2. In the circuit shown there are five resistors. The battery is not an ideal battery and has an internal resistance of 25 0. A) What is the power dissipated by the 300 resistor below? B) What is the energy lost in the battery in 10 minutes of operation? C) If the 300 resistor burns out is the energy lost in the battery the same? Why or why not? 2002 50V 50 m 3002 100 600

In the figure R₁-3.41R, the ammeter resistance is zero, and the battery is ideal. What multiple of e/R gives the current in the ammeter? EH Number Units *E/R

In step 3, what is the potential difference across D2 after the jumper wire is connected

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