Mastering Numerical Problems in Electricity for CBSE Class 10 Science

In CBSE Class 10 Science, Chapter 11 (Electricity) contains the highest density of mathematical numericals in the entire science syllabus. From calculating the equivalent resistance of intricate resistor bridges to finding the current in individual parallel branches, determining Joule heating in electric appliances, and computing 30-day household electricity bills, numerical mastery is essential for scoring top marks.

Many students memorize formulas like $V = IR$ and $P = VI$, yet struggle when faced with composite circuit networks or multi-step questions.

In this guide, we break down the four essential calculation archetypes in electricity and provide step-by-step solutions with examiner tips.

What You Will Learn

• Formula reference compendium for electric circuits
• Archetype 1: Mixed Series-Parallel Resistor Network Analysis
• Archetype 2: Calculating Current and Potential Drop across individual components
• Archetype 3: Joule's Law of Heating and Fuse Rating selection ($H = I^2Rt$)
• Archetype 4: Electric Power and Monthly Electricity Bills in Kilowatt-Hours (kWh)
• Units, conversions, and common arithmetic traps

1. Essential Formula Compendium

    1. Current:             I = Q / t                     (1 A = 1 C/s)
    2. Potential Drop:      V = W / Q                     (1 V = 1 J/C)
    3. Ohm's Law:           V = I × R                     (R = V / I)
    4. Resistance:          R = ρ × (l / A)               (Unit: Ω·m)
    5. Series Resistors:    Rs = R1 + R2 + R3             (Current is SAME)
    6. Parallel Resistors:  1/Rp = 1/R1 + 1/R2            (Voltage is SAME, Rp = Product/Sum)
    7. Joule's Heating:     H = I² R t = V I t = (V²/R) t (Unit: Joules)
    8. Electric Power:      P = V I = I² R = V² / R       (Unit: Watts)
    9. Commercial Energy:   E (kWh) = (Watts × Hours × Days) / 1000

2. High-Yield Solved Board Examination Problems

Solved Problem 1: Mixed Series-Parallel Network

Problem: In the circuit diagram shown below, an arrangement of five resistors is connected to a $12\text{ V}$ battery:

• $R_1 = 10\ \Omega$ and $R_2 = 40\ \Omega$ are in parallel.
• $R_3 = 30\ \Omega, R_4 = 20\ \Omega,$ and $R_5 = 60\ \Omega$ are in parallel.
• These two parallel blocks are connected in series. Calculate: (a) the total resistance of the circuit, and (b) the total current flowing in the circuit.

                  +--- R1 (10 Ω) ---+               +--- R3 (30 Ω) ---+
                  |                 |               |                 |
        Battery --+--- R2 (40 Ω) ---+---------------+--- R4 (20 Ω) ---+-- Ground
        (12 V)          Block A                     |                 |
                                                    +--- R5 (60 Ω) ---+
                                                          Block B

Solution:

1. Calculate Equivalent Resistance of Block A ($R_A$): $R_1$ and $R_2$ are in parallel: $\frac{1}{R_A} = \frac{1}{10} + \frac{1}{40} = \frac{4 + 1}{40} = \frac{5}{40} = \frac{1}{8} \implies \mathbf{R_A = 8\ \Omega}$ (Using product over sum: $\frac{10 \times 40}{10 + 40} = \frac{400}{50} = 8\ \Omega$).
2. Calculate Equivalent Resistance of Block B ($R_B$): $R_3, R_4,$ and $R_5$ are in parallel: $\frac{1}{R_B} = \frac{1}{30} + \frac{1}{20} + \frac{1}{60}$ Taking LCM of 30, 20, and 60 (which is 60): $\frac{1}{R_B} = \frac{2 + 3 + 1}{60} = \frac{6}{60} = \frac{1}{10} \implies \mathbf{R_B = 10\ \Omega}$
3. Calculate Total Circuit Resistance ($R_{\text{total}}$): Block A and Block B are connected in series: $R_{\text{total}} = R_A + R_B = 8\ \Omega + 10\ \Omega = \mathbf{18\ \Omega}$
4. Calculate Total Circuit Current ($I$): By Ohm's Law ($I = V / R_{\text{total}}$): $I = \frac{12\text{ V}}{18\ \Omega} = \frac{2}{3}\text{ A} = \mathbf{0.67\text{ Ampere}}$
5. Therefore:
   • <u>(a) The total resistance of the circuit is $18\ \Omega$</u>.
   • <u>(b) The total current flowing in the circuit is rac{2}{3} ext{ A} (or $0.67 ext{ A}$)</u>.

Solved Problem 2: Joule Heating and Energy in a Toaster

Problem: An electric heater of resistance $8\ \Omega$ draws $15\text{ A}$ from the service mains for $2\text{ hours}$. Calculate the rate at which heat is developed in the heater, and the total heat produced in Joules.

Solution:

1. Analyze Given Data:
   • Resistance $R = 8\ \Omega$.
   • Current $I = 15\text{ A}$.
   • Time $t = 2\text{ hours} = 2 \times 3600 = \mathbf{7200\text{ seconds}}$.
2. Part (a): Rate at Which Heat is Developed:

   Important: <u>The 'rate at which heat is developed' means HEAT PER SECOND, which is ELECTRIC POWER ($P = H/t$)!</u> $P = I^2 R = (15)^2 \times 8 = 225 \times 8 = \mathbf{1800\text{ Joules/second (Watts)}}$

3. Part (b): Total Heat Produced in 2 Hours: $H = P \times t = 1800\text{ J/s} \times 7200\text{ s} = \mathbf{1.296 \times 10^7\text{ Joules}}$
4. Therefore:
   • <u>The rate at which heat is developed is $1800 ext{ W}$ (or $1800 ext{ J/s}$)</u>.
   • <u>The total heat produced in 2 hours is $1.296 imes 10^7 ext{ Joules}$</u>.

Solved Problem 3: Household Monthly Electricity Bill

Problem: A household uses the following electric appliances:

1. Refrigerator of rating $400\text{ W}$ for $10\text{ hours/day}$.
2. Two electric fans of rating $80\text{ W}$ each for $12\text{ hours/day}$.
3. Six electric LED bulbs of rating $18\text{ W}$ each for $6\text{ hours/day}$. Calculate the electricity bill of the household for the month of June ($30\text{ days}$) if the cost of electrical energy is ₹$4.50\text{ per unit (kWh)}$.

Solution:

1. Calculate Daily Energy Consumption for Each Appliance:
   • Refrigerator: $400\text{ W} \times 10\text{ h} = \mathbf{4000\text{ Wh}}$.
   • 2 Fans: $(2 \times 80\text{ W}) \times 12\text{ h} = 160 \times 12 = \mathbf{1920\text{ Wh}}$.
   • 6 Bulbs: $(6 \times 18\text{ W}) \times 6\text{ h} = 108 \times 6 = \mathbf{648\text{ Wh}}$.
2. Total Energy Consumed in ONE Day: $\text{Daily Energy} = 4000 + 1920 + 648 = \mathbf{6568\text{ Wh}}$
3. Total Energy Consumed in June (30 Days): $\text{Total Wh} = 6568 \times 30 = \mathbf{197040\text{ Wh}}$
4. Convert to Commercial Units (kWh): $\text{Total Units (kWh)} = \frac{197040}{1000} = \mathbf{197.04\text{ kWh (Units)}}$
5. Calculate Total Bill Cost at ₹$4.50$ per Unit: $\text{Total Cost} = 197.04 \times 4.50 = \mathbf{₹\,886.68}$
6. Therefore, <u>the electricity bill for the month of June is ₹$886.68$</u>.

3. Summary and Examination Tips

Exam Tip: In questions asking for "Rate of heat generation", DO NOT multiply by time! Rate of heat means Power ($P = H/t = I^2R$ in Watts). Multiplying by time calculates total heat, not rate!

Common Mistake: Forgetting the number of days in specific months. If the question states "month of February in a leap year", multiply by $29$ days; for April, June, September, November, multiply by $30$ days; for January, March, May, July, August, October, December, multiply by $31$ days!