A series circuit having R= 40 ohm, and inductance of 0.3H suddenly connected across a 150V supply through a switch. Find;
I. The rate of change of current after closing the switch
II. The steady state value of the current.
III. The the rate of change of current I(t) = V/R – V/R . 1.
III. The value of current after 7 milliseconds
IV. The time for current to attain half it’s final value
Solutions
R = 40 ohm, V = 150V, L = 0.3H
The rate of change of current I(t) = V/R – V/R . e–¹³³·³t
1. Rate of change of current after closing the switch
di(t)/ d(t), when t = 0
di(t)/ d(t) = 3.75 – 3.75e–¹³³·³t
Then differentiate,
0 – 3.75×-133.3e–¹³³·³<⁰>
0 – 499.88e⁰
0 – 499.88 × 1
0 – 499.88
= 499.88 A/S (Amperes per second)
2. The steady state value of the current
Base current (Ib) = V/R = 150/40
= 3.75A
3. The value of current after 7 milliseconds
The rate of change of current (It) = 3.75 – 3.75e–¹³³·³t
when t = 0.007seconds
The rate of change of current (It) = 3.75 – 3.75e–¹³³·³×⁰·⁰⁰⁷ (It) = 3.75 – 3.75e–⁰·⁹³
(It) = 3.75 – 3.75e–⁰·⁹³NOTE: e–⁰·⁹³ Means exponential of 0.93 (how to get it, press (in) in your calculator with 0.93. You’ll have 0.117)
(It) = 3.75 – 3.75(0.117)
(It) = 3.75 – 0.44
(It) = 3.31A
4.The time for current to attain half its final value
Base current (Ib) = 3.75A,
Ib/2 » 3.75/2 » 1.875A
1.875 = 3.75-3.75e–¹³³·³t
Collect like terms
3.75e–¹³³·³t = 3.75-1.875
3.75e–¹³³·³t = 1.875
e–¹³³·³t = 1.875/3.75
e–¹³³·³t = 0.5
-133.3t = in(0.5)
t = [-1/133.3 ] in(0.5)
t = -7.5 × 10–³ × -0.69
t = 5.2 × 10–³ seconds » 5.2milliSeconds. [final answer]
