1A
VD = 0V
VA = 12 V
VB – VC = 4V
apply KCL at node B-C, assume all currents into node
I1 + I2 + I3 = 0
I1 = (12-B)/4
I2 = (0-C)/12
I3 = (0-C)/6
I1 + I2 + I3 = 0
(12-B)/4 + (0-C)/12 + (0-C)/6 = 0
B + C = 12 –> VB + VC = 12V
VB – VC = 4V
VB + VC = 12V
2VB = 16V
VB = 8V
VC = 4V
I1 = (12-B)/4 = (12V-VB)/4Ω = (12V-8V)/4Ω = (4V)/4Ω = 1A
I2 = (0-C)/12 = (-VC)/12Ω = (-4V)/12Ω = -1/3 A
I3 = (0-C)/6= (-VC)/6Ω = (-4V)/6Ω = -2/3 A
1B
P = IV3 V = IR
∴P = I·(IR) = I2R
P = (IA)2·(4Ω) = 4W = 4 J/s
E = PΔt = 4 J/s · 20s = 80J (this equation is from the conservation of energy law)
1C
P = IV = I1 · (4V) = IA · (4V) = 4W
(Note: the power is absorbed as the current is flowing from +Ve → -Ve
2A
T = RC : [ V/A · C/V = C/A ]
T = (10×10<sup>3</sup> Ω) · (100 x 10<sup>-6</sup> F) = 1000 x 10<sup>-3</sup> ΩF
Ω = V/A F = (A·S)/V Ω·F = V/A · (A·S)/V = S
T=1s
2B
I(t) = I<sub>0</sub>(e<sup>-t/2</sup>)
Ohms law: I<sub>0</sub> = V/R = 20/(10×10<sup>6</sup>) = 2 x 10<sup>-6</sup> A = 2mA
“find (t) when I(t) = .5I<sub>0</sub>”
I(t) = .5I<sub>0</sub> = I<sub>0</sub>(e<sup>-t/x</sup>) → .5 = e<sup>(-t/x)</sup>
ln.5 = lne<sup>(-t/x)</sup> =ln.5 = -t/x
Tln.5 = t = 1(ln.5) = ln·5s
t = -(ln·5)s