1A

V_D = 0V

V_A = 12 V

V_B – V_C = 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   –> V_B + V_C = 12V

V_B – V_C = 4V

V_B + V_C = 12V

2V_B = 16V

V_B = 8V

V_C = 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 +V_e → -V_e

2A

T = RC : [ V/A · C/V = C/A ]

T = (10×10³ Ω) · (100 x 10^-6 F) = 1000 x 10^-3 ΩF

Ω = V/A    F = (A·S)/V     Ω·F = V/A  · (A·S)/V  = S

T=1s

2B

I(t) = I₀(e^-t/2)

Ohms law:  I₀ = V/R = 20/(10×10⁶) = 2 x 10^-6 A = 2mA

“find (t) when I(t) = .5I₀”

I(t) = .5I₀ = I₀(e^-t/x) → .5 = e^(-t/x)

ln.5 = lne^(-t/x) =ln.5 = -t/x

Tln.5 = t = 1(ln.5) = ln·5s

t = -(ln·5)s