Question

When 100V D.C. is applied across a coil, a current of 1A flows through it. When 100V A.C. of frequency 50 Hz is applied to the same coil only 0.5A current flows through it. Calculate resistance, impedance and self inductance of the coil.
(Ans. R = 100Ω , Z = 200Ω , L = 0.55H)

Answer 1

heya...

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

It may help you.... ☺☺

Answer 2

Hey mate ^_^

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Answer:
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DC:

=> Voltage V dc :100DC.

=> Current through the solenoid I dc :1A.

According to Ohms’s law:

V dc = I dc R

R=V dc /I dc

R=100/1=100Ω


The AC:

=> Voltage V ac = 100AC.

=> Current I ac = 0.5AC.

Frequency of AC source =50Hz.


Z=V 0 /i 0 =√[R^ 2 +(X L – X C )^ 2 ]

R here is 100Ω. (as calculated from above)


Z=V AC /I AC

=100/0.5=2000 Ω (I AC = 0.5A given)


X L= ω L

=2πvL

= 2 × 3.14 × 50 × L

= 314L

X L ^2 = 314 ^2 *L ^2 = 98596L ^2

R ^2 = 100 ^2 = 10000Ω

Z ^2 = 40000Ω

We get,

200 = √ (10000+98596L ^2 )

Squaring both sides we get,

40000 = 10000 + 98596 L ^2

L = 0.55H


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