In Figure (a), both batteries have emf = 1.40 V and the external resistance R is a variable resistor. Figure (b) gives the electric potentials V between the terminals of each battery as functions of R: Curve 1 corresponds to battery 1, and curve 2 corresponds to battery 2. The horizontal scale is set by R$ = 0.400 Q2. What is the internal resistance of (a) battery 1 and (b) battery 2? 0.70-

For the circuit below, determine the resistor voltage. The source is 5 volts, the Zener potential is 5.2 volts and the resistor is 1k ohms.

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…

The conductance of a 9230 resistance is a) 1.08 * 10-3 mho b) 1.08 * 10-4 mho c) 1.02 * 10-3 mho d) 1.02 * 104 mho