An RL circuit has a V=7V battery, a R=15000Ohm resistor, and a L=0.7H, connected

Question

An RL circuit has a V=7V battery, a R=15000Ohm resistor, and a L=0.7H, connected in series with a switch. The switch is left open for several seconds and then closed at time t = 0.

(a) What is the current at time t = 0?

(b) What is the current several seconds later? This is the maximum current.

(c) At what time was the current 95.0% of the maximum current?

(d) At what time was the current 63.2% of the maximum current?

(e) What percentage of the maximum current flows through the circuit at time t = 0.09333milliseconds?

Explanation / Answer

a)

i = (/R) (1 - e^(-Rt/L))

= (7/15000) * (1 - e^(-15000*0/0.7))

= (7/15000) * (1 - 1)

= 0 A

b)

i = /R = 7/15000 = 4.67 * 10^-4 A

c)

i = imax (1 - e^(-Rt/L))

>>>> i/imax = 1 - e^(-15000*t/0.7)

>>>> 0.95 = 1 - e^(-15000*t/0.7)

>>>> t = 1.40 * 10^-4 s

d)

i = imax (1 - e^(-Rt/L))

>>>> i/imax = 1 - e^(-15000*t/0.7)

>>>> 0.632 = 1 - e^(-15000*t/0.7)

>>>> t = 4.67 * 10^-5 s

e)

i = imax (1 - e^(-Rt/L))

>>>> i/imax = 1 - e^(-15000*0.09333e-3/0.7) * 100

>>>> i/imax = 86.5 %

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