In the figure is shown a battery having an emf of 12 Volts and an internal resistance of 0.40 Ohm. A resistor R, of 2.6 Ohms is connected across terminals ab. Determine (a) resistance of R its which, when connected across terminals ab in E = 12 V R,= 2.62 parallel with R, will make the power delivered by the battery a maximum; (b) current in R, and R; (c) terminal voltage of battery; (d) power in R, and in R; (e) power lost in battery r = 0.40 Q- Hint: To maximize power in R, use differentiation

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An example of Simple Resistive Circuits is ,

Find the equivalent resistance Rab for this circuit

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…