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Home/CLASS 10/science/electricity
Curated PYQ Collection

Top 50 Most Repeated ELECTRICITY PYQs | CLASS 10

A curated collection of the most important questions from ELECTRICITY, fully solved with step-by-step concepts to prepare for CLASS 10.

Question #1

Practice Question

A.0.48 kWh
B.4.8 kWh
C.48 kWh
D.0.048 kWh

Concept Applied

Energy = Power × Time. Convert time to hours: 8 h. So, \(E = 60 W × 8 h = 480 Wh = 0.48 kWh\). The conversion factor to MJ is not needed for the answe...

Read Full Step-by-Step Solution →

Question #2

Practice Question

A.10 \(\Omega\)
B.12 \(\Omega\)
C.14 \(\Omega\)
D.15 \(\Omega\)

Concept Applied

First find the resistance of the parallel part: \(R_{p}=\frac{6\times3}{6+3}=2\,\Omega\). Then add the series resistor: \(R_{eq}=12\,\Omega+2\,\Omega=...

Read Full Step-by-Step Solution →

Question #3

Practice Question

A.To prevent short circuits
B.To prevent overloading
C.To provide a safe path for current to flow to the ground
D.To increase the voltage

Concept Applied

Earthing is a safety measure that provides a path for current to flow to the ground in case of a fault, preventing electrical shock....

Read Full Step-by-Step Solution →

Question #4

Practice Question

A.200 W
B.300 W
C.400 W
D.500 W

Concept Applied

Power = Energy / Time = 1.2 kWh / 4 h = 0.3 kW = 300 W....

Read Full Step-by-Step Solution →

Question #5

Practice Question

A.East
B.West
C.North
D.South

Concept Applied

With the field (first finger) pointing north→south (down) and current (second finger) upward, the thumb (force) points to the left, i.e., west....

Read Full Step-by-Step Solution →

Question #6

Practice Question

A.Regulate voltage fluctuations
B.Prevent damage due to overloading or short circuit
C.Reduce power consumption
D.Improve efficiency of appliances

Concept Applied

An electric fuse is a safety device that melts and breaks the circuit when excessive current flows due to overloading or short circuit. This prevents ...

Read Full Step-by-Step Solution →

Question #7

Practice Question

Concept Applied

Use $Q = I \times t$. $t = 120$ s, $I = 0.5$ A → $Q = 60$ C....

Read Full Step-by-Step Solution →

Question #8

Practice Question

Concept Applied

Charge $Q = I \times t = 2 \times 180 = 360$ C....

Read Full Step-by-Step Solution →

Question #9

Practice Question

A.Electric heater
B.Electric iron
C.Electric fuse
D.Electric motor

Concept Applied

Joule's heating effect refers to heat produced due to resistance in a conductor. Electric heaters, irons, and fuses use this effect. However, an elect...

Read Full Step-by-Step Solution →

Question #10

Practice Question

A.0.2 A
B.0.5 A
C.2 A
D.12 A

Concept Applied

Using Ohm's law: $V = IR$. Rearranging, $I = \frac{V}{R} = \frac{6}{12} = 0.5$ A. Common mistake: using $I = VR$ gives 72 A (not in options), or confu...

Read Full Step-by-Step Solution →

Question #11

Practice Question

A.A fundamental principle of Science.
B.A complex derivation in School Level syllabus.
C.An experimental observation.
D.A theoretical assumption.

Concept Applied

This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Electricity....

Read Full Step-by-Step Solution →

Question #12

Practice Question

A.0.48 kWh
B.4.8 kWh
C.48 kWh
D.0.048 kWh

Concept Applied

Energy = Power × Time. Convert time to hours: 8 h. So, \(E = 60 W × 8 h = 480 Wh = 0.48 kWh\). The conversion factor to MJ is not needed for the answe...

Read Full Step-by-Step Solution →

Question #13

Practice Question

A.10 \(\Omega\)
B.12 \(\Omega\)
C.14 \(\Omega\)
D.15 \(\Omega\)

Concept Applied

First find the resistance of the parallel part: \(R_{p}=\frac{6\times3}{6+3}=2\,\Omega\). Then add the series resistor: \(R_{eq}=12\,\Omega+2\,\Omega=...

Read Full Step-by-Step Solution →

Question #14

Practice Question

A.To prevent short circuits
B.To prevent overloading
C.To provide a safe path for current to flow to the ground
D.To increase the voltage

Concept Applied

Earthing is a safety measure that provides a path for current to flow to the ground in case of a fault, preventing electrical shock....

Read Full Step-by-Step Solution →

Question #15

Practice Question

A.200 W
B.300 W
C.400 W
D.500 W

Concept Applied

Power = Energy / Time = 1.2 kWh / 4 h = 0.3 kW = 300 W....

Read Full Step-by-Step Solution →

Question #16

Practice Question

A.East
B.West
C.North
D.South

Concept Applied

With the field (first finger) pointing north→south (down) and current (second finger) upward, the thumb (force) points to the left, i.e., west....

Read Full Step-by-Step Solution →

Question #17

Practice Question

A.Regulate voltage fluctuations
B.Prevent damage due to overloading or short circuit
C.Reduce power consumption
D.Improve efficiency of appliances

Concept Applied

An electric fuse is a safety device that melts and breaks the circuit when excessive current flows due to overloading or short circuit. This prevents ...

Read Full Step-by-Step Solution →

Question #18

Practice Question

Concept Applied

Use $Q = I \times t$. $t = 120$ s, $I = 0.5$ A → $Q = 60$ C....

Read Full Step-by-Step Solution →

Question #19

Practice Question

Concept Applied

Charge $Q = I \times t = 2 \times 180 = 360$ C....

Read Full Step-by-Step Solution →

Question #20

Practice Question

A.Electric heater
B.Electric iron
C.Electric fuse
D.Electric motor

Concept Applied

Joule's heating effect refers to heat produced due to resistance in a conductor. Electric heaters, irons, and fuses use this effect. However, an elect...

Read Full Step-by-Step Solution →

Question #21

Practice Question

A.0.2 A
B.0.5 A
C.2 A
D.12 A

Concept Applied

Using Ohm's law: $V = IR$. Rearranging, $I = \frac{V}{R} = \frac{6}{12} = 0.5$ A. Common mistake: using $I = VR$ gives 72 A (not in options), or confu...

Read Full Step-by-Step Solution →

Question #22

Practice Question

A.A fundamental principle of Science.
B.A complex derivation in School Level syllabus.
C.An experimental observation.
D.A theoretical assumption.

Concept Applied

This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Electricity....

Read Full Step-by-Step Solution →

Question #23

Practice Question

A.0.48 kWh
B.4.8 kWh
C.48 kWh
D.0.048 kWh

Concept Applied

Energy = Power × Time. Convert time to hours: 8 h. So, \(E = 60 W × 8 h = 480 Wh = 0.48 kWh\). The conversion factor to MJ is not needed for the answe...

Read Full Step-by-Step Solution →

Question #24

Practice Question

A.10 \(\Omega\)
B.12 \(\Omega\)
C.14 \(\Omega\)
D.15 \(\Omega\)

Concept Applied

First find the resistance of the parallel part: \(R_{p}=\frac{6\times3}{6+3}=2\,\Omega\). Then add the series resistor: \(R_{eq}=12\,\Omega+2\,\Omega=...

Read Full Step-by-Step Solution →

Question #25

Practice Question

A.To prevent short circuits
B.To prevent overloading
C.To provide a safe path for current to flow to the ground
D.To increase the voltage

Concept Applied

Earthing is a safety measure that provides a path for current to flow to the ground in case of a fault, preventing electrical shock....

Read Full Step-by-Step Solution →

Question #26

Practice Question

A.200 W
B.300 W
C.400 W
D.500 W

Concept Applied

Power = Energy / Time = 1.2 kWh / 4 h = 0.3 kW = 300 W....

Read Full Step-by-Step Solution →

Question #27

Practice Question

A.East
B.West
C.North
D.South

Concept Applied

With the field (first finger) pointing north→south (down) and current (second finger) upward, the thumb (force) points to the left, i.e., west....

Read Full Step-by-Step Solution →

Question #28

Practice Question

A.Regulate voltage fluctuations
B.Prevent damage due to overloading or short circuit
C.Reduce power consumption
D.Improve efficiency of appliances

Concept Applied

An electric fuse is a safety device that melts and breaks the circuit when excessive current flows due to overloading or short circuit. This prevents ...

Read Full Step-by-Step Solution →

Question #29

Practice Question

Concept Applied

Use $Q = I \times t$. $t = 120$ s, $I = 0.5$ A → $Q = 60$ C....

Read Full Step-by-Step Solution →

Question #30

Practice Question

Concept Applied

Charge $Q = I \times t = 2 \times 180 = 360$ C....

Read Full Step-by-Step Solution →

Question #31

Practice Question

A.Electric heater
B.Electric iron
C.Electric fuse
D.Electric motor

Concept Applied

Joule's heating effect refers to heat produced due to resistance in a conductor. Electric heaters, irons, and fuses use this effect. However, an elect...

Read Full Step-by-Step Solution →

Question #32

Practice Question

A.0.2 A
B.0.5 A
C.2 A
D.12 A

Concept Applied

Using Ohm's law: $V = IR$. Rearranging, $I = \frac{V}{R} = \frac{6}{12} = 0.5$ A. Common mistake: using $I = VR$ gives 72 A (not in options), or confu...

Read Full Step-by-Step Solution →

Question #33

Practice Question

A.A fundamental principle of Science.
B.A complex derivation in School Level syllabus.
C.An experimental observation.
D.A theoretical assumption.

Concept Applied

This is a placeholder question to ensure comprehensive syllabus coverage. The correct answer highlights the fundamental nature of Electricity....

Read Full Step-by-Step Solution →
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