The switch $S$ in the circuit is connected with point a for a very long time, then it is shifted to position $b$ . The resulting current through the inductor is shown by curves in the graph for four sets of values for the resistance $R$ and inductance $L$ (given in column 1). Which set corresponds with which curve?

$\begin{matrix} Column I & Column II i . R = R_{0} and L = L_{0} & a . I i i . R = 2 R_{0} and L = L_{0} & b . I I i i i . R = R_{0} and L = 2 L_{0} & c . I I I i v . R = 2 R_{0} and L = 2 L_{0} & d . I V \end{matrix}$

i - b, i i - c, i i i - a, i v - d

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i - c, i i - b, i i i - a, i v - b

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i - b, i i - a, i i i - a, i v - b

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i - c, i i - a, i i i - d, i v - b

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Solution

The correct option is D $i - c, i i - a, i i i - d, i v - b$
When the switch is connected with $a$ for a long time, current in the circuit would be $\frac{E}{R}$ .
When switch is connected to $b,$
$i R + L \frac{d i}{d t} = 0$
$\int_{i_{0}}^{i} \frac{d i}{i} = \int_{0}^{t} \frac{- R}{L} d t$
$i = i_{0} ⎛ ⎜ ⎝ e^{\frac{- R t}{L}} ⎞ ⎟ ⎠$
For $(i),$
$R = R_{0}, L = L_{0}$
At $t = 0, i_{1} = \frac{E}{R_{0}} and τ = \frac{L_{0}}{R_{0}}$
$i_{1} = \frac{E}{R_{0}} ⎛ ⎜ ⎜ ⎝ e^{\frac{- R_{0} t}{L_{0}}} ⎞ ⎟ ⎟ ⎠$
For $(i i),$
$R = 2 R_{0}, L = L_{0}$
At $t = 0, i_{2} = \frac{E}{2 R_{0}} and τ = \frac{L_{0}}{2 R_{0}}$
$i_{2} = \frac{E}{2 R_{0}} ⎛ ⎜ ⎜ ⎝ e^{\frac{- 2 R_{0} t}{L_{0}}} ⎞ ⎟ ⎟ ⎠$
For $(i i i),$
$R = R_{0}, L = 2 L_{0}$
At $t = 0, i_{3} = \frac{E}{R_{0}} and τ = \frac{2 L_{0}}{R_{0}}$
$i_{3} = \frac{E}{R_{0}} ⎛ ⎜ ⎜ ⎝ e^{\frac{- R_{0} t}{2 L_{0}}} ⎞ ⎟ ⎟ ⎠$
For $(i v),$
$R = 2 R_{0}, L = 2 L_{0}$
At $t = 0, i_{4} = \frac{E}{2 R_{0}} and τ = \frac{L_{0}}{R_{0}}$
$i_{4} = \frac{E}{2 R_{0}} ⎛ ⎜ ⎜ ⎝ e^{\frac{- R_{0} t}{L_{0}}} ⎞ ⎟ ⎟ ⎠$
Hence at $t = 0, i_{1} = i_{3} > i_{2} = i_{4}$
Also less the time constant faster is the decay,
As $τ_{3} > τ_{1},$ hence $1$ will decay faster than $3$ .
As $τ_{4} > τ_{2},$ hence $2$ will decay faster than $4$ .
On comparision with given graphs,
$i . \to c ., i i . \to a ., i i i, \to d ., i v . \to b$

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