Most Important electricity class 10 numericals with Solutions for the JKBOSE and CBSE NCERT Books.
Electricity Class 10 Numericals with solutions
Practice numericals with solutions for CBSE Class 10 Physics
Key Electricity Formulas
Important Values:
- Charge on electron (\(e\)) = \(1.6 \times 10^{-19}\,\text{C}\)
- 1 kWh = \(3.6 \times 10^6\,\text{J}\)
- Resistivity unit = Ωm
Basic Numericals
Q1. Calculate the resistance of an electric bulb which draws 2 A current when connected to a 220 V supply.
Solution:
Q2. Calculate the current flowing through a 60 W bulb rated for 220 V.
Solution:
Q3. Find the resistance of a conductor if 10 A current flows through it when a voltage of 5 V is applied.
Solution:
Q4. Calculate the energy consumed in kWh when a 100 W electric bulb is used for 10 hours.
Solution:
Q5. A heater of 500 W is used for 5 hours. Find the electrical energy consumed.
Solution:
Q6. An electric iron of resistance 44 Ω is operated at 220 V. Find the current and power consumed.
Solution:
Q7. A wire of resistance 5 Ω is bent to form a closed circle. What is the resistance between any two diametrically opposite points?
Solution:
Q8. Find the resistance of an electric lamp if it draws 0.5 A current when connected to a 100 V supply.
Solution:
Q9. A resistor of 10 Ω is connected to a battery of 6 V. Find the current and power.
Solution:
Q10. A 200 W electric heater is used for 3 hours. Calculate the energy consumed in joules.
Solution:
Q1. A current of 0.5 A flows through a conductor when a potential difference of 2 V is applied across its ends. Calculate the resistance of the conductor.
Solution:
Q2. Calculate the current flowing through a 60 W bulb rated for 220 V.
Solution:
Intermediate Numericals
Q3. Three resistors of 2 Ω, 3 Ω and 5 Ω are connected in series to a 10 V battery. Calculate:
(a) Equivalent resistance
(b) Current through each resistor
(c) Potential difference across the 3 Ω resistor
Solution:
Q4. A copper wire has diameter 0.5 mm and resistivity \(1.6 \times 10^{-8}\,\Omega m\). What length of this wire will make a resistance of 10 Ω?
Solution:
Resistivity (\(\rho\)) = \(1.6 \times 10^{-8}\,\Omega m\), Resistance (\(R\)) = 10 Ω
Power and Energy Calculations
Q5. An electric heater rated 1500 W operates 2 hours daily. Calculate:
(a) Energy consumed per day in kWh
(b) Cost of operating it for 30 days at ₹5 per kWh
Solution:
Cost = 90 × 5 = ₹450
Q6. A 100 W bulb is used for 6 hours daily, and a 60 W fan is used for 8 hours daily. Calculate:
(a) Total energy consumed in 30 days
(b) Total cost at ₹4 per kWh
Solution:
Bulb: 0.1 × 6 = 0.6 kWh
Fan: 0.06 × 8 = 0.48 kWh
Total per day = 1.08 kWh
For 30 days = 1.08 × 30 = 32.4 kWh
Circuit Problems
Q7. Calculate the equivalent resistance between points A and B:
Solution:
Q8. In the given circuit, calculate:
(a) Total resistance
(b) Current through 6 Ω resistor
(c) Potential difference across 4 Ω resistor
Solution:
Advanced Numericals
Q9. A wire of resistance 5 Ω is stretched to double its length. Calculate its new resistance (assuming volume remains constant).
Solution:
Q10. In a house, 4 bulbs of 60 W each operate for 5 hours, 2 fans of 80 W each operate for 10 hours, and a refrigerator of 300 W operates for 24 hours daily. Calculate:
(a) Total energy consumed in 30 days
(b) Electricity bill at ₹6 per kWh
Solution:
Bulbs: \(4 \times 60 \times 5 = 1200\,\text{Wh} = 1.2\,\text{kWh}\)
Fans: \(2 \times 80 \times 10 = 1600\,\text{Wh} = 1.6\,\text{kWh}\)
Refrigerator: \(300 \times 24 = 7200\,\text{Wh} = 7.2\,\text{kWh}\)
Total per day = 1.2 + 1.6 + 7.2 = 10 kWh
For 30 days = 10 × 30 = 300 kWh
Q11. In the given circuit, calculate:
(a) Total current drawn from the battery
(b) Potential difference across 8 Ω resistor
(c) Power dissipated in 4 Ω resistor
Solution:
Total current = V/R = 12/8.67 = 1.38 A
Q12. Calculate the equivalent resistance between points A and B:
Solution:
Resistance and Resistivity
Q13. A wire of length 2 m and diameter 0.5 mm has resistance 5 Ω. Calculate:
(a) Resistivity of the material
(b) Resistance of another wire of same material with length 4 m and diameter 1 mm
Solution:
Area A = πr² = 3.14 × (2.5 × 10⁻⁴)² = 1.96 × 10⁻⁷ m²
Using R = ρl/A → ρ = RA/l = (5 × 1.96 × 10⁻⁷)/2 = 4.9 × 10⁻⁷ Ωm
Since R ∝ l/A → R’ = R × (l’/l) × (A/A’) = 5 × (4/2) × (1/4) = 2.5 Ω
Q14. The resistance of a wire is 10 Ω. If it is stretched to three times its original length, calculate its new resistance (assuming uniform cross-section and volume remains constant).
Solution:
New: l₂ = 3l, Volume constant → A₂ = A/3
R₂ = 9 × 10 = 90 Ω
Mixed Problems
Q15. An electric iron consumes energy at 1100 W when connected to 220 V. Calculate:
(a) Current drawn
(b) Resistance of the heating element
(c) Energy consumed in 2 hours
Solution:
Q16. A battery of 6 V is connected to two resistors of 3 Ω and 6 Ω in parallel. Calculate:
(a) Total current from battery
(b) Current through each resistor
(c) Power dissipated in the 6 Ω resistor
Solution:
Current through 6 Ω: I₂ = V/R = 6/6 = 1 A
Challenging Problems
Q17. In the given circuit, calculate the equivalent resistance between points A and B:
Solution:
Q18. A household uses the following appliances daily:
– 4 bulbs of 60 W for 6 hours
– 2 fans of 80 W for 10 hours
– 1 refrigerator of 300 W for 24 hours
– 1 TV of 100 W for 4 hours
Calculate the monthly bill (30 days) at ₹7 per kWh.
Solution:
Bulbs: 4 × 60 × 6 = 1440 Wh = 1.44 kWh
Fans: 2 × 80 × 10 = 1600 Wh = 1.6 kWh
Refrigerator: 300 × 24 = 7200 Wh = 7.2 kWh
TV: 100 × 4 = 400 Wh = 0.4 kWh
Total per day = 1.44 + 1.6 + 7.2 + 0.4 = 10.64 kWh
Bill amount = 319.2 × 7 = ₹2234.40
Q19. A wire of resistance R is cut into five equal parts which are then connected in parallel. Calculate the equivalent resistance of this combination.
Solution:
Q20. A circuit consists of three resistors – 2 Ω, 3 Ω and 6 Ω connected in parallel, connected to a 12 V battery. Calculate:
(a) Total resistance
(b) Current through each resistor
(c) Total power consumed
Solution:
I₁ = V/R₁ = 12/2 = 6 A
I₂ = V/R₂ = 12/3 = 4 A
I₃ = V/R₃ = 12/6 = 2 A
Electricity Numerical Problems and Solutions
25 Electricity Numericals with Solutions
Numerical 1: Electric Current
A charge of 10 C passes through a conductor in 2 seconds. Calculate the electric current flowing through the conductor.
Solution:
We know the formula for electric current:
I = Q / t
Where,
- Q = Charge = 10 C
- t = Time = 2 seconds
Substituting the values:
I = 10 C / 2 s = 5 A
The electric current flowing through the conductor is 5 Amperes.
Numerical 2: Potential Difference
Find the potential difference between two points if 15 J of work is done in moving a charge of 3 C from one point to another.
Solution:
The formula for potential difference is:
V = W / Q
Where,
- W = Work done = 15 J
- Q = Charge = 3 C
Substituting the values:
V = 15 J / 3 C = 5 V
The potential difference between the two points is 5 Volts.
Numerical 3: Ohm’s Law
A resistor of 10 Ω is connected to a potential difference of 20 V. Calculate the current flowing through the resistor.
Solution:
According to Ohm’s Law:
V = I × R
Where,
- V = Potential Difference = 20 V
- R = Resistance = 10 Ω
Rearranging the formula for current:
I = V / R
Substituting the values:
I = 20 V / 10 Ω = 2 A
The current flowing through the resistor is 2 Amperes.
Numerical 4: Electric Power
An electrical appliance operates with a current of 3 A at a voltage of 220 V. Calculate the power consumed by the appliance.
Solution:
The formula for electric power is:
P = V × I
Where,
- V = Voltage = 220 V
- I = Current = 3 A
Substituting the values:
P = 220 V × 3 A = 660 W
The power consumed by the appliance is 660 Watts.
Numerical 5: Joule’s Law of Heating Effect
Calculate the heat produced in a resistor of 5 Ω when a current of 4 A flows through it for 10 seconds.
Solution:
According to Joule’s Law of Heating:
H = I2 × R × t
Where,
- I = Current = 4 A
- R = Resistance = 5 Ω
- t = Time = 10 seconds
Substituting the values:
H = (4 A)2 × 5 Ω × 10 s = 16 × 5 × 10 = 800 J
The heat produced in the resistor is 800 Joules.
Numerical 6: Electric Current
If 5 C of charge passes through a conductor in 0.5 seconds, what is the electric current?
Solution:
The formula for electric current is:
I = Q / t
Where,
- Q = Charge = 5 C
- t = Time = 0.5 s
Substituting the values:
I = 5 C / 0.5 s = 10 A
The electric current is 10 Amperes.
Numerical 7: Potential Difference
How much work is needed to move a charge of 2 C through a potential difference of 12 V?
Solution:
The formula for work done is:
W = V × Q
Where,
- V = Potential Difference = 12 V
- Q = Charge = 2 C
Substituting the values:
W = 12 V × 2 C = 24 J
The work done is 24 Joules.
Numerical 8: Ohm’s Law
A 15 Ω resistor is connected to a battery with a potential difference of 30 V. Find the current.
Solution:
Using Ohm’s Law:
V = I × R
Where,
- V = Voltage = 30 V
- R = Resistance = 15 Ω
Rearranging the formula:
I = V / R
Substituting the values:
I = 30 V / 15 Ω = 2 A
The current through the resistor is 2 Amperes.
Numerical 9: Resistance
Calculate the resistance of a wire with a potential difference of 40 V and a current of 8 A.
Solution:
From Ohm’s Law:
R = V / I
Where,
- V = Voltage = 40 V
- I = Current = 8 A
Substituting the values:
R = 40 V / 8 A = 5 Ω
The resistance of the wire is 5 Ohms.
Numerical 10: Resistivity
A wire of resistivity 1.68 × 10-8 Ω·m has a length of 2 m and a cross-sectional area of 1 × 10-6 m2. Calculate its resistance.
Solution:
The formula for resistance is:
R = ρ × (L / A)
Where,
- ρ = Resistivity = 1.68 × 10-8 Ω·m
- L = Length = 2 m
- A = Cross-sectional area = 1 × 10-6 m2
Substituting the values:
R = (1.68 × 10-8 Ω·m) × (2 m / 1 × 10-6 m2) = 0.0336 Ω
The resistance of the wire is 0.0336 Ohms.
Numerical 11: Conductance
A conductor has a resistance of 20 Ω. Calculate its conductance.
Solution:
The formula for conductance is:
G = 1 / R
Where,
- R = Resistance = 20 Ω
Substituting the values:
G = 1 / 20 Ω = 0.05 S
The conductance of the conductor is 0.05 Siemens.
Numerical 12: Conductivity
The resistivity of a material is 5 × 10-7 Ω·m. Calculate its conductivity.
Solution:
The formula for conductivity is:
σ = 1 / ρ
Where,
- ρ = Resistivity = 5 × 10-7 Ω·m
Substituting the values:
σ = 1 / (5 × 10-7 Ω·m) = 2 × 106 S/m
The conductivity of the material is 2 × 106 Siemens per meter.
Numerical 13: Electric Power
A 60 W bulb operates with a current of 0.5 A. Calculate the voltage across the bulb.
Solution:
The formula for power is:
P = V × I
Where,
- P = Power = 60 W
- I = Current = 0.5 A
Rearranging the formula for voltage:
V = P / I
Substituting the values:
V = 60 W / 0.5 A = 120 V
The voltage across the bulb is 120 Volts.
Numerical 14: Joule’s Law of Heating
Find the heat produced in a resistor of 10 Ω when a current of 2 A flows through it for 5 seconds.
Solution:
The formula for heat produced is:
H = I2 × R × t
Where,
- I = Current = 2 A
- R = Resistance = 10 Ω
- t = Time = 5 seconds
Substituting the values:
H = (2 A)2 × 10 Ω × 5 s = 4 × 10 × 5 = 200 J
The heat produced is 200 Joules.
Numerical 15: Charge
If a current of 6 A flows through a conductor for 10 seconds, calculate the total charge passed through it.
Solution:
The formula for charge is:
Q = I × t
Where,
- I = Current = 6 A
- t = Time = 10 s
Substituting the values:
Q = 6 A × 10 s = 60 C
The total charge passed through the conductor is 60 Coulombs.
Numerical 16: Electric Power
An electric heater operates at 500 W and is connected to a 230 V supply. Calculate the current through the heater.
Solution:
Using the power formula:
P = V × I
Where,
- P = Power = 500 W
- V = Voltage = 230 V
Rearranging for current:
I = P / V
Substituting the values:
I = 500 W / 230 V = 2.17 A
The current through the heater is 2.17 Amperes.
Numerical 17: Resistivity
A 1 m long copper wire has a resistivity of 1.72 × 10-8 Ω·m and a cross-sectional area of 0.5 × 10-6 m2. Calculate the resistance of the wire.
Solution:
The formula for resistance is:
R = ρ × (L / A)
Where,
- ρ = Resistivity = 1.72 × 10-8 Ω·m
- L = Length = 1 m
- A = Cross-sectional area = 0.5 × 10-6 m2
Substituting the values:
R = (1.72 × 10-8 Ω·m) × (1 m / 0.5 × 10-6 m2) = 0.0344 Ω
The resistance of the copper wire is 0.0344 Ohms.
Numerical 18: Charge
How much charge flows through a circuit when a current of 2.5 A is maintained for 2 minutes?
Solution:
The formula for charge is:
Q = I × t
Where,
- I = Current = 2.5 A
- t = Time = 2 minutes = 120 seconds
Substituting the values:
Q = 2.5 A × 120 s = 300 C
The total charge that flows through the circuit is 300 Coulombs.
Numerical 19: Electric Power
A machine uses 750 W of power at a voltage of 250 V. Find the current flowing through the machine.
Solution:
Using the power formula:
P = V × I
Where,
- P = Power = 750 W
- V = Voltage = 250 V
Rearranging the formula:
I = P / V
Substituting the values:
I = 750 W / 250 V = 3 A
The current through the machine is 3 Amperes.
Numerical 20: Ohm’s Law
Find the voltage across a 25 Ω resistor when a current of 4 A flows through it.
Solution:
Using Ohm’s Law:
V = I × R
Where,
- I = Current = 4 A
- R = Resistance = 25 Ω
Substituting the values:
V = 4 A × 25 Ω = 100 V
The voltage across the resistor is 100 Volts.
Numerical 21: Joule’s Law of Heating
If a 15 Ω resistor carries a current of 3 A for 4 seconds, calculate the heat energy produced.
Solution:
The formula for heat produced is:
H = I2 × R × t
Where,
- I = Current = 3 A
- R = Resistance = 15 Ω
- t = Time = 4 seconds
Substituting the values:
H = (3 A)2 × 15 Ω × 4 s = 9 × 15 × 4 = 540 J
The heat energy produced is 540 Joules.
Numerical 22: Resistance
A wire of length 4 m and cross-sectional area 2 × 10-6 m2 has a resistance of 0.1 Ω. Calculate its resistivity.
Solution:
The formula for resistance is:
R = ρ × (L / A)
Rearranging the formula for resistivity:
ρ = R × (A / L)
Where,
- R = Resistance = 0.1 Ω
- L = Length = 4 m
- A = Cross-sectional area = 2 × 10-6 m2
Substituting the values:
ρ = 0.1 Ω × (2 × 10-6 m2 / 4 m) = 5 × 10-8 Ω·m
The resistivity of the wire is 5 × 10-8 Ω·m.
Numerical 23: Electric Power
An electric iron uses 600 W of power and operates at 220 V. Calculate the resistance of the iron.
Solution:
Using the power formula:
P = V × I
We can express current as:
I = P / V
Also, from Ohm’s Law:
V = I × R
Substituting I from the power formula:
V = (P / V) × R
Rearranging for resistance:
R = V2 / P
Where,
- P = Power = 600 W
- V = Voltage = 220 V
Substituting the values:
R = (220 V)2 / 600 W = 80.67 Ω
The resistance of the electric iron is 80.67 Ohms.
Numerical 24: Conductivity
The conductivity of a material is 1.25 × 107 S/m. Calculate its resistivity.
Solution:
The formula for resistivity is:
ρ = 1 / σ
Where,
- σ = Conductivity = 1.25 × 107 S/m
Substituting the values:
ρ = 1 / (1.25 × 107 S/m) = 8 × 10-8 Ω·m
The resistivity of the material is 8 × 10-8 Ω·m.
Numerical 25: Electric Current
If 12 C of charge passes through a conductor in 3 seconds, what is the electric current?
Solution:
The formula for electric current is:
I = Q / t
Where,
- Q = Charge = 12 C
- t = Time = 3 s
Substituting the values:
I = 12 C / 3 s = 4 A
The electric current is 4 Amperes.
