Electricity Solved Examples (Class 10 Physics)
Electricity involves the flow of charge through conductors and calculations of current, resistance, and power. Understanding Ohm's law and the relationship
TL;DR: Electricity involves the flow of charge through conductors and calculations of current, resistance, and power. Understanding Ohm's law and the relatio…
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Electricity involves the flow of charge through conductors and calculations of current, resistance, and power. Understanding Ohm's law and the relationship
Electricity — Solved Numerical Examples (Step by Step)
Example 1: A wire has a resistance of 5 ohms and a current of 2 amperes flows through it. Calculate the potential difference across the wire.
Solution: Given:
Resistance R = 5 ohms
Current I = 2 amperes
Using Ohm's law: V = I × R
V = 2 × 5
V = 10 volts
Example 2: Calculate the current flowing through a resistor of 10 ohms when a potential difference of 50 volts is applied across it.
Solution: Given:
Resistance R = 10 ohms
Potential difference V = 50 volts
Using Ohm's law: I = V / R
I = 50 / 10
I = 5 amperes
Example 3: A bulb has a power rating of 40 watts at 220 volts. Find the resistance of the bulb.
Solution: Given:
Power P = 40 watts
Voltage V = 220 volts
Using the formula: R = V^2 / P
R = (220)^2 / 40
R = 48400 / 40
R = 1210 ohms
Example 4: Three resistors of 2 ohms, 3 ohms, and 5 ohms are connected in series. Find the total resistance.
Solution: Given:
R1 = 2 ohms
R2 = 3 ohms
R3 = 5 ohms
Connected in series
For series connection: Rtotal = R1 + R2 + R3
Rtotal = 2 + 3 + 5
Rtotal = 10 ohms
Example 5: Two resistors of 6 ohms and 12 ohms are connected in parallel. Calculate the equivalent resistance.
Solution: Given:
R1 = 6 ohms
R2 = 12 ohms
Connected in parallel
For parallel connection: 1/Req = 1/R1 + 1/R2
1/Req = 1/6 + 1/12
1/Req = 2/12 + 1/12
1/Req = 3/12 = 1/4
Req = 4 ohms
Example 6: Calculate the power consumed by a device carrying 3 amperes of current through a resistance of 8 ohms.
Solution: Given:
Current I = 3 amperes
Resistance R = 8 ohms
Using the formula: P = I^2 × R
P = (3)^2 × 8
P = 9 × 8
P = 72 watts
Example 7: A device consumes 500 joules of energy in 10 seconds. Calculate the power of the device.
Solution: Given:
Energy W = 500 joules
Time t = 10 seconds
Using the formula: P = W / t
P = 500 / 10
P = 50 watts
Tips
- Always use Ohm's law V = IR when given any two electrical quantities (voltage, current, resistance).
- In series circuits, resistance adds up. In parallel circuits, the reciprocals of resistance add up.
- Power can be calculated using P = VI, P = I^2R, or P = V^2/R depending on which quantities are given.
Frequently Asked Questions
What is the difference between series and parallel circuits?
In a series circuit, all components are connected one after another in a single path, so current flows through each component sequentially. In a parallel circuit, components are connected between the same two points, so current divides among the paths. Series increases total resistance, while parallel decreases it.
How do I find resistance if I only know power and voltage?
Use the formula R = V^2 / P. For example, if P = 100 watts and V = 200 volts, then R = (200)^2 / 100 = 40000 / 100 = 400 ohms.
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