Question

A) at what time t_a will the current in the circuit reach 1/2 of its maximum value?

B) at what time t_b will the energy stored in the inductor reach 1/2 of its maximum value?

Problem 5 (10 POINTS) A battery, ε-30.0V, a resistor, R = 40.00, and an inductor, L-1.5mH, are con- nected in series with an open switch. The switch is then closed at time t = 0. (a) At what time ta will the current in the circuit reach 1/2 of its maximum value? (b) At what time th will the energy stored in the inductor reach 1/2 of its maximum value?

Answer 1

a) The current as a function of time is given by:

where $I_0= \frac{V}{R}$

the time ta when the current is 1/2 of its maximum value:

$\frac{I_0}{2}= I_0(1-e^{-Rt_a/L})$

solving for ta we have:

$t_a = \frac{L}{R}l\ln(2) = \frac{1.5\times 10^{-3}}{40.0}\ln(2)=2.599\times 10^{-5}s$

b) The time tb when the energy stotred in the inductor reach 1/2 of its maximum value:

$U = \frac{1}{2}LI^2$

as a function of time:

for tb

$\frac{U_0}{2} = U_0(1-e^{-Rt_b/L})^2$

$\frac{1}{2} = (1-e^{-Rt_b/L})^2$

solving for tb we have:

$t_b =\frac{L}{R} \ln(\sqrt{2})=\frac{1.5\times 10^{-3}}{40.0} \ln(\sqrt{2})=4.605\times 10^{-5} s$