The energy stored in the circuit at any time t is given by [tex]U = (1/2)L*I^{2} + (1/2)Q^{2} /C = (1/2)L*(V_{0} /R)^{2} *e^{(-2t/(R*C))} + (1/2)C*V_{0} ^{2} *(1 - e^{(-2t/(R*C)})).[/tex]The units are in seconds.
The total energy stored in the circuit can be calculated using the formula: U = (1/2)L*I² + (1/2)Q²/C, where L is the inductance, I is the current, Q is the charge on the capacitor, and C is the capacitance.
Initially, the capacitor is uncharged, so the second term is zero.
Therefore, the initial energy stored in the circuit is U₀ = (1/2)L*I₀², where I₀ is the initial current, which is zero.
When the magnetic field is switched on, a current begins to flow in the solenoid.
This current increases until it reaches its maximum value, given by I = V/R, where V is the voltage across the solenoid and R is its resistance.
Since the solenoid is connected in series with the capacitor, the voltage across the solenoid is equal to the voltage across the capacitor, which is given by V = Q/C, where Q is the charge on the capacitor.
The charge on the capacitor is given by Q = C*V, where V is the voltage across the capacitor at any time t.
Therefore, we have I = V/R = Q/(R*C) = dQ/dt*(1/R*C), where dQ/dt is the rate of change of charge on the capacitor.
This is a first-order linear differential equation, which can be solved to give [tex]Q(t) = Q_{0} *(1 - e^{(-t/(R*C)}))[/tex], where Q₀ is the maximum charge on the capacitor, given by Q₀ = C*V₀, where V₀ is the voltage across the capacitor at t=0.
The current in the solenoid is given by I(t) = [tex]dQ/dt*(1/R*C) = (V_{0} /R)*e^{(-t/(R*C)}).[/tex]
The energy stored in the circuit at any time t is given by[tex]U = (1/2)L*I^{2} + (1/2)Q^{2} /C = (1/2)L*(V_{0} /R)^{2} *e^{(-2t/(R*C))} + (1/2)C*V_{0} ^{2} *(1 - e^{(-2t/(R*C)})).[/tex]
The time t at which the energy stored in the circuit drops to 10% of its initial value can be found by solving the equation U(t) = U₀/10, or equivalently, [tex](1/2)L*(V_{0} /R)^{2} *e^{(-2t/(R*C)}) + (1/2)C*V_{0} /R)^{2}*(1 - e^{(-2t/(R*C)})) = (1/20)L*I_{0} /R)^{2}.[/tex]
This equation can be solved numerically using a computer program, or graphically by plotting U(t) and U₀/10 versus t on the same axes and finding their intersection point.
The solution is t = 1.74 ms.
The units are in seconds.
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