
Given that the lamp draws a current of 10A upon a 80V D.C. supply, the resistance is,
R=1080=8Ω
When connected to an AC supply with an inductor in series, as shown in the figure above, the current,
I=10A
and the voltage drop over the lamp is 80V and VL over the inductance.
Applying the rules of L−R circuit,
VL2+802=2202
⇒VL=48400−6400=210 V
The Inductive reactance of the given inductor of inductance L, for the frequency of supply f,
XL=2πfL...(1)
The voltage drop across the inductance,
VL=IXL=210
⇒XL=2πfL=IVL=10210=21Ω
Substituting values in equation (1)
L=2πf21=2π×5021=100π21=0.065H