E=8.3V and R=5 ohms a) Find the current in the 2 ohm resistor (in mA) b) Find the potential difference between points a and b (in V)

A resistance of 10 ohms and C=0.1F is connected in series with a 12V battery. Find the charge at any time and the charge current 0.4 seconds after switch is closed.

C -Q (A) 0.25 uF 10V 2uF 4uF (B) Figure 3: 3. In Figure 3 the capacitors were discharged before being connected to the battery. (a) In Figure 3A the net positive charge delivered by the battery is +Q1 to C1 and a net negative charge -[Q2 + Q3] is delivered to C2 and C3 [-Q2 is delivered to C2 and -Q3 to C3]. From conservation of charge[net charge is zero since both the battery and the capacitors have a zero net excess charge] and path independence of the electric field[sum of voltages must add up to zero for a closed path or conservation of energy]: Q1 + (-Q2) + (-Q3) = 0 conservation of charge +%-왕-용 (True, False) = 0 conservation of energy +V - = 0 conservation of energy (b) If (a) is true then the charge Q1, Q2 and Q3 follow from above three equations as C1(C2 + C3) V Q2 = Ci + C2 + C3 C1 C3 Ci + C2 + C3 Q1 = Q3 = V, (True, False) C1 + C2 + C3 (c) In Figure 3A the net positive charge delivered by the battery is Qtot = Q1 = C1(C2+C3) v which implies an equivalent capacitance of Qtot…