A potential difference of 6V is applied to a coil inductance 0.5H and a resistance of 4 ohm connected in series. The time taken for the current to reach half the maximum value is (in seconds)

\frac{l o g_{e} 2}{8}

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\frac{l o g_{e} 2}{4}

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\frac{l o g_{e} 4}{8}

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\frac{l o g_{e} 4}{4}

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Solution

The correct option is A $\frac{l o g_{e} 2}{8}$
The current growth in L-R circuit is given by
$i = i_{0} [1 - e^{- \frac{R}{L} t}]$
$\frac{i_{0}}{2} = i_{0} [1 - e^{- \frac{4}{0.5} t}]$
$e^{- 8 t} = 1 - \frac{1}{2} = 2^{- 1}$
$8 t = l o g_{e} 2$ ; $t = \frac{l o g_{e} 2}{8}$

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A potential difference of 6V is applied to a coil inductance 0.5H and a resistance of 4 ohm connected in series. The time taken for the current to reach half the maximum value is (in seconds)

The correct option is A loge28The current growth in L-R circuit is given by i=i0[1−e−RLt] i02=i0[1−e−40.5t] e−8t=1−12=2−1 8t=loge2; t=loge28

The correct option is A $\frac{l o g_{e} 2}{8}$
The current growth in L-R circuit is given by
$i = i_{0} [1 - e^{- \frac{R}{L} t}]$
$\frac{i_{0}}{2} = i_{0} [1 - e^{- \frac{4}{0.5} t}]$
$e^{- 8 t} = 1 - \frac{1}{2} = 2^{- 1}$
$8 t = l o g_{e} 2$ ; $t = \frac{l o g_{e} 2}{8}$