Question

Solve this:

Solve this: In an A.C. sub-circuit as shown in figure, the resistance R = 0.2 Q. At a certain instant V 0.5 V A1 - 8 AIS / = 0.5 A, and current is increasing at the rate of — - T e inductance of the coil is (2) 0.02 H (4) 05 H (3) 0.01 H 0.05 H

Answer 1

From the question it is clear that the circuit is a LR circuit.

Consider the circuit is a series LR circuit as shown in the fig.

Given R= 0.2Ω,

The change in potential at terminal A and B is given,

V₁-V₂= 0.5V= ΔV

Circuit current. i= 0.5 A

Rate of change of current, di/dt= 8 A (given)

Discharging voltage,  ΔV= e-iR,  (where e= emf in the inductor)

    =>0.5= e- 0.5×0.2

    => e= 0.5-0.1= 0.4

We also know that, e= -L(di/dt)

    0.4=-L×8

     L= 0.05H

Answer 2

Answer:

V=e+iR 0.5=e+(0.2)(0.5) e=0.4 e=L(di/dt) L=0.4/8 L=0.05H