2
Vab = 4x(E-Ir) = IR ð 4E - 4Ir = IR
4Ir = 4E - IR
r = (4E - IR)/4I = (E - 1/4(IR)/I = 0.33W
3
Vab = E - Ir ð r = E - Vab/I = 0.048 W
Vab = IR ð R = Vab/I = 0.11 W
6
a. Total resistance in series
Rtotal = 3*45 + 3*75 = 360 W
b. Total resistance in parallel
1/Rtotal = 3/45 + 3/75 ð Rtotal = 9.4 W
24
We can apply Kirchoff's loop rule to calculate the current. This rule, based on the conservation of energy, states that the sum of hte changes in potential around any closed path of a circuit must be zero. We assume that the current is flowing clockwise.
-I(1.0 W) = 18V - I(6.6 W) - 12 V - I(2.0 W) = 0 ð I = 0.625 A
We can now find the terminal voltage for each battery by taking the sum of the potential differences across the internal resistance and EMF - left to right. Note thoudh that for the 12 V battery, there is a voltage gain.
For the 18 V battery: Vterminal = -I(1.0 ohm) + 18 V = 17.4 V
For the 12 V battery: Vterminal = I(2.0 ohm) + 12 V = 13.3 V
36
The maximum capacitance is found by connecting them in paralllel and the minimum is fond by connecting them in series
Cmax = C1 + C2 + C3 = 2.07x10-8 F
Cmin = (1/C1 + 1/C2 + 1/C3)-1 = 1.83x10-9 F
50
a. The product RC is equal to the time constant
t = RC ð C = t/R = 2.33x10-9 F
b. We are given that the battery has an emf of 24.0 V and the voltage across the resistor is 16.0 V.
The voltage across the capacitor will then be
24 - 16 = 8 V.
VC = E(1-e-t/RC) = E(1-e-t/t ) ð e-t/t = (1 - VC/E)
t/t = ln(1 - VC/E) = 1.4 x 10-5 s