Electricity — Class 10 Physics NCERT Solutions (Free)
Free step-by-step NCERT solutions for Class 10 Physics chapter "Electricity" — 10 important questions with detailed answers for CBSE board exam preparation.
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TL;DR: Free step-by-step NCERT solutions for Class 10 Physics chapter "Electricity" — 10 important questions with detailed answers for CBSE board exam prepar…
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Key Questions Covered:
- A current of 0.5 A is drawn by a filament of an electric bulb for 10 minutes.…
- How much work is done in moving a charge of 2 Coulombs across two points havi…
- An electric heater draws a current of 5 A when connected to a 220 V supply. W…
- A copper wire has a diameter of 0.5 mm and resistivity of 1.6 × 10⁻⁸ Ω m. Wha…
- Three resistors of 5 Ω, 8 Ω, and 12 Ω are connected in series to a 6 V batter…
- Two resistors of 6 Ω and 12 Ω are connected in parallel to a 6 V battery. Cal…
- + 4 more questions in the full chapter
Solutions Summary:
| Question | Status |
|---|---|
| A current of 0.5 A is drawn by a filament of an electric … | ✓ Solved |
| How much work is done in moving a charge of 2 Coulombs ac… | ✓ Solved |
| An electric heater draws a current of 5 A when connected … | ✓ Solved |
| A copper wire has a diameter of 0.5 mm and resistivity of… | ✓ Solved |
| Three resistors of 5 Ω, 8 Ω, and 12 Ω are connected in se… | ✓ Solved |
| Two resistors of 6 Ω and 12 Ω are connected in parallel t… | ✓ Solved |
Showing 6 of 10 questions
Q1: A current of 0.5 A is drawn by a filament of an electric bulb for 10 minutes. Find the amount of electric charge that flows through the circuit.
To find the amount of electric charge, we use the relationship between current, charge, and time.
Given:
Current (I) = 0.5 A
Time (t) = 10 minutes
First, convert the time from minutes to seconds, as the SI unit of time is seconds:
t = 10 minutes × 60 seconds/minute = 600 seconds
The formula relating current, charge, and time is:
I = Q / t
Where Q is the electric charge.
To find Q, we rearrange the formula:
Q = I × t
Now, substitute the given values:
Q = 0.5 A × 600 s
Q = 300 C
Therefore, t...
Q2: How much work is done in moving a charge of 2 Coulombs across two points having a potential difference of 12 V?
To find the work done, we use the definition of potential difference.
Given:
Charge (Q) = 2 C
Potential difference (V) = 12 V
The formula for potential difference is:
V = W / Q
Where W is the work done.
To find W, we rearrange the formula:
W = V × Q
Now, substitute the given values:
W = 12 V × 2 C
W = 24 J
Therefore, 24 Joules of work is done in moving the charge.
Q3: An electric heater draws a current of 5 A when connected to a 220 V supply. What is the resistance of the heater element?
We can find the resistance using Ohm's Law, which relates voltage, current, and resistance.
Given:
Voltage (V) = 220 V
Current (I) = 5 A
According to Ohm's Law:
V = I × R
Where R is the resistance.
To find R, we rearrange the formula:
R = V / I
Now, substitute the given values:
R = 220 V / 5 A
R = 44 Ω
Therefore, the resistance of the heater element is 44 Ohms.
Q4: A copper wire has a diameter of 0.5 mm and resistivity of 1.6 × 10⁻⁸ Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?
Part 1: Calculate the length of the wire.
Given:
Diameter (d) = 0.5 mm = 0.5 × 10⁻³ m
Resistivity (ρ) = 1.6 × 10⁻⁸ Ω m
Desired Resistance (R) = 10 Ω
First, calculate the cross-sectional area (A) of the wire.
The radius (r) = d/2 = (0.5 × 10⁻³ m) / 2 = 0.25 × 10⁻³ m
Area (A) = πr² = π (0.25 × 10⁻³ m)²
A = 3.14159 × (0.0625 × 10⁻⁶ m²)
A ≈ 0.19635 × 10⁻⁶ m²
Now, use the formula for resistance:
R = ρL / A
Where L is the length of the wire.
To find L, rearrange the formula:
L = R × A / ρ
Substit...
Q5: Three resistors of 5 Ω, 8 Ω, and 12 Ω are connected in series to a 6 V battery. Calculate: (a) The total resistance of the circuit. (b) The current flowing through the circuit. (c) The potential difference across the 8 Ω resistor.
Given:
R₁ = 5 Ω
R₂ = 8 Ω
R₃ = 12 Ω
Battery Voltage (V) = 6 V
(a) Total resistance of the circuit (Rs) for series combination:
In a series circuit, the total resistance is the sum of individual resistances.
Rs = R₁ + R₂ + R₃
Rs = 5 Ω + 8 Ω + 12 Ω
Rs = 25 Ω
(b) The current flowing through the circuit (I):
Using Ohm's Law (V = I × Rs) for the entire circuit:
I = V / Rs
I = 6 V / 25 Ω
I = 0.24 A
(c) The potential difference across the 8 Ω resistor (V₂):
In a series circuit, the current is the sam...
Q6: Two resistors of 6 Ω and 12 Ω are connected in parallel to a 6 V battery. Calculate: (a) The equivalent resistance of the parallel combination. (b) The total current drawn from the battery. (c) The current flowing through the 6 Ω resistor.
Given:
R₁ = 6 Ω
R₂ = 12 Ω
Battery Voltage (V) = 6 V
(a) The equivalent resistance of the parallel combination (Rp):
For resistors in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of individual resistances.
1/Rp = 1/R₁ + 1/R₂
1/Rp = 1/6 Ω + 1/12 Ω
1/Rp = 2/12 Ω + 1/12 Ω
1/Rp = 3/12 Ω
1/Rp = 1/4 Ω
Rp = 4 Ω
(b) The total current drawn from the battery (I_total):
Using Ohm's Law (V = I_total × Rp) for the entire circuit:
I_total = V / Rp
I_total = 6 V / 4 Ω
I_...
Showing 6 of 10 questions. Visit the full page for complete solutions.