The self-inductance of a circuit is numerical | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The energy stored in an inductor of self-inductance $L$ carrying a current $I$ is given by $U = \frac{1}{2} L I^2$.If we consider the work done to

Vedclass · Accepted Answer

twice the work done in establishing the magnetic flux associated with unit current in the circuit.

Vedclass · Answer

(B) The energy stored in an inductor of self-inductance $L$ carrying a current $I$ is given by $U = \frac{1}{2} L I^2$.If we consider the work done to establish the magnetic flux,it is equivalent to the energy stored in the magnetic field.For a unit current $(I = 1 \ A)$,the energy stored is $U = \frac{1}{2} L (1)^2 = \frac{L}{2}$.Therefore,$L = 2U$.This means the self-inductance $L$ is numerically equal to twice the work done (or energy stored) in establishing the magnetic flux associated with a unit current in the circuit.

The self-inductance of a circuit is numerically equal to

Similar Questions

$A$ graph of magnetic flux ( $\phi$ ) versus current $(I)$ is shown for four inductors $A, B, C, D$. The smaller value of self-inductance is for inductor:

$A$ coil of wire of radius $r$ has $600$ turns and self-inductance of $108 \ mH$. The self-inductance of a coil with the same radius and $500$ turns is (in $mH$)

In which circuit will the current be maximum immediately after closing the switch $S$?

An average induced emf of $0.5 \,V$ appears in a coil when the current in it is changed from $10 \,A$ in one direction to $10 \,A$ in the opposite direction in $0.8 \,s$. The self-inductance of the coil is: (in $\,mH$)

When a current in the conducting coil is changed from $5 \ A$ in one direction to $5 \ A$ in the opposite direction in $0.5 \ s$,an average induced e.m.f. in the coil is $2 \ V$. The self-inductance of the coil is (in $mH$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module