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example: 6.3-1
A DC voltage of 10 V is switched on to a 10
W
resistor at t 
=
0. Find and plot p(t) and E(t) for t 
≥
0.
Solution
p t
t
E t
dt
t
V
R
t
( )
( )
=
=
≥
=
+ =
∫
2
10
0
10
0 10
0
W for 
and
J.
The waveforms of applied instantaneous 
power and the energy dissipated are shown Fig. 6.3-2. 
20
1
2
3 4
5 6 7
t
(s)
t
(s)
p
(
t
) (W)
10
E
(
t
) (J)
40
60
80
100
120
1
2
3 4
5 6 7
Fig. 6.3-2 
Instantaneous power and energy in Example 6.3-1 
example: 6.3-2
A symmetric square wave of 100 Hz frequency and 
±
10V amplitude is switched on to a resistor of 
10
W
at t 
=
0 at the positive-going zero crossing of the square wave. Find and plot the instantaneous 
power and energy dissipation in the resistor.
Solution
p(t) 
=
v(t) i(t) 
=
[v(t)]
2
/R
When a symmetric square wave is squared, it becomes a constant quite irrespective of its frequency. 
Therefore, [v(t)]
2
=
100 V
2
for t 
≥
0.
\
p(t) 
=
10 W for t 
≥
0 and E(t) 
=
10 t J. The plots are same as in Fig. 6.3-1.
Therefore, we conclude that applying a symmetric square wave voltage of amplitude V and 
frequency f to a resistive load has the same effect as applying a DC voltage of V to it as far as the 
instantaneous power and energy dissipation function are concerned.