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

Concept: A practical inductor possesses both inductance $L$ and internal resistance $R$. For a DC supply, the inductive reactance becomes zero because the frequency is zero. Therefore, only the resistance opposes the current. \[ R=\frac{V}{I} \] For an AC supply, the inductor behaves as an $R-L$ series circuit whose impedance is \[ Z=\sqrt{R^2+X_L^2} \] where \[ X_L=\omega L \] is the inductive reactance.

Step 1: Calculate the resistance of the inductor from DC data Given, \[ V_{dc}=200\text{ V} \] \[ I_{dc}=5\text{ A} \] Using Ohm's law, \[ R=\frac{V_{dc}}{I_{dc}} \] \[ R=\frac{200}{5} \] \[ R=40\Omega \]

Step 2: Calculate the impedance from AC data Given, \[ V_{ac}=200\text{ V} \] \[ I_{ac}=4\text{ A} \] Therefore, \[ Z=\frac{V_{ac}}{I_{ac}} \] \[ Z=\frac{200}{4} \] \[ Z=50\Omega \]

Step 3: Determine the inductive reactance Using \[ Z^2=R^2+X_L^2 \] Substituting the values, \[ 50^2=40^2+X_L^2 \] \[ 2500=1600+X_L^2 \] \[ X_L^2=900 \] \[ X_L=30\Omega \]

Step 4: Calculate the inductance The inductive reactance is given by \[ X_L=\omega L \] Given, \[ \omega=300\text{ rad s}^{-1} \] Thus, \[ 30=300L \] \[ L=\frac{30}{300} \] \[ L=0.1\text{ H} \] Converting into millihenry, \[ L=0.1\times1000 \] \[ L=100\text{ mH} \] Hence, the inductance of the inductor is \[ \boxed{100\text{ mH}} \]