When a coil is connected across a 20 V dc supply, it draws a current of 5 A. When it is connected across 20 V, 50 Hz ac supply, it draws a current of 4 A. The self inductance of the coil is ______ mH.
(Take \(\pi\) = 3)
(Take \(\pi\) = 3)
Correct Answer: 10
Approach Solution - 1
To find the self-inductance \( L \) of the coil, we begin by analyzing the conditions given.
1. DC Condition:
When connected to a 20 V dc supply, the coil draws a current of 5 A. Ohm’s Law (\( V = IR \)) gives us the resistance \( R \) of the coil:
\[ R = \frac{V}{I} = \frac{20\, V}{5\, A} = 4\, \Omega \]
2. AC Condition:
When the coil is connected to a 20 V, 50 Hz ac supply, it draws a 4 A current. In this case, the impedance \( Z \) in the coil is:
\[ Z = \frac{V}{I} = \frac{20\, V}{4\, A} = 5\, \Omega \]
The impedance in an ac circuit with inductance is given by:
\[ Z = \sqrt{R^2 + (X_L)^2} \] where \( X_L \) is the inductive reactance.
3. Calculate Inductive Reactance:
Rearrange the impedance formula to find \( X_L \):
\[ X_L = \sqrt{Z^2 - R^2} = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = 3\, \Omega \]
The inductive reactance \( X_L \) is also defined as \( X_L = 2\pi f L \). Solve for \( L \):
\[ 3 = 2 \pi f L \implies L = \frac{3}{2 \times 3 \times 50} = \frac{3}{300} = 0.01 \, \text{H} = 10 \, \text{mH} \]
4. Validation:
The calculated self-inductance \( L = 10 \) mH is within the specified range of 10,10.
Thus, the self-inductance of the coil is 10 mH.
Approach Solution -2
Case 1: DC Circuit When the coil is connected to a DC supply, the inductive reactance is 0, so:
\[ I = \frac{V}{R} \implies R = \frac{20}{5} = 4 \, \Omega. \]
Case 2: AC Circuit When the coil is connected to an AC supply:
\[ I = \frac{V}{Z} \implies Z = \frac{20}{4} = 5 \, \Omega, \]
where \( Z \) is the impedance of the coil, given by:
\[ Z = \sqrt{R^2 + X_L^2}. \]
Substitute \( R = 4 \, \Omega \):
\[ 5 = \sqrt{4^2 + X_L^2} \implies X_L^2 = 25 - 16 = 9 \implies X_L = 3 \, \Omega. \]
The inductive reactance \( X_L \) is related to the self-inductance \( L \) by:
\[ X_L = 2 \pi f L \implies L = \frac{X_L}{2 \pi f}. \]
Substitute \( X_L = 3 \, \Omega \), \( f = 50 \, \text{Hz} \), and \( \pi = 3 \):
\[ L = \frac{3}{2 \cdot 3 \cdot 50} = \frac{3}{300} = 0.01 \, \text{H} = 10 \, \text{mH}. \]
Final Answer: 10 mH.
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