An emf of 20; V is applied at time t=0 to a c | vedclass

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(B) The current in an $RL$ circuit at time $t$ is given by $i(t) = i_0(1 - e^{-Rt/L})$,where $i_0 = V/R$ is the steady-state current at $t = \infty$.G

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$1.15$

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(B) The current in an $RL$ circuit at time $t$ is given by $i(t) = i_0(1 - e^{-Rt/L})$,where $i_0 = V/R$ is the steady-state current at $t = \infty$.Given $V = 20\; V$,$R = 5\; \Omega$,and $L = 10\; mH = 10 	imes 10^{-3}\; H = 0.01\; H$.The time constant $	au = L/R = 0.01 / 5 = 0.002\; s = 2\; ms$.At $t = 40\; ms$,the exponent is $-Rt/L = -t/	au = -40\; ms / 2\; ms = -20$.Wait,checking the time provided: if $t = 40\; ms$,then $t/	au = 20$. If $t = 40\; s$,then $t/	au = 20000$. Assuming the question implies $t = 4\; ms$ to match the $e^2$ hint: $t/	au = 4\; ms / 2\; ms = 2$.Then $i(4\; ms) = i_0(1 - e^{-2})$.The ratio $i(\infty) / i(4\; ms) = i_0 / [i_0(1 - e^{-2})] = 1 / (1 - 1/e^2) = e^2 / (e^2 - 1)$.Using $e^2 = 7.389$,the ratio is $7.389 / (7.389 - 1) = 7.389 / 6.389 \approx 1.156$.

An emf of $20\; V$ is applied at time $t=0$ to a circuit containing in series $10\; mH$ inductor and $5\; \Omega$ resistor. The ratio of the currents at time $t=\infty$ and at $t=40\; ms$ is close to: (Take $e^{2}=7.389$)

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