Question : A current of 2A flows in a coil when connected to a 10V dc source. If the same coil is connected to a 10 V, 50/π Hz ac source a current of 1A flows in the circuit. The inductance of the coil is
a) √3/20 H
b) √3/10 H
c) √3/5 H
d) √3/2 H
Doubt by Yashika
Solution :
Case I
In DC circuit, the coil will only offer resistance and no inductance.
I = 2A
I = 2A
V = 10 V
We know,
V=IR
R = V/I
R = 10/2
R = 5 Ω
Hence, the resistance offered by the coil is 5Ω.
We know,
V=IR
R = V/I
R = 10/2
R = 5 Ω
Hence, the resistance offered by the coil is 5Ω.
Case II
In AC circuit the coil will offer both resistance and inductance to the flow of AC current.
In AC circuit the coil will offer both resistance and inductance to the flow of AC current.
ε = 10
f=50/π Hz
I = 1 A
f=50/π Hz
I = 1 A
Inductive Reactance of the coil (XL)
XL = ωL
XL=2πfL
XL=2π(50/π)L
XL=100L
Resistance of the coil (R) = 5Ω
Impedance of the circuit (Z) = ε/I
Z=10/1
Z=10Ω
Also, Impedance of LR circuit (Z)
Z=√(R2+XL2)
SBS
XL = ωL
XL=2πfL
XL=2π(50/π)L
XL=100L
Resistance of the coil (R) = 5Ω
Impedance of the circuit (Z) = ε/I
Z=10/1
Z=10Ω
Also, Impedance of LR circuit (Z)
Z=√(R2+XL2)
SBS
Z2=R2+XL2
(10)2=(5)2+(100L)2
100=25+10000L2
100-25=10000L2
75=10000L2
75/10000=L2
3/400=L2
L=√(3/400)
L=√3/20 H
(10)2=(5)2+(100L)2
100=25+10000L2
100-25=10000L2
75=10000L2
75/10000=L2
3/400=L2
L=√(3/400)
L=√3/20 H
Hence, a) would be the correct option.
