Most Important Electricity Numericals with Solutions Class 10th essential for the JKBOSE and CBSE NCERT Books.
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.
