Question

When a coil is connected to a D.C source of emf 12V, a current of 4 amp flows in it. If same coil is connected to 12V, 50Hz AC source, the current is 2.4A. The self inductance of the coil is

• A. $\frac{1}{20\pi }$
• B. $\frac{1}{10\pi }$
• C. $\frac{1}{25\pi }$
• D. $\frac{1}{5\pi }$

Answer 1

The correct option is C $\frac{1}{25\pi }$

$R=\frac{V_{DC}}{I_{DC}}=\frac{12}{4}=3\Omega$
$Z=\frac{V_{rm}}{I_{rms}}=\frac{12}{24}=5$
$R^2+(2w{)}^2=25;Lw=\sqrt{25-9}=\sqrt{16}=4$
$L=\frac{4}{w}=\frac{4}{2\pi f}=\frac{4}{100\pi }$
$L=\frac{1}{25\pi }$