5) If the current in a solenoid changes from 5 A to 1 A in 2 seconds, emf generated of 4 Volt. If it was connected with an alternating source of frequency 70 Hz , calculate its inductive reactance. $$ \bar{L}=5 $$ $$ t=2 $$

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We start by using the formula relating the induced emf in a solenoid and its self‐inductance. In a solenoid the induced emf is given by

$$
\varepsilon = L\,\frac{dI}{dt}
$$

It is given that the current changes from 5 A to 1 A in 2 seconds. Thus, the change in current is

$$
\Delta I = 5\,A - 1\,A = 4\,A,
$$

and the rate of change of current is

$$
\frac{\Delta I}{\Delta t} = \frac{4\,A}{2\,s} = 2\,A/s.
$$

If we were to use the measured emf of 4 V, then

$$
4\,V = L \times (2\,A/s)
$$

would give

$$
L = \frac{4\,V \times s}{2\,A} = 2\,H.
$$

However, the problem provides the value

$$
\bar{L} = 5\,H,
$$

which we take as the self‐inductance for the next part of the problem.

When the solenoid is connected to an AC source of frequency

$$
f = 70\,Hz,
$$

its inductive reactance is given by

$$
X_L = 2\pi f L.
$$

Substitute $$ L = 5\,H $$ and $$ f = 70\,Hz $$:

$$
X_L = 2\pi \times 70 \times 5 = 700\pi\,\Omega.
$$

Thus, the inductive reactance is

$$
X_L \approx 700\pi \approx 2199\,\Omega.
$$
The inductive reactance is approximately 2199 Ω.

5) If the current in a solenoid changes from 5 A to 1 A in 2 seconds, emf generated of 4 Volt. If it was connected with an alternating source of frequency 70 Hz , calculate its inductive reactance. \( \bar{L}=5 \) \( t=2 \)

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