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A uniform coil of self-inductance $1.8 \times 10^{-4} \mathrm{H}$ and resistance $6 \Omega$ is broken up into two identical coils. These two coils are then connected in parallel across a $12 \mathrm{~V}$ battery. The circuit time constant and steady state current through the battery respectively are
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$30 \mu \mathrm{s}, 8 \mathrm{~A}$
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
$I=\frac{V}{R^{\prime \prime}}=\frac{12}{9 / 6}=8 \mathrm{~A}$
$R^{\prime}=\frac{R}{2}=\frac{6}{2}=3 \Omega$
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
$I=\frac{V}{R^{\prime \prime}}=\frac{12}{9 / 6}=8 \mathrm{~A}$
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