Electric Resistance and Ohm’s Law Nootan Solutions, Nageen Prakashan
Electric Resistance and Ohm’s Law Nootan Solutions, ISC Class-12 Physics Nageen Prakashan Chapter-5 Solved Numericals. Step by step Solutions of Kumar and Mittal ISC Physics Class-12 Nageen Prakashan Numerical’s Questions. Visit official Website CISCE for detail information about ISC Board Class-12 Physics.
Electric Resistance and Ohm’s Law Nootan Solutions, ISC Class-12 Physics Nageen Prakashan Chapter-5 Solved Numericals of Kumar and Mittal
| Class: | 12 |
| Subject: | Physics |
| Chapter 3: |
Electric Resistance and Ohm’s Law |
| Board | ISC |
| Writer /Publications | Nootan / Nageen Prakashan/Kumar and Mittal |
| Topics | Solved Numericals of page 221,222,223,224 |
Definition of Electrical Conductivity
It is the inverse of specific resistance for a conductor whereas the specific resistance is the resistance of unit cube of the material of the conductor.
SI Unit of Conductivity:
The SI unit of conductivity is mhom-1
Electrical Conductivity
When a conducting substance is brought under the influence of an electric field ,free charges (e.g. free electrons in metals) move under the influence of this field in such a manner, that the current density
due to their motion is proportional to the applied electric field.
Significance of Ohm’s Law:
Ohm’s law is obeyed by many substances, but it is not a fundamental law of nature. It fails if
- V depends on I non- linearly. Example is when ρ increases with I (even if the temperature is kept fixed).
- The relation between V and I depends on the sign of V for the same absolute value of V.
- The relation between V and I is non- unique. For e.g., GaAs
Kirchhoff’s First Rule:
At any junction of several circuit elements, the sum of currents entering the junction must be equal the sum of currents leaving it.
In the above junction, current I enters it and currents I1 and I2 leave it. Then, I = I1 + I2.
This is a consequence of charge conservation and assumption that currents are steady, that is no charge piles up at the junction.
Kirchhoff’s Second Rule:
The algebraic sum of changes in potential around any closed resistor loop must be zero. This is based on the principle that electrostatic forces alone cannot do any work in a closed loop, since this work is equal to the potential difference, which is zero, if we start at one point of the loop and come back to it.
Nageen Prakashan Chapter-5 Solved Numericals of Electric Resistance and Ohm’s Law Nootan Solutions Kumar and Mittal, ISC Class-12 Physics
( Page – 221 )
Question 1
A current ………………….the wire .
Answer
Question 2
A current ………………………..conductor
(i)………………………
(ii)……………………….
Answer
( Page – 222 )
Question 3
The cross ………………the wire ?
Answer
Question 4
90 C ……………………the wire .
Answer
Question 5
Estimate ………………………numbers …………
Answer
Question 6
Calculate …………….. cross section ……………..
Answer
Question 7
A 3.0 m …………………………..copper is …………..
Answer
Question 8
Find out …………………….applied on it .
Answer
Question 9
A very …………………..conductance ?
Answer
Question 10
What length ………………………10 ohms ?
Answer
Question 11
The specific …………………cross section .
Answer
Question 12
A piece ………………………respectively .
Answer
Question 13
A thick ……………………………resistance .
Answer
Question 14
A wire …………..new wire .
Answer
Question 15
A 1 kg ……………these wires .
Answer
Question 16
Two coppers …………………….wire B .
Answer
Question 17
Two wires ………………respectively .
Answer
Question 18
A 200 V ………………………..filament .
Answer
Question 19
A rectangular ……………………………..temperature ………
Answer
Question 20
Two wires ………………………materials ?
Answer
Question 21
A 100 …………………….the Bulb .
Answer
Question 22
The maximum ………………….maximum current .
Answer
Question 23
How much ………………………..minute ?
Answer
Question 24
Compute the …………………….20 ohms ..
Answer
Question 25
A heating …………… the element .
Answer
Question 26
An electric ……………….the motors .
Answer
Question 27
An electric ……………………….the press .
Answer
Question 28
A current …………….. wire ?
Answer
Question 29
A bulb ………………filament ?
Answer
Question 30
Find the …………………… supply line .
Answer
( Page – 223 )
Question 31
Find the ……………..filament .
Answer
Question 32
Find the ……………..filament .
Answer
Question 33
A lamp illumination . find
(i)………………………
(ii)……………………….
(iii)……………………….
Answer
Question 34
In a ……………..operating .
(i)………………………
(ii)……………………….
(iii)……………………….
Answer
Question 35
We have …………….for it .
Answer
Question 36
Two cities …………..the wires .
Answer
Question 37
A 500 watt ………………liberated .
Answer
Question 38
An electric ………………….its element .
Answer
Question 39
Calculate ……………….glowing .
Answer
Question 40
Two bulbs ……………the bulbs .
Answer
Question 41
Two resistor ………………….resistor R2 .
Answer
Question 42
A 60 watts …………….. become 120/(48 +6 ) A .
Answer
Question 43
How can …………………….4 ohms .
Answer
Question 44
Join three ……………2 ohms .
Answer
Question 45
A 5 ohms ……………….find n .
Answer
Question 46
Three resistor ………….find .
(a)………………………
(b)……………………….
Answer
Question 47
Three resistor ………….find .
(a)………………………
(b)……………………….
(c)…………………
Answer
Question 48
A 150 ohms …………………mains .
Answer
Question 49
An arc ………………same current .
Answer
Question 50
A filament ………………supply .
Answer
Question 51
Each of ………………..of them ?
Answer
( Page – 224 )
Question 52
Three resistors …………………..A and B .
Answer
Question 53
Four resistors ……………………..corners .
Answer
Question 54
Calculate …………………resistors .
Answer
Question 55
Calculate …………………resistors .
Answer
Question 56
A wire …………………..it read ?
Answer
Question 57
The resistance …………………of silver ?
Answer
Question 58
The resistance ……………………temperature .
Answer
Question 59
A heating ……………………….involved is .
Answer
Question 60
The resistance ………………….carbon is ……………
Answer
Question 61
A metallic …………………..its material .
Answer
— End of Electric Resistance and Ohm’s Law Numericals :–
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