A resistance R and an inductor L are connected in a series with a battery and a switch S the switch S is closed at t=0 find the time at which the energy stored in the inductor is 1/4th of the maximum value
To find the time at which the energy stored in the inductor is ( \frac{1}{4} ) of its maximum value, we start by analyzing the circuit consisting of a resistor ( R ), an inductor ( L ), and a battery with voltage ( V ).
The maximum current ( I_{\text{max}} ) in the circuit, when the switch is closed, can be calculated using Ohm's law:
[ I_{\text{max}} = \frac{V}{R} ]
The energy ( U ) stored in the inductor when the current reaches its maximum value is given by the formula:
[ U = \frac{1}{2} L I^2 ]
Substituting ( I_{\text{max}} ):
[ U_{\text{max}} = \frac{1}{2} L \left( \frac{V}{R} \right)^2 = \frac{1}{2} \frac{L V^2}{R^2} ]
The current ( I(t) ) in the circuit as a function of time after the switch is closed is given by:
[ I(t) = I_{\text{max}} \left( 1 - e^{-\frac{R}{L} t} \right) = \frac{V}{R} \left( 1 - e^{-\frac{R}{L} t} \right) ]
The energy stored in the inductor at time ( t ) is:
[ U(t) = \frac{1}{2} L I(t)^2 ]
Substituting ( I(t) ):
[ U(t) = \frac{1}{2} L \left( \frac{V}{R} \left( 1 - e^{-\frac{R}{L} t} \right) \right)^2 ]
Expanding this:
[ U(t) = \frac{1}{2} L \frac{V^2}{R^2} \left( 1 - e^{-\frac{R}{L} t} \right)^2 ]
We want to find ( t ) such that:
[ U(t) = \frac{1}{4} U_{\text{max}} = \frac{1}{4} \left( \frac{1}{2} \frac{L V^2}{R^2} \right) = \frac{L V^2}{8 R^2} ]
Setting the two expressions equal:
[ \frac{1}{2} L \frac{V^2}{R^2} \left( 1 - e^{-\frac{R}{L} t} \right)^2 = \frac{L V^2}{8 R^2} ]
Dividing both sides by ( \frac{1}{2} L \frac{V^2}{R^2} ):
[ \left( 1 - e^{-\frac{R}{L} t} \right)^2 = \frac{1}{4} ]
Taking the square root of both sides:
[ 1 - e^{-\frac{R}{L} t} = \frac{1}{2} \quad \text{or} \quad 1 - e^{-\frac{R}{L} t} = -\frac{1}{2} ]
The second equation does not yield a valid solution since ( 1 - e^{-\frac{R}{L} t} ) cannot be negative. Thus, we have:
[ 1 - e^{-\frac{R}{L} t} = \frac{1}{2} ]
Rearranging gives:
[ e^{-\frac{R}{L} t} = \frac{1}{2} ]
Taking the natural logarithm of both sides:
[ -\frac{R}{L} t = \ln\left(\frac{1}{2}\right) ]
Thus,
[ t = -\frac{L}{R} \ln\left(\frac{1}{2}\right) = \frac{L}{R} \ln(2) ]
The time at which the energy stored in the inductor is ( \frac{1}{4} ) of the maximum value is:
[ t = \frac{L}{R} \ln(2) ]