Browse Questions

# A uniformly would soleacidal (ie) of self inductance $8 \times 10^{-4}\;H$ and resistance $45\;L$ is broken into two identical coils. These colis are then connected in parallel across a battery of $10V$ of negligible resistance the time constant for this circuit is

$(a)\;4 \times 10^{-4}\;S \\ (b)\;\frac{1}{2} \times 10^{-4}\;S \\(c)\;2 \times 10^{-4}\;S \\(d)\;1 \times 10^{-4}\;S$

When a srteroid is cut it two parts the self inductance and resistance will become half.
$L$ for each part $= \large\frac{8 \times 10^{-4}}{2} $$=4 \times 10^{-4}\;H R for each part =2 \Omega When they are connected in parallel the effective inductance L= \large\frac{4 \times 10^{-4}}{2} \qquad= 2 \times 10^{-4}\;H effective R=\large\frac{2}{2}$$ \;\Omega=1 \Omega$
The time constant $T= \large\frac{L}{R}$
$\qquad= \large\frac{2 \times 10^{-4}}{1}$
$\qquad= 2 \times 10^{-4}$
Hence c is the correct answer.