The figure below shows a circuit with two batteries and three resistors. (Assume R₁ = 1.500, R₂ = 4.80 , and R3 = 6.000.) a R₁ www 6.00 V + R3 - R₂ ww + 9.00 V b (a) How much current flows through the 1.50- resistor? A (b) What is the potential difference between points a and b in the circuit? V

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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…

Question 2 4 for > 0 as shown in Figure 1 below. What is the energy? The independent current source in the circuit generates zero current for < 0 and a pulse of O a. g0r e Ob.30 e 20ke 2_-10t b.20,-10 c. 90re J 2--10t