Question

When \(100\,V\) DC is applied across a solenoid, current is \(1\,A\). When \(100\,V\) AC is applied, current is \(0.5\,A\). Frequency \(=50\,Hz\). Find inductance \(= x\,mH\).

Hint:

Use DC to find resistance and AC to find impedance. Then \(X_L = \sqrt{Z^2 - R^2}\).

Answer 1

Correct Answer: 550

Step 1: Understanding the Concept:
A solenoid has both resistance (\(R\)) and inductance (\(L\)). In a DC circuit, only the resistance opposes the current. In an AC circuit, the total opposition is called impedance (\(Z\)), which depends on both resistance and inductive reactance (\(X_{L}\)).
Step 2: Key Formula or Approach:
1. Resistance from DC: \(R = \frac{V_{DC}}{I_{DC}}\)
2. Impedance from AC: \(Z = \frac{V_{AC}}{I_{AC}}\)
3. Relation: \(Z^{2} = R^{2} + X_{L}^{2}\)
4. Inductive Reactance: \(X_{L} = 2\pi fL\)
: Detailed Explanation:
From the DC case:
\[ R = \frac{100\text{ V}}{1\text{ A}} = 100\text{ }\Omega \]
From the AC case:
\[ Z = \frac{100\text{ V}}{0.5\text{ A}} = 200\text{ }\Omega \]
Now, calculate the inductive reactance (\(X_{L}\)):
\[ X_{L} = \sqrt{Z^{2} - R^{2}} = \sqrt{200^{2} - 100^{2}} \]
\[ X_{L} = \sqrt{40000 - 10000} = \sqrt{30000} \approx 173.205\text{ }\Omega \]
Using the formula for \(X_{L}\):
\[ 173.205 = 2 \times 3.14 \times 50 \times L \]
\[ L = \frac{173.205}{314} \approx 0.5516\text{ H} \]
Converting to millihenry (mH):
\[ L \approx 551.6\text{ mH} \]
According to the provided solution key, the value taken is \(550\).
Step 3: Final Answer:
The value of \(x\) is \(550\).