# Basic Electrical Engineering Questions and Answers – Ferromagnetic Cored Inductor in a DC Circuit

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This set of Basic Electrical Engineering online quiz focuses on “Ferromagnetic Cored Inductor in a DC Circuit”.

1. When a ferromagnetic core is inserted into an inductor, what happens to the flux linkage?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero

Explanation: When a ferromagnetic core is introduced into an inductor, its flux increases because the number of magnetic field lines increases due to the introduction of magnetic field within the coil.

2. What happens to the current when a ferromagnetic material is introduced within an inductor?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero

Explanation: When a ferromagnetic is introduced within an inductor, the current remains fairly constant. This is because the current does not depend on the magnetic field.

3. What is the relation between the flux and the magnetizing current when a ferromagnetic core is introduced within the inductor?
a) Directly proportional
b) Inversely proportional
c) Not proportional
d) Current is double of flux

Explanation: When a ferromagnetic core is introduced within an inductor the flux changes rapidly, whereas the current changes at the same pace. Hence the two are not proportional.

4. What happens to the effective inductance when a ferromagnetic core is introduced?
a) Increases
b) Decreases
c) Remains the same
d) Becomes zero

Explanation: The effective inductance increases when a ferromagnetic core is introduced within an inductor because of the flux increases. Inductance varies directly with the flux hence it increases.

5. A laminated steel ring is wound with 200 turns. When the magnetizing current varies between 5 and 7 A, the magnetic flux varies between 760 and 800 Wb. Calculate the inductance of the coil.
a) 40H
b) 4 H
c) 4000H
d) 0.004 H

Explanation: From the formula of incremental inductance, we know that:
L=(Change in flux/Change in current)*Number of turns
Substituting the values from the given question, we get L = 4000 H.

6. Calculate the number of turns in an inductor with a ferromagnetic core when the inductance is 4000 H, the current changes from 5A to 7A and the flux changes from 760 to 800 Wb.
a) 100
b) 200
c) 300
d) 400

Explanation: From the formula of incremental inductance, we know that:
L=(Change in flux/Change in current)*Number of turns
Substituting the values from the given question, we get N=200.

7. Calculate the change in current in an inductor having inductance 4000H, number of turns is 200 and the flux changes from 760 to 800 Wb.
a) 2A
b) 4A
c) 6A
d) 8A

Explanation: From the formula of incremental inductance, we know that:
L=(Change in flux/Change in current)*Number of turns
Substituting the values from the given question, we get a change in current = 2A.

8. Calculate the initial current in an inductor having inductance 4000 H, number of turns is 200 and the flux changes from 760 to 800 Wb. Current changes to 7A.
a) 10A
b) 2A
c) 3A
d) 5A

Explanation: From the formula of incremental inductance, we know that:
L=(Change in flux/Change in current)*Number of turns
Substituting the values from the given question, we get change in current = 2A.
Change in current = final current- initial current.
2=7-initial current.
Initial current = 5A.

9. Calculate the change in flux of an inductor having inductance 4000 H, number of turns is 200 and the current changes from 5A to 7A.
a) 20 Wb
b) 40 Wb
c) 60 Wb
d) 80 Wb

Explanation: From the formula of incremental inductance, we know that:
L=(Change in flux/Change in current)*Number of turns
Substituting the values from the given question, we get change in flux = 40 Wb.

10. Calculate the final flux in an inductor having inductance 4000 H, number of turns is 200 and the current changes from 5A to 7A. The initial flux is 760 Wb.
a) 200 Wb
b) 400 Wb
c) 600 Wb
d) 800 Wb

Explanation: From the formula of incremental inductance, we know that:
L=(Change in flux/Change in current)*Number of turns
Substituting the values from the given question, we get a change in flux = 40 Wb.
Change in flux = final flux- initial flux.
Thus final flux = 800 Wb.

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