Question

When an inductor is connected to a $200\text{ V}$ dc supply, the current through it is $5\text{ A}$ and when the same inductor is connected to a $200\text{ V}$ ac supply of angular frequency $300\text{ rad s}^{-1}$, the current through it is $4\text{ A}$. The inductance of the inductor is:

• A. $100\text{ mH}$
• B. $200\text{ mH}$
• C. $50\text{ mH}$
• D. $75\text{ mH}$

Hint:

For a practical inductor, first use the DC data to find the resistance $R$. Then use the AC data to find the impedance $Z$. Finally apply \[ Z^2=R^2+X_L^2 \] to obtain the inductive reactance and hence the inductance.

Answer 1

The Correct Option is A

Step 1: Treat the inductor as having resistance too.
A real inductor has both inductance $L$ and a series resistance $R$. On DC the reactance is zero, so only $R$ limits the current; on AC the impedance combines $R$ and $X_L$.
Step 2: Get R from the DC test.
With $200\,V$ DC giving $5\,A$, $R = \dfrac{200}{5} = 40\,\Omega$.
Step 3: Get the impedance from the AC test.
With $200\,V$ AC giving $4\,A$, $Z = \dfrac{200}{4} = 50\,\Omega$.
Step 4: Extract the reactance.
Since $Z^2 = R^2 + X_L^2$, $X_L^2 = 50^2 - 40^2 = 2500 - 1600 = 900$, so $X_L = 30\,\Omega$.
Step 5: Solve for the inductance.
$X_L = \omega L$ with $\omega = 300\,rad\,s^{-1}$ gives $L = \dfrac{30}{300} = 0.1\,H$.
Step 6: Convert and conclude.
$L = 0.1\,H = 100\,mH$, which is option (1).
\[ \boxed{L = 100\,mH} \]