NIOS Class 12 Physics Chapter 17 Electric Current Solutions English Medium As Per New Syllabus to each chapter is provided in the list so that you can easily browse throughout different chapters NIOS Class 12 Physics Chapter 17 Electric Current Notes in English and select need one. NIOS Class 12 Physics Solutions English Medium Download PDF. NIOS Study Material of Class 12 Physics Notes Paper Code: 312.
NIOS Class 12 Physics Chapter 17 Electric Current
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Electric Current
Chapter: 17
| Module-V: Electricity and Magnetism |
INTEXT QUESTIONS 17.1
1. (a) A current I is established in a copper wire of length l. If the length of the wire is doubled, calculate the current due to the same cell.
Ans: Given: Same cell (same potential difference V)
same material (same resistivity ρ)
From Ohm’s law: V = IR,
so I = V/R
From resistance formula: R = ρl/A
Length doubled:
Original resistance: R₁ = ρl/A
New length: l₂ = 2l
New resistance: R₂ = ρ(2l)/A
= 2 2R₁
New current: I₂ = V/R₂
= V/(2R₁)
= I₁/2
Current reduces to half (I₂ = I/2)
(b) What happens to current in an identical copper wire if the area of cross-section is decreased to half of the original value?
Ans: Original resistance: R₁ = ρl/A
New area: A₂ = A/2
New resistance: R₂ = ρl/(A/2)
= 2ρl/A
= 2R₁
New current: I₂ = V/R₂
= V/(2R₁)
= I₁/2
Current reduces to half (I₂ = I/2)
2. The resistivity of a wire of length l and area of cross-section A is 2 × 10⁻⁸ Ωm. What will be the resistivity of the same metallic wire of length 2l and area of cross-section 2A?
Ans: Resistivity (ρ) is a material property that depends only on:
(i) Nature of the material.
(ii) Temperature.
(iii) Pressure.
Resistivity does NOT depend on:
(i) Length of the wire.
(ii) Area of cross-section.
(iii) Shape of the conductor.
Since the material remains the same (copper wire), the resistivity remains unchanged.
ρ = 2 × 10⁻⁸ Ωm
3. A potential difference of 8 V is applied across the ends of a conducting wire of length 3m and area of cross-section 2cm². The resulting current in the wire is 0.15A. Calculate the resistance and the resistivity of the wire.
Ans: Given: V = 8 V
l = 3 m
A = 2 cm²
= 2 × 10⁻⁴ m²
I = 0.15 A
(i) Resistance:
Using Ohm’s law: R = V/I
R = 8/0.15
= 53.33 Ω
(ii) Resistivity:
Using: R = ρl/A
Therefore: ρ = RA/l
ρ = (53.33 × 2 × 10⁻⁴)/3
ρ = (106.66 × 10⁻⁴)/3
ρ = 3.55 × 10⁻³ Ωm
Resistance = 53.33 Ω,
Resistivity = 3.55 × 10⁻³ Ωm
4. Do all conductors obey Ohm’s law? Give examples to support your answer.
Ans: No, not all conductors obey Ohm’s law.
Ohmic Conductors: (Obey Ohm’s law – V ∝ I relationship is linear).
(i) Pure metals (copper, silver, aluminum) at constant temperature.
(ii) Metallic alloys (constantan, manganin).
(iii) Carbon resistors.
(iv) Electrolytic Anss (at low concentrations).
Non-ohmic Conductors:
(i) Semiconductor diodes.
(ii) Vacuum diodes (thermionic valves).
(iii) Transistors.
(iv) Filament bulbs (tungsten filament).
Non-ohmic devices show non-linear V-I characteristics because their resistance changes with voltage, current, or temperature. No. Examples: Ohmic – metals, alloys; Non-ohmic – diodes, transistors, bulb filaments.
5. 5 × 10¹⁷ electrons pass through a cross-section of a conducting wire per second from left to right. Determine the value and direction of current.
Ans: Given: Number of electrons per second: n = 5 × 10¹⁷ s⁻¹
Charge of electron: e = 1.6 × 10⁻¹⁹ C
Electron flow direction: Left to right
Current is the rate of flow of charge:
I = Total charge/Time
= (Number of electrons × Charge per electron)/Time
I = ne = 5 × 10¹⁷ × 1.6 × 10⁻¹⁹
I = 8.0 × 10⁻² A
= 0.08 A
Direction of current:
(i) Electrons move from left to right.
(ii) Conventional current direction is opposite to electron flow.
(iii) Therefore, current flows from right to left.
Current = 0.08 A, Direction: Right to left
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