Answer:
A) 30 mH
B ) 10-ohm
Explanation:
resistor = 10-ohm
Inductor = 30mH ( l )
L = inductance
R = resistance
r = internal resistance
values of the original Inductors
Note : inductor = constant time (t) case 1
inductor + 10-ohm resistor connected in series = constant time ( t/2) case2
inductor + 10-ohm resistor + 30 mH inductance in series = constant time (t) case3
From the above cases
case 1 = time constant ( t ) = L / R
case 2 = Req = R + r hence time constant t / 2 = L / R + r therefore
t = [tex]\frac{2L}{R+r}[/tex]
case 3 = Leq = L + l , Req = R + r . constant time ( t )
hence Z = [tex]\frac{L + l}{R + r}[/tex] = t
A) Inductance
To calculate inductance equate case 1 to case 3
[tex]\frac{L}{R}[/tex] = [tex]\frac{L + l}{R + r}[/tex] = L / 10 = (L + 30) / ( 10 + 10 )
= 20 L = 10 L + 300 mH
10L = 300 m H
therefore L = 30 mH
B ) The internal resistance
equate case 1 to case 2
[tex]\frac{L}{R}[/tex] = [tex]\frac{2L}{R + r}[/tex]
R + r = 2 R therefore ( r = R ) therefore internal resistance = 10-ohm
