Physics - EMI/AC Question with Solution | TestHub

When coil is broken into two identical parts, then resistance of each part $R^{\prime }=\frac{R}{2}=\frac{6}{2}=3\Omega$ Inductance of each part $L^{\prime }=\frac{L}{2}=\frac{1.8}{2}\times 10^{-4}=0.9\times 10^{-4}\mathrm{H}$ Now, time constant $\tau =\frac{L^{\prime }}{R^{\prime }}$ $\tau =\frac{L}{2\times \frac{R}{2}}=\frac{L}{R}=\frac{1.8\times 10^{-4}}{6}=30\mu \mathrm{s}$ Now, effective resistance when both coils are connected in parallel $R^{\prime \prime }=\frac{R^{\prime }\times R^{\prime }}{R^{\prime }+R^{\prime }}=\frac{3\times 3}{3+3}=\frac{9}{6}\Omega$ So, maximum current drawn from battery $I=\frac{V}{R^{\prime \prime }}=\frac{12}{9/6}=8\mathrm{A}$