Download Application of Derivatives in Electrical Engineering

ENGR 1990 Engineering Mathematics
Application of Derivatives in Electrical Engineering
The diagram shows a typical element (resistor,
capacitor, inductor, etc.) in an electrical circuit. Here,
v (t ) represents the voltage across the element, and i (t )
represents the current flowing through the element. Both
are generally functions of time, t . For any such element,
the following equations apply.
dw
v (t ) =
dq
i (t ) =
dq
dt
p (t ) =
v (t )
+
circuit
element
−
i (t )
⎧ v(t ) is the voltage (volts)
⎪
⎨ w is the energy (joules)
⎪q is the charge (coulombs)
⎩
⎧i (t ) is the current (amps)
⎨
⎩ t is the time (sec)
dw ⎛ dw ⎞ ⎛ dq ⎞
=⎜
⎟ ⎜ ⎟ = v(t ) ⋅ i (t )
dt ⎝ dq ⎠ ⎝ dt ⎠
{ p(t ) is the power (joules/sec)
or (watts)
This last equation is an application of the chain rule.
Example 1:
Given: The charge and voltage for a given circuit element are given by the following
1 sin(250 π t ) (coulombs) and v(t ) = 100sin(250π t ) (volts) .
equations: q(t ) = 50
Find: a) i (t ) the current passing through the element, b) p (t ) the power dissipated by the
element, and c) pmax the maximum power dissipated by the element.
Solution:
dq d 1
1 × cos(250 π t ) × 250 π = 5π cos(250 π t ) (amps)
= ( 50 sin(250 π t ) ) = 50
dt dt
b) p (t ) = v(t ) ⋅ i (t ) = (100sin(250π t ) )( 5π cos(250 π t ) )
a) i (t ) =
= 500π sin(250π t )cos(250 π t ) (watts)
c) To find maximum power, we use the trigonometric identity: 2sin(θ )cos(θ ) = sin(2θ )
p(t ) = 500π sin(250π t )cos(250π t ) = 12 ( 500π sin(500π t ) ) = 250π sin(500π t ) (watts)
⇒ pmax = 250 π ≈ 785 (watts)
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Current–Voltage Relationships for Resistors, Capacitors and Inductors
The voltage across and the current through a resistor are related
simply by its resistance.
R
v (t )
v(t ) = Ri (t ) or i (t ) = v(t ) R
Given a voltage v (t ) applied to a capacitor, the corresponding
current i (t ) can be calculated as
i (t ) = C
i (t )
dv
dt
i (t )
v (t )
C
Given the current i (t ) passing through an inductor, the
corresponding voltage v (t ) can be calculated as
v (t ) = L
di
dt
i (t )
L
v (t )
Example 2:
Given: A voltage v (t ) = 110cos(120π t ) (volts) is applied to a
capacitor with C = 100 (μ f ) .
Find: a) i (t ) , the current through the capacitor, and b) the
maximum power pmax .
i (t )
v (t )
C
Solution:
a) The current may be found by differentiating the voltage.
dv
d
= (100 × 10−6 ) (110cos(120 π t ) )
dt
dt
= (100 × 10−6 ) (110 )(120 π )( − sin(120 π t ) )
i (t ) = C
= −4.147sin(120 π t ) (amps)
b) p (t ) = v(t ) ⋅ i(t ) = (110cos(120 π t ) )( −4.147sin(120 π t ) )
= −456sin(120 π t )cos(120 π t )
= −456 × 12 sin(240 π t )
= −228sin(240 π t ) (watts)
⇒ pmax = 228 (watts)
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Alternate Solution for part (a): (without using derivatives)
We could have solved this problem using complex numbers.
Z C = − j ωC = − j (120 π ) (100 × 10−6 ) = − j (106 ) (120 π )(100 )
= − j 26.526 (ohms) = 26.526∠ ( −90o )
I=
110∠ ( 0o )
26.526∠ ( −90
o
)
= 4.147∠ ( 90o )
i (t ) = 4.147cos(120 π t + 90o ) (amps)
⎛
⎞
= 4.147 ⎜ cos(120 π t )cos(90o ) − sin(120 π t )sin(90o ) ⎟
⎟
⎜
0
1
⎝
⎠
= −4.147sin(120 π t ) (amps)
Example 3:
Given: A voltage v (t ) = 110 e−10 t cos(120π t ) (volts) is applied to a
capacitor with C = 100 (μf ) .
i (t )
v (t )
Find: i (t ) , the current through the capacitor
C
Solution:
To find the current we differentiate the voltage using the product and chain rules.
dv
d
i (t ) = C = (100 × 10−6 ) (110 e −10t cos(120 π t ) )
dt
dt
d
⎡⎛ d
⎞ ⎛
⎞⎤
= (100 × 10−6 ) (110 ) ⎢⎜ ( e −10t ) × ( cos(120 π t ) ) ⎟ + ⎜ ( e −10t ) × ( cos(120 π t ) ) ⎟ ⎥
dt
⎠ ⎝
⎠⎦
⎣⎝ dt
= 0.011 ⎡⎣( −10e −10t cos(120 π t ) ) + ( −120 π e −10t sin(120 π t ) ) ⎤⎦
= −0.011e −10t [10cos(120 π t ) + 120 π sin(120 π t ) ] (amps)
The term in square brackets can be reduced to a single, phase-shifted sine or cosine
function. For example, 120 π sin(120 π t ) + 10cos(120 π t ) = M cos(120 π t + φ ) where
M = (120π )2 + 102 ≈ 377.1 and φ = tan −1 (−120π / 10) = −88.48o = −1.544 (rad) .
So,
i (t ) = −4.15 e−10 t cos(120 π t − 1.544) (amps)
3/4
Example 4:
Given: A current i (t ) = 5 t e−10 t (amps) is applied to an inductor
with L = 250 (mh) .
i (t )
v (t )
L
Find: v (t ) , the voltage across the inductor.
Solution:
To find the voltage we differentiate the current using the product and chain rules.
v (t ) = L
di
d
d
⎛d
⎞
⎛
⎞
= 0.25 ( 5 t e−10 t ) = 14 ⎜ ( 5t ) ⎟ ( e−10 t ) + 14 ⎜ ( 5t ) ( e−10 t ) ⎟
dt
dt
dt
⎝ dt
⎠
⎝
⎠
(
)
= 14 ( 5e−10 t ) + 14 5t ( −10e−10 t ) = 14 ( 5e−10 t − 50 t e−10 t )
=
5 e−10 t
4
(1 − 10 t ) (volts)
4/4