Question

What happens to the inductive reactance and the current in a purely inductive circuit if the frequency is halved?

• A. Both, including reactance and current, will be doubled
• B. Both, inductive reactance and current will be halved
• C. Inductive reactance will be halved and current will be doubled
• D. Inductive reactance will be doubled and current will be halved

Answer 1

The correct option is C

Inductive reactance will be halved and current will be doubled

Explanation for correct option

Option C

Inductive reactance will be halved and current will be doubled.

1. Inductive reactance will be halved and current will be doubled as the frequency is halved.
2. Formula for inductance current is $i=\frac{V_0}{X_L}$
3. The inductive reactance in a purely inductive circuit is written as: $X_L=\omega L$, Where, substitute in the above equation, $\omega$ represents the angular frequency, $L$represents the inductance $f$represents the frequency, $X_L$represents the inductance reactance.
4. , $$\begin{gathered} i=\frac{V_0}{X_L} \\ heref=f/2 \\ i=2\left(\frac{V_0}{2\mathrm{\pi f}L}\right) \end{gathered}$$ Where $V_0$ represents the output voltage ,$f$represents the frequency, $i$ represents the current
6. Therefore, if the frequency is halved, the inductive reactance is also halved:
7. Hence, if the frequency is halved the value of the current is doubled.

Hence, option C is the correct option.