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Question : 32
Total: 35
(a) (i) Define coefficient of self-induction. Obtain an expression for self-inductance of a long solenoid of length l N
(ii) Calculate the self-inductance of a coil using the following data of obtained when an AC source of frequency(
) Hz
OR
(b) (i) With the help of a labelled diagram, describe the principle and working of an ac generator. Hence, obtain an expression for the instantaneous value of the emf generated.
(ii) The coil of an ac generator consists of 100 turns of wire, each of area0.5 m 2 100 Ω 0.8 T
(ii) Calculate the self-inductance of a coil using the following data of obtained when an AC source of frequency
| AC Source | ||
|---|---|---|
| S.No. | V(Volts) | I(A) |
| 1 | 3.0 | 0.5 |
| 2 | 6.0 | 1.0 |
| 3 | 9.0 | 1.5 |
| DC Source | ||
|---|---|---|
| S.No. | V(Volts) | I(A) |
| 1 | 4.0 | 1.0 |
| 2 | 6.0 | 1.5 |
| 3 | 8.0 | 2.0 |
OR
(b) (i) With the help of a labelled diagram, describe the principle and working of an ac generator. Hence, obtain an expression for the instantaneous value of the emf generated.
(ii) The coil of an ac generator consists of 100 turns of wire, each of area
Solution: 👈: Video Solution
(a) (i) Coefficient of self induction: Coefficient of self induction of a coil is defined as the emf induced in the coil per unit rate of change of current in the same coil.
Self-inductance of a long solenoid:
Length of the solenoid= l
Total number of turns= N
Area of cross section= A
Current flowing through the solenoid= i
Magnetic flux associated with each turn= µ 0 µ r n i A
Magnetic flux at any point on the axis
= µ 0 µ r (
) i
Magnetic flux associated with each turn
= µ 0 µ r (
) i A
Total magnetic flux associated with the solenoid= ϕ
= µ 0 µ r (
) i A N
= µ 0 µ r (
) i A
Self inductance of the solenoid= L =
= µ 0 µ r (
) A Henry
(ii) Calculation of self inductance:
X L = √ Z 2 − R 2
Or, 2 π f L = √ 6 2 − 4 2
Or, 2 π f L = √ 20
Or, 2 π ×
× L = √ 20
∴ L =
√ 5 Henry
OR
(b) (i) Refer the answer of Q28 of CBSE 12 DELHI, Set-1.
(ii) Given, N = 100
A = 0.5 m 2
ω = 60 radians ∕ second
B = 0.8 T
r = 100 Ω
Maximum emf generated = NBA ω
Or, Maximum emf generated= 100 × 0.8 × 0.5 × 60
∴ = E 0 = 2400 V
Maximum current= I 0 =
=
= 24 A
Power dissipation= E 0 I 0 = 2400 × 24 = 57600
Self-inductance of a long solenoid:
Length of the solenoid
Total number of turns
Area of cross section
Current flowing through the solenoid
Magnetic flux associated with each turn
Magnetic flux at any point on the axis
Magnetic flux associated with each turn
Total magnetic flux associated with the solenoid
Self inductance of the solenoid
(ii) Calculation of self inductance:
| DC SOURCE | ||||
|---|---|---|---|---|
| S. No. | V(Volts) | I(Ampere) | Resistance (Ohms) | Average resistance value (R) |
| 1 | 4.0 | 1.0 | 4.0 | |
| 2 | 6.0 | 2.0 | 4.0 | |
| 3 | 8.0 | 2.0 | 4.0 | |
| AC SOURCE | ||||
|---|---|---|---|---|
| S. No. | V(Volts) | I(Ampere) | Impedance (Ohms) | Average Impedance value (Z) |
| 1 | 3.0 | 0.5 | 6.0 | |
| 2 | 6.0 | 1.0 | 6.0 | |
| 3 | 9.0 | 1.5 | 6.0 | |
OR
(b) (i) Refer the answer of Q28 of CBSE 12 DELHI, Set-1.
(ii)
Or, Maximum emf generated
Maximum current
Power dissipation
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