When 100 volts dc is supplied across a solenoid, a current of 1.0 amperes flows in it. When 100 volts ac is applied across the same coil, the current drops to 0.5 ampere. If the frequency of ac source is 50 Hz, then the impedance and inductance of the solenoid are
Answers
Answer:
L : 0.55H
Explanation:
Here is your answer..
In the problem we are given that :
DC voltage V dc :100DC.
DC current through the solenoid I dc :1A.
Therefore, according to Ohms’s law:
We know that:
V dc = I dc R
R=V dc /I dc
R=100/1=100Ω
We are also given that:
The AC voltage V ac =100AC.
The AC current I ac =0.5AC.
Frequency (v)of AC source =50Hz.
From the formula of an RLC circuit ,
We know that:
Z=V 0 /i 0 =√[R 2 +(X L – X C ) 2 ]
Since in this numerical we do not have capacitance ,X C =0.
The formula therefore becomes:
Z=√(R 2 +X L 2 )
Where,
Z=impedence (i.e the total effective resistance of the RL circuit)
R=resistance of DC source.
X L =Inductive reactance.
R here is 100Ω. (as calculated from above)
Z=V AC /I AC
=100/0.5=200Ω……………………………………….(I AC = 0.5A given)
We know that:
X L= ω L
=2πvL…………………………………….
.(where ω = 2πv , and v is the frequency which is given as 50 Hz)
= 2*3.14*50*L
=314L
X L 2 = 314 2 *L 2 = 98596L 2
R 2 = 100 2 = 10000Ω…………………..( value of R is calculated above as 100Ω)
Z 2 = 40000Ω
Putting the value of Z, R 2 ,X L 2 in the equation of the R-L circuit we get,
200 = √ (10000+98596L 2 )
Squaring both sides we get,
40000 = 10000 + 98596 L 2
30000 = 98596L 2
L = 0.55H