**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

**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**]