 1/R₁ + 1/R₂ + …; the total is less than the smallest resistance.

   3. Now connect this 2 Ω combination in series with the remaining 2 Ω resistor.
      why?

      In series the same current flows through every resistor, so the resistances add: R = R₁ + R₂ + R₃ …

   4. The total resistance is 2 Ω + 2 Ω = 4 Ω.
   5. (b) Connect all three resistors (2 Ω, 3 Ω, and 6 Ω) in parallel.
      why?

      In parallel every resistor has the same potential difference, so the currents add, giving 1/R = 1/R₁ + 1/R₂ + …; the total is less than the smallest resistance.

   6. The reciprocal of the total resistance is (1 ÷ 2) + (1 ÷ 3) + (1 ÷ 6).
   7. Taking 6 as the common denominator, this becomes (3 + 2 + 1) ÷ 6 = 6 ÷ 6 = 1.
   8. Therefore, the total resistance is 1 ÷ 1 = 1 Ω.

   Answer: (a) 4 Ω; (b) 1 Ω

   Report an error in this solution

5. In-text (page 188) Q5

   What are (a) the highest and (b) the lowest total resistance possible with coils of 4 Ω, 8 Ω, 12 Ω and 24 Ω?

   1. (a) To get the highest total resistance, connect all four coils in series.
      why?

      In series the same current flows through every resistor, so the resistances add: R = R₁ + R₂ + R₃ …

   2. The equivalent resistance in series is the sum of individual resistances: R = 4 + 8 + 12 + 24.
      why?

      In series the same current flows through every resistor, so the resistances add: R = R₁ + R₂ + R₃ …

   3. Calculating the sum gives R = 48 Ω.
   4. (b) To get the lowest total resistance, connect all four coils in parallel.
      why?

      In parallel every resistor has the same potential difference, so the currents add, giving 1/R = 1/R₁ + 1/R₂ + …; the total is less than the smallest resistance.

   5. The reciprocal of the equivalent resistance is the sum of the reciprocals of individual resistances: 1/R = 1/4 + 1/8 + 1/12 + 1/24.
   6. Finding a common denominator of 24, the sum becomes (6 + 3 + 2 + 1)/24 = 12/24.
   7. Simplifying the fraction gives 1/R = 1/2.
   8. Therefore, the equivalent resistance R = 2 Ω.

   Answer: (a) 48 Ω; (b) 2 Ω

   Report an error in this solution

← Electricity: notes, practice and all exercises