Download Determine and plot as a function of time the current

70302
Determine and plot as a function of time the current through a component if the
voltage across it has the waveform shown in
Figure P4.17 and the component is a:
a) Resistor R = 7 Ω
b) Capacitor C = 0.5 μF
c) Inductor L = 7 mH
Answer:
a)
In case of a resistor the Ohm’s law as gives the
current
I = V/R
For the first interval of time 0-5 ms the
current is increasing with voltage from
I = 0 at t = 0
to
I = V/R = 15/7 = 2.143 A
linearly and then remains constant with
voltage for next 5 ms. After that as the
voltage becomes zero the current
becomes zero and hence the graph is
given as shown.
I(A)
3
2
1
5
10
15 t(ms)
-------------------------------------------------------b)
(Here no resistance is given means we have to consider a pure capacitance)
The potential difference across a capacitor is given by V = q/C where q is charge on
the capacitor. As the voltage increases, charge q on the capacitor will increase and
the hence there through the circuit is given as
I = dq/dt = d (CV)/dt = C dV/dt
For the first interval of 5 ms the voltage is changing at a constant rate
dV/dt = 15/(5*10-3) = 3000 V/s
and hence the current is constant given
by
I = C (dV/dt)
= 0.5*10-6*3000 = 1500*10-6 A
= 1500 A
For the second interval of time the voltage
is constant and hence the rate of change
voltage dV/dt is zero and no current will
flow through the circuit.
I(A)
2000
1500
1000
5
10
15 t (ms)
At t = 10 ms the voltage across the capacitor becomes zero suddenly, means the
charge on the capacitor suddenly becomes zero. The charge already stored in the
capacitor will flow back in an infinitely short time (as no resistance is there) and after
that the current becomes zero. That pulse of negative current cannot be shown on
the graph as the time interval is infinitely small as compare to the given scale. The
graph is shown as in figure.
---------------------------------------c)
The voltage across the inductor is equal to the induced emf because of change in
current in it.
Faraday’s law gives the relation between the induced emf and the current in an
inductor as
e
d B
dI (t )
 L
dt
dt
In the circuit for the loop net voltage must be zero hence V – e = 0 (Krichoffs’s law}
This gives
Or
Or
dI
dt
V
dI  dt
L
1
I   V (t )dt......(1)
L
V L
Now the voltage is changing with time linearly from t = 0 to 5 ms, and thereafter it
is constant. Hence the during the first interval
dV V
15


 3000 Volts/s.
dt
t 5 *10 3
Hence the voltage across the inductor as a function of time t, in the first interval of
time is given by
V  3000 * t
Hence the current at the end of first interval is given by using equation (1) as
1
I1 
L
5*103
5*103
3000  t 2 
0 3000 * t * dt  L  2 
0
3000(5 *10 3 ) 2

 5.357 A
7 *10 3 * 2
The current is square function of time the curve will be a parabola.
Now the voltage in the second interval is constant (15 V) hence the current I2
increases linearly and at time t = 10 s is given by using the same equation (1) as
15
I 2  I1 
7 *10 3
10*103
15
 dt  7 *10 10 *10
5*10 3
3
3

 5 *10 3  10.714 A
Hence the current at 10 ms is given by
I2 = I1 + 10.714
= (5.357 + 10.714)
= 16.071 A.
After 10 ms the voltage across the
inductor is suddenly becomes zero hence
induced emf in the inductor is zero, the
rate of change of the current is zero and
the current remains constant thereafter =
16.071 A.
I(A)
15
10
5
The graph is shown in the figure.
------------------------------------------------------------
5
10
15 t (ms)