Question

A circuit consists of a coil that has a self-inductance equal to 4.3 mH and an internal resistance equal to 16 Ω, an ideal 9 V battery, and an open switch--all connected in series. At t = 0 the switch is closed. Find the time when the rate at which energy is dissipated in the coil equals the rate at which magnetic energy is stored in the coil.

Answer 1

Answer:

t = 186.2 μs

Explanation:

Current in LR series circuit

$I(t) = I_{s}( 1 - e^{-Rt/L)}$----(1)

steady current =  I_{s} = V/R

time constant = τ =$L/R =4.3 * 10^{-3} / 16$

                                              = 0.268 ms

magnetic energy stored in coil = $U_{L} = \frac{1}{2}LI^{2}$

rate at which magnetic energy stored in coil= $\frac{d}{dt}U_{L} =\frac{d}{dt} \frac{1}{2}LI^{2} \\ = LI\frac{dI}{dt}$----(2)

rate at which power is dissipated in R:

                                       $P = I^{2}R$---(3)

To find the time when the rate at which energy is dissipated in the coil equals the rate at which magnetic energy is stored in the coil equate (2) and (3)

$I^{2}R=LI \frac{dI}{dt}$

[/tex]I=\frac{L}{R}\frac{dI}{dt}[/tex]----(4)

differentiating (1) w.r.to t

$I(t)=I_{f} (1-e^{\frac{Rt}{L} })$

$\frac{dI}{dt} = I_{f}\frac{d}{dt}(1-e^{\frac{-Rt}{L} } )$

$\frac{dI}{dt}= I_{f}(-\frac{R}{L} e^{\frac{-Rt}{L} } )$---(5)

substituting (5) in (4)

$I=I_{f}e^{-\frac{Rt}{L} }$----(6)

equating (1) and (6)

$I_{f}( 1- e^{-\frac{Rt}{L} } ) = I_{f}e^{-\frac{Rt}{L} }$

$1 - e^{-\frac{Rt}{L} } = e^{-\frac{Rt}{L} }$

$\frac{1}{2}= e^{-\frac{Rt}{L} }$

$t= -\frac{L}{R}ln\frac{1}{2}$

L= 4.3 mH

R= 16 Ω

t = 186.2 μs