Question

An a.c. source of voltage V and of frequency 50Hz is connected to an inductor of 2 H and negligible
resistance. A current of r.m.s value I flows in the coil. When the frequency of the voltage is changed to 400 Hz keeping the magnitude of the same,
the current is now :-
(1) 8 I in phase with V
(2) 4 I and leading by 90°from V
(3)I/4and lagging by 90° from V
(4) I/8 and lagging by 90° from V​

Answer 1

$\mathfrak{\huge{\pink{\underline{\underline{AnSwEr:-}}}}}$

⭐Actually Welcome to the Concept of the Alternating Current

⭐option 4)

Answer 2

$\dfrac{I}{8}$ and lagging by 90° from V.

(4) is correct option.

Explanation:

Given that,

Frequency = 50 Hz

Inductor = 2 H

Voltage = V

New frequency = 400 Hz

We need to calculate the inductance reactance

Using formula of inductance reactance

$X_{L}=\omega L$

$X_{L}=2\pi fL$

Put the value into the formula

For 50 Hz,

$X_{L}=2\times\pi\times50\times2$

$X_{L}=628.3\ \Omega$

For 400 Hz,

$X_{L}=2\times\pi\times400\times2$

$X_{L}=5026.5\ \Omega$

We need to calculate the current of r.m.s value

Using formula of current

For 50 Hz,

$I=\dfrac{V}{X_{L}}$

$I=\dfrac{V}{628.3}$.....(I)

For 400 Hz,

$I=\dfrac{V}{X_{L}}$

$I=\dfrac{V}{5026.5}$....(II)

Divided equation (I) by equation (II)

$\dfrac{I}{I'}=\dfrac{5026.5}{628.3}$

$\dfrac{I}{I'}=8$

$I'=\dfrac{I}{8}$

Hence, $\dfrac{I}{8}$ and lagging by 90° from V.

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