- #1

DevonZA

- 181

- 6

## Homework Statement

In the inductive circuit shown in Figure 2, the switch S1 is closed 0.75s after switch S2 is closed. Calculate the instantaneous current in the coil 1.5s after switch S1 is closed.

## Homework Equations

T=L/R

I=V/R

## The Attempt at a Solution

When S2 is closed:

T=L/R

=2.4/7.2

=0.333

=333ms

IMAX=V/R

=24/7.2

=3.33A

At time 0.75s S1 is closed, the current is:

i1=IMAX(1-e^-t/T)

=24/3.6(1-e^-750/333)

=5.97A

When S1 is closed current flows through the 3.6ohm resistor and decays:

Time constant is:

T=L/R

=2.4/3.6

=0.666

=666ms

Current after 1.5s after S1 is closed:

i2=i1e^-t/T

=5.97e^-(1.5x10^-3/666)

=5.97A

This cannot be correct?