A coil of inductance 200 H and resistance $20 Ω$ is connected to a constant voltage source. How soon in seconds will the coil current reach 50% of the steady state value?(ln2 = 0.693). Write upto two digits after the decimal point.

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Solution

$i = i_{0} (1 - e^{- R t / L})$
$\Rightarrow \frac{i_{0}}{2} = i_{0} (1 - e^{- R t / L})$
$e^{- R t / L} = \frac{1}{2} \Rightarrow e^{R t / L} = 2 \Rightarrow \frac{R t}{L} = l n (2)$
$t = \frac{L}{R} l n (2) = \frac{200}{20} \times 0.693$
= $6.93 s e c$

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A coil of inductance 200 H and resistance 20Ω is connected to a constant voltage source. How soon in seconds will the coil current reach 50% of the steady state value?(ln2 = 0.693). Write upto two digits after the decimal point.

i=i0(1−e−Rt/L) ⇒i02=i0(1−e−Rt/L) e−Rt/L=12⇒eRt/L=2⇒RtL=ln(2) t=LRln(2)=20020×0.693 = 6.93 sec

A coil of inductance 200 H and resistance $20 Ω$ is connected to a constant voltage source. How soon in seconds will the coil current reach 50% of the steady state value?(ln2 = 0.693). Write upto two digits after the decimal point.

$i = i_{0} (1 - e^{- R t / L})$
$\Rightarrow \frac{i_{0}}{2} = i_{0} (1 - e^{- R t / L})$
$e^{- R t / L} = \frac{1}{2} \Rightarrow e^{R t / L} = 2 \Rightarrow \frac{R t}{L} = l n (2)$
$t = \frac{L}{R} l n (2) = \frac{200}{20} \times 0.693$
= $6.93 s e c$