Ideal Voltage and Current Sources

Ideal Voltage Source--Voltage at terminals is NOT a function of current through source
Ideal Current Source--Current is independent of Voltage.
Independent sources represented in circuits as circle with V, +, - or I and arrow showning current direction
Dependent sources are represented in circuits with a diamond (same labeling as independent sources) and the functional relationship with the source's "basis"
Can't have contradiction of two independent sources providing different measure to same points in circuit. Example was two different independent voltage sources simply connected together negative to negative, positive to positive. One source was 20v, one 10v. Thus, at the connection each should have caused a voltage of a different amount.

Ohm's Law

v = iR
Current enters at terminal labeled +.
G = i/R, conductance, siemens
V and I are no longer independent. (Now we know how V and I are related.)
P = vi = (iR)i = i²R
= v(v/R) = v²/R

Kirchhoff's Laws

Current Law--Algebraic sum of currents entering (or leaving) a node = 0
Voltage Law--Algebraic sum of voltages around any path = 0.
The points do not have to be connected!

Example

V_s + node d + R_d + node c - R_c + node b + R₁ + node a - V_s.
i_s from V_s +
i₁ from - R₁ to - V_s
i_c from - R_c to - R_d
i_d from - R_d to - R_c
v₁, v_c, v_d over corresponding resistors.
Unknowns: i_s, i₁, i_c, i_d, v₁, v_c, v_d
From Ohm's Law:
v₁ = i₁R₁
v_c = i_cR_c
v_d = i_dR_d
need four more equations
From Kirchhoff's Current Law:
a: i_s - i₁ = 0
b: i₁ + i_c = 0
c: -i_c - i_d = 0
d: -i_s + i_d = 0
We find
i_s = i_d = -i_c = i₁
Writing Kirchhoff's Voltage Law around the circuit:
-v_s + v_d - v_c + v₁ = 0.
-v_s + i_dR_d - i_cR_c + i₁R₁ = 0
-v_s + i_sR_d +i_sR_c +i_sR₁ = 0
i_s = v_s/(R_d + R_c + R₁)

Example

120v + 10W + node a - 6A + node b - 120v
node a + 50 W + node b
i₀ from + 120v to + 10W
i₁ from node a to + 50W to node b
v₀ over 10W
v₁ over 50W
From Kirchhoff's Current Law:
(1) i₀ - i₁ + 6 = 0
From Kirchhoff's Voltage Law:
(Do not include the path with the Ideal current source, because the voltage over the source is only defined by the circuit connected to the ideal source.
-120 + v₀ + v₁ = 0
Using Ohm's Law
(2) -120 + 10i₀ + 50i₁ = 0
Solving (1) and (2):
i₀ = -3A
i₁ = +3A
Power provided and Power Absorbed
P120 = +120(3) = +360W (absorbing)
P10 = i₀²(10) = 9(10) = 90W Abs
P50 = i₁²50 = 9(50) = 450W Abs
P6A = -v₁(6) = -3(50)(6) = -900W Provided