EE 201 ELECTRIC CIRCUITS
CLASS 2
Class Notes of the online team for EE-201 (Dr Jamid, EE, KFUPM)
The material covered in this class will be as follows:
⇒
Energy and Power in Electric Circuits.
⇒
Ohm’s Law.
⇒
Kirchoff’s Current Law.
At the end of this class you should be able to:
⇒
Understand the relation between power and energy.
⇒
Understand the passive sign convention.
⇒
Use the passive sign convention in power calculation.
⇒
Determine if the power is actually absorbed or delivered by an electric element.
⇒
Verify power conservation.
⇒
Use the passive sign convention in Ohm’s law.
⇒
Apply Kirchoff’s Current Law at a node.
⇒
Apply Kirchoff’s Current Law at a super node.
Electric Energy and Power:
The power p (t ) absorbed by an electric element and the energy w(t ) in the same element are
related by:
p (t ) = dw(t ) / dt
Unit of w is Joule [J]
Unit of p is Watt [W].
Unit of t is second [s].
If the energy w(t ) increases with time [ w(t ) has a positive slope], then:
dw(t ) / dt > 0 ⇒
p (t ) > 0
⇒
power is being actually absorbed by the element
If w(t ) decreases with time [ w(t ) has a negative slope], then:
dw(t ) / dt < 0 ⇒
p (t ) < 0
⇒ power is being actually delivered by the element
w(t ) increases
⇔
power being absorbed
decreases
⇔
power being delivered
w(t )
The power p (t ) can be expressed in terms of i (t ) and v (t ) :
p=
dw dq
= vi
dq dt
⇒
p (t ) = v (t )i (t )
The above relation p = iv applies only when the current enters the element the (+) terminal and
leaves the (-) terminal.
a
i(t)
+
Electric
Element
v(t)
-
b
Figure 1
If the current enters the element the (-) terminal and leaves the (+) terminal, then we have:
p = −iv
In this case, it is necessary to insert a minus sign in the power expression, in order to have consistent
results.
Figure 2
Example 1:
i)
Calculate the power absorbed by each element in the given circuit.
ii)
State whether the power is actually absorbed or actually delivered by the element.
Figure 3
Solution:
a) p = +iv = + (5) × (10) = +50 W
⇒
p>0
⇒
actually absorbed
b) p = −iv = −(−4) × (3) = +12 W
⇒
p>0
⇒
actually absorbed
p<0
⇒
actually delivered
p<0
⇒
actually delivered
c) p = −iv = −(−12) × (−10) = −120 W
d) p = +iv = + (60) × ( −6) = −360 W ⇒
The following statements are equivalent:
Power absorbed by the element
Power delivered to the element
⇒
Power dissipated by the element
Power consumed by the element.
The following statements are also equivalent:
Power delivered by the element
Power generated by the element.
The symbol p will be reserved for the power absorbed by the element.
Example 2:
i)
Calculate the power absorbed by each element in the given circuit.
ii)
Show that the total power dissipated equals to the total power generated.
Ω
Ω
Figure 4
Solution:
i) p6V = +iv = +(7 / 4)(6) = 10.5W
⇒ dissipated
p3 A = −iv = −(3)(20) = −60W
⇒ generated
p16 Ω = +iv = +(5 / 4)(20) = 25W
⇒ dissipated
p8Ω = −iv = −(7 / 4)(−14) = 24.5W
⇒ dissipated
ii)
∑p
dis
= 10.5 + 25 + 24.5 = 60W
∑p
gen
= 60W
∴ ∑ pdis = ∑ pgen = 60W
Ohm’s law:
The voltage v (t ) and the current i (t ) in a resistor R are related by:
v = iR
The above relation is valid only if i (t ) enters the (+) terminal and leaves the (-) terminal.
Figure 5
If i (t ) enters the (-) terminal and leaves the (+) terminal, then Ohm’s law must be changed to:
v = −iR
+
R
v(t)
i(t)
Figure 6
The passive sign convention:
The use of the ± signs in Ohm’s law and the power expression is known as the passive sign
convention.
i (t ) enters the (+) terminal ⇒
p = +iv &
v = +iR
⇒
p = −iv &
v = −iR
i (t ) enters the (-) terminal
Example 3:
Calculate the unknown quantities in the following circuit.
R
6
V
4
Ω
=
2
-3
2V
-
V
16
Ω
8
Ω
Figure 7
Solution:
v1 = +(6)(2) = 12V
v2 = −(6)(6) = −36V
i3 = +(16) /(8) = 2 A
R = −(v4 ) / i3 = −(−32) /(2) = 16Ω
Kirchoff’s Current Law (KCL).
The sum of currents entering a node is equal to the sum of currents leaving that node.
i1 + i4 = i2 + i3 + i5
Equivalent statement of KCL:
The algebraic sum of currents entering a node is equal to zero.
i1 − i2 − i3 + i4 − i5 = 0
Figure 8
Example 4:
Calculate the unknown currents in the following circuits.
Ω
Ω
Ω
Ω
Ω
Ω
Ω
Ω
Ω
Ω
Figure 9
Solution:
a) KCL at node (a)
⇒
i1 = 2 + 4 = 6 A
b) KCL at node (a)
⇒
3 + i1 = 1
⇒
i1 = −2 A
⇒
3 + i1 − 1 = 0
⇒
i1 = −2 A
⇒
i1 − 4 + 2 = 0
⇒
i1 = 2 A
0.5 − i2 − 2 = 0
⇒
i2 = −1.5 A
Alternatively
KCL at node (a)
c) KCL at node (b)
KCL at node (c)
⇒
Check KCL at node (a)
⇒
−i1 + 4 + i2 − 0.5 = −(2) + 4 + (−1.5) − 0.5 = −4 + 4 = 0
KCL is also applicable to a closed area (super node).
The algebraic sum of currents entering a super node is equal to zero.
i1 + i2 − i3 + i4 − i5 = 0
Figure 10
Example:
Calculate the currents i1 and i2 in the circuit shown below:
Ω
Ω
Ω
Ω
Ω
Ω
Ω
Ω
Figure 11
Solution:
KCL at super node 1
⇒
3 − i1 = 0
⇒
i1 = 3 A
KCL at super node 2
⇒
i2 = 0
⇒
i2 = 0 A
Ω
Ω
Ω
Ω
Figure 12