A coil of self-inductance `50` henry is joined to the terminals of a battery of e.m.f. `2` volts through a resistance of `10ohm` and a steady current is flowing through the circuit. If the battery is now disconnected, the time in which the current will decay to `1//e` of its steady value is

Options

A. (a)500 seconds
B. (b)50 seconds
C. (c)5 seconds
D. (d)`0.5`seconds

Answer

To solve the problem step by step, we will follow the concepts of self-inductance and the decay of current in an inductor when the circuit is disconnected from the power source.

### Step 1: Understand the Circuit
We have a coil with a self-inductance \( L = 50 \, \text{H} \) connected to a battery of EMF \( V = 2 \, \text{V} \) through a resistor \( R = 10 \, \Omega \).

### Step 2: Calculate the Steady-State Current
In a steady state, the current flowing through the circuit can be calculated using Ohm's law:
\[ I_0 = \frac{V}{R} \]
Substituting the values:
\[ I_0 = \frac{2 \, \text{V}}{10 \, \Omega} = 0.2 \, \text{A} \]

### Step 3: Understand Current Decay in an Inductor
When the battery is disconnected, the current through the inductor will decay exponentially. The equation for the current \( I(t) \) at time \( t \) after the disconnection is given by:
\[ I(t) = I_0 e^{-\frac{t}{\tau}} \]
where \( \tau \) is the time constant of the circuit.

### Step 4: Calculate the Time Constant \( \tau \)
The time constant \( \tau \) for an inductor is given by:
\[ \tau = \frac{L}{R} \]
Substituting the values:
\[ \tau = \frac{50 \, \text{H}}{10 \, \Omega} = 5 \, \text{s} \]

### Step 5: Find the Time for Current to Decay to \( \frac{I_0}{e} \)
We need to find the time \( t \) when the current decays to \( \frac{I_0}{e} \):
\[ I(t) = \frac{I_0}{e} \]
Substituting into the decay equation:
\[ \frac{I_0}{e} = I_0 e^{-\frac{t}{\tau}} \]
Dividing both sides by \( I_0 \):
\[ \frac{1}{e} = e^{-\frac{t}{\tau}} \]
Taking the natural logarithm of both sides:
\[ -1 = -\frac{t}{\tau} \]
Thus:
\[ t = \tau \]
Substituting the value of \( \tau \):
\[ t = 5 \, \text{s} \]

### Final Answer
The time in which the current will decay to \( \frac{1}{e} \) of its steady value is \( 5 \, \text{s} \).

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